Introduction to Group Theory

MAT 2143

Winter 2019

Instructor: Hadi Salmasian

These lecture notes were produced using my course notes from Winter 2016 and Winter 2019. They are

loosely based on the following texts:

Thomas W. Judson, Abstract Algebra, Theory and Applications, Annual Edition 2018.

W. Keith Nicholson, Introduction to Abstract Algebra, Third Edition, . John Wiley & Sons Inc., 2007.

INTEGERS

Mathematical induction. There is a general proof technique in mathemat-

ics which is extremely useful when we want to prove a certain statement

for all values of n P N. Recall that N “ t1, 2, 3, . . .u is the set of natural

numbers.

Example 1.1. We want to prove that 1 ` ¨ ¨ ¨ ` n “ npn`1q2 for every n P N.

The statement is easily verified for n “ 1. Now we assume it is true for

some value n “ n0, and using this assumption, we prove that it holds for

n “ n0 ` 1:

1` ¨ ¨ ¨ ` n0 ` pn0 ` 1q “ p1` ¨ ¨ ¨ ` n0q ` pn0 ` 1q

“n0pn0 ` 1q

2` pn0 ` 1q

“ pn0 ` 1q´n02` 1

¯

“pn0 ` 1qpn0 ` 2q

2.

Principle of induction. Let Spnq be a predicate that depends on a variable

n P N. Assume that

(i) Spn0q is true for some n0 P N.

(ii) We can prove that for any k P N where k ě n0, if Spkq is true, then

Spk ` 1q is also true.

Then Spnq is true for every n P N satisfying n ě n0.

Before giving another example of application of the induction principle,

we recall the definition of divisibility.

Definition 1.2. Given m,n P Z, we say m divides n (or m is a divisor of n; or

n is divisible by m) and we write m|n, if n “ md for some d P Z.

We writem - n ifm does not divide n (or equivalently, if n is not divisible

by m).

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Example 1.3. We want to prove that for every n P N, the integer 4n`1`4n`1

is divisible by 3. For n “ 1 we have

42 ` 41 ` 1 “ 21

which is divisible by 3. Now assume the statement is true for some n “ k.

Then for n “ k ` 1 we have

4k`2 ` 4k`1 ` 1 ““

4p4k`1 ` 4k ` 1` 1q ´ 4‰

` 1 “ 4p4k`1 ` 4k ` 1q ´ 3.

By the induction hypothesis, 4k`1 ` 4k ` 1 is divisible by 3, i.e.,

4k`1 ` 4k ` 1 “ 3q for some integer q.

It follows that

4p4k`1 ` 4k ` 1q ´ 3 “ 3q ´ 3 “ 3pq ´ 1q,

so that 4p4k`1 ` 4k ` 1q ´ 3 is also divisible by 3.

Principle of induction (strong version). Let Spnq be a predicate that de-

pends on a variable n P N. Assume that

(i) Spn0q is true for some n0 P N.

(ii) We can prove that for every k P N where k ě n0, if Spn0q, . . . , Spkq

are true then Spk ` 1q is also true.

Then Spnq is true for every n P N satisfying n ě n0.

Both of the induction principles are equivalent to the following state-

ment:

Well-ordering principle. Every nonempty subset of N has a (unique)

smallest member.

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Theorem 1.4. (Division Algorithm) Let n, d P Z such that d ě 1. Then thereexists a unique pair of integers q, r P Z such that n “ qd` r and 0 ď r ă d.

Proof. Existence:

Step 1. Set X :“ tk P Z : k “ n ´ td for some t P Z and k ě 0u. Clearly

X Ď NY t0u.

Step 2.: We show that X ‰ ∅:

‚ If n ě 0 then we can set t “ 0. It follows that n P X .

‚ If n ă 0, then we set t “ n. It follows that np1 ´ dq “ n ´ nd P X .

Note that np1´ dq ě 0 because 1´ d ď 0.

Step 3. By the well-ordering principle, there exists a smallest element

r P X . Thus r “ n´ t0d for some t0 P Z. Set q :“ t0.

Step 4. From r P X it follows that r ě 0. Next we show r ă d. Assume on

the contrary, that r ě d, so that r ´ d ě 0. Then

r ´ d “ pn´ t0dq ´ d “ n´ pt0 ` 1qd P X,

which contradicts the fact that r “ minX .

Uniqueness: Assume that q1, r1 is another pair such that n “ q1d ` r1 and

0 ď r ă d. We want to show q “ q1 and r “ r1. We have

n “ qd` r “ q1d` r1 ñ dpq1 ´ qq “ r ´ r1.

If q “ q1, then it follows that r ´ r1 “ 0, hence r “ r1, and the proof is

complete. Now assume that q ‰ q1. Without loss of generality we can

assume q ă q1 (the case q ą q1 is similar). Thus

r ´ r1 “ dpq1 ´ qq ě d, so that r ě r1 ` d ě d.

This contradicts the assumption r ă d. �

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Example 1.5. Here are some explicit examples of the division algorithm:

(i) Let us assume n “ 19 and d “ 4. We want to find q, r. We have

X “ t. . . , 19, 15, 11, 7, 3,´1,´5, . . .u,

where the gray entries are not in X . It follows that r “ minX “ 3

and q “ 4.

(ii) n “ ´13 and d “ 7. Then X “ t. . . ,´13,´6,1, 8, . . .u, where the gray

entries are not in X . It follows that r “ minX “ 1 and q “ ´2.

Exercises.

Let m,n, d be integers. Prove the following statements.

(i) n|n.

(ii) If m ‰ 0, d P N and d|m then d ď |m|.

(iii) If d|m and m|n then d|n.

(iv) If d|n and n|d then d “ ˘n.

(v) If d|m and d|n then d|xm` yn for every x, y P Z.

Definition 1.6. Let m and n be integers. Assume that at least one of m and

n is nonzero. An integer d ě 1 is called the greatest common divisor (or the

gcd) of m and n, if it satisfies the following properties:

(i) d is a common divisor of m and n, i.e., d|m and d|m.

(ii) For every common divisor d1 of d, we have d1|d.

From the above definition, it is not immediately obvious that the gcd of

m and n exists. If it does, then it is also the largest common divisor of m

and n; because for every d1 such that d1|m and d1|n, we have d1|d, hence

d1 ď |d| “ d.

Remark 1.7. (i) If gcdpm,nq exists, then it is unique. This is because if

d ě 1 and d1 ě 1 are two possible gcd’s form and n, then in particular

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we have

d|d1 and d1|d,

so that d “ ˘d1. Since d, d1 ě 1, it follows that d “ d1.

(ii) By convention, gcdp0, 0q is not defined.

(iii) If either m “ 0 or n “ 0, then gcdpm,nq “ maxt|m|, |n|u.

Theorem 1.8. (Bezout’s Theorem) Let m and n be integers such that eitherm ‰ 0 or n ‰ 0. Then there exist x, y P Z such that

gcdpm,nq “ xm` yn.

In particular, gcdpm,nq exists.

Proof. Step 1. Consider the set

S :“ tk P N : k “ xm` yn for some x, y P Zu .

Clearly S ‰ ∅. Thus, by the well-ordering principle, S has a smallest

member. Set d :“ minS, so that

d “ x0m` y0n where x0, y0 P Z.

Step 2. We prove that d|m and d|n. We use the method of proof by contra-

diction. Assume, on the contrary, that d - m. By Theorem 1.4 we have

m “ dq ` r for 0 ă r ă d.

Thus

r “ m´ dq “ m´ px0m` y0nq “ p1´ x0qm` p´y0qn P S and r ă d,

which contradicts the assumption that d “ minS. The argument for d|n is

the same.

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Step 3. Suppose that d1 P Z is a common divisor of m and n (i.e., d1|m and

d1|n). We prove that d1|d. We have

d1|m ñ d1|x0m and d1|n ñ d1|y0n.

Finally, we have

d1|x0m and d1|y0n ñ d1|x0m` y0n ñ d1|d. �

Definition 1.9. Two integers m and n are called co-prime if gcdpm,nq “ 1.

Bezout’s Theorem implies that for co-prime integers m ann, there exist

x, y P Z such that xm` yn “ 1.

Remark 1.10. A digresstion about definitions: it is customary in mathe-

matical writing to use “if” (rather than “if and only if”). This is because a

definition is not a logical implication. It is just a sentence that describes a

meaning of a new concept.

Corollary 1.11. Let d,m, n P Z be integers, such that either m ‰ 0 or n ‰ 0.Assume that d “ gcdpm,nq. Then gcdpmd ,

ndq “ 1.

Proof. It is enough to show that if d1 P Z satisfies d1|md and d1|nd , then d1 “ ˘1.

By Bezout’s Theorem, there exist x, y P Z such that mx ` ny “ d. Thus

we havem

dx`

n

dy “ 1.

Note that md ,

nd P Z. Now

d1|m

dand d1|

n

dimply that d1|

´m

dx`

n

dy¯

,

hence d1|1, and therefore d1 “ ˘1. �

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The Euclidean algorithm. Theorem 1.8 proves the existence of x, y P Zsuch that xm` yn “ gcdpm,nq. But does there exist a method for comput-

ing x, y? The Euclidean algorithm gives such a method.

Start with integers m ě n ą 0. Set r0 “ m, r1 “ n, and define ri for i ě 2

by

ri´2 :“ qiri´1 ` ri where 0 ď ri ă ri´1,

where qi, ri are obtained by the division algorithm (Theorem 1.4). We have

r1 ą r2 ą r3 ą ¨ ¨ ¨ ě 0.

Set k :“ minti : ri`1 “ 0u.

Proposition 1.12. rk “ gcdpm,nq.

Proof. Step 1. Set d :“ gcdpm,nq. We have d|r0 and d|r1. Next we note that:

For every i ě 2, if d|ri´2 and d|ri´1 then d|pri´2 ´ qiri´1q “ ri.

It follows by induction that d|ri for every i ě 1. In particular, d|rk.

Step 2. To complete the proof, it is enough to show that rk|m and rk|n.

Indeed when these relations are proved if follows that rk is a common

divisor of m and n, hence rk|d. Now

d|rk and rk|d ñ d “ ˘rk.

Since d, rk ą 0, we obtain d “ rk.

Step 3. We now prove the claim of Step 2. To this end, we show that rk|rifor every 1 ď i ď k. This is done by “reverse” induction:

(i) We have rk|rk (trivial) and rk|rk´1 because rk´1 “ qk`1rk ` rk`1 “

qk`1rk.

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(ii) For i ě 1, if rk|ri and rk|ri´1 then

rk|ri and rk|ri´1qi ñ rk|pqiri´1 ` riq “ ri´2.

From (i) and (ii) above it follows that rk|ri for all 1 ď i ď k. In particular,

rk|r0 “ m and rk|r1 “ n. �

To compute x0, y0 P Z such that gcdpm,nq “ x0m ` y0n, we use the fol-

lowing reverse calculation:

gcdpm,nq “ rk “ rk´2 ´ qkrk´1

“ rk´2 ´ qkprk´3 ´ qk´1rk´2q

“ p1` qkqk´1qrk´2 ´ qkrk´3

“ ¨ ¨ ¨

Each time we replace the remaining ri with the largest index by ri´2´qiri´1,

so that we obtain a linear combination of ri´1 and ri´2.

Example 1.13. Let us compute gcdp108, 63q. We set r0 “ m “ 108 and

r1 “ n “ 63. Then

108 “ 1ˆ 63` 45 r2 “ 45

63 “ 1ˆ 45` 18 r3 “ 18

45 “ 2ˆ 18` 9 r4 “ 9

18 “ 2ˆ 9` 0 r5 “ 0.

Therefore gcdp108, 63q “ r4 “ 9. We also have

9 “ 45´ 2ˆ 18

“ 45´ 2ˆ p63´ 1ˆ 45q

“ 3ˆ 45´ 2ˆ 63

“ 3ˆ p108´ 1ˆ 63q ´ 2ˆ 63

“ 3ˆ 108´ 5ˆ 63.

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Therefore for x0 “ 3 and y0 “ ´5 we have 9 “ x0m` y0n.

Definition 1.14. An integer p is called prime if it satisfies the following two

properties:

(i) p ą 1.

(ii) The only divisors of p are ˘1 and ˘p.

An integer n ą 1 which is not prime is called a composite number. An

important property of prime numbers is the following:

Lemma 1.15. (i) Let a, b P Z, and let p be a prime number. Assume that p|ab.Then either p|a or p|b.

(ii) Let a1, . . . , ak P Z for some k P N. Let p be a prime number such thatp|a1 ¨ ¨ ¨ ak. Then there exists some 1 ď i ď k such that p|ai.

(iii) Let d,m, n P Z such that d|mn and gcdpd,mq “ 1. Then d|n.

Proof. (i) Assume that p - a. We will show that p|b. Indeed, set d :“

gcdpa, pq. Since the only positive divisors of p are 1 and p, we should have

d P t1, pu. But if d “ p then we have p|a. Thus we can assume that d “ 1.

Now by Bezout’s theorem there exist x, y P Z such that xa ` yp “ 1. We

have

b “ bpxa` ypq “ abx` pby.

Since p|ab, we have p|abx. Also, p|pby. Thus

p|abx` pby and abx` pby “ b, and consequently p|b.

(ii) For k “ 1 this is trivial and for k “ 2 this is part (i). Now we prove

the statement by induction on k. Assume that it holds for some k P N. We

need to show that for every m1, . . . ,mk`1 P Z, if p is a prime number such

that p|m1 ¨ ¨ ¨mk`1, then p|mi for some 1 ď i ď k ` 1. There are two cases to

consider:

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‚ Case I: p|m1 ¨ ¨ ¨mk. Then by induction hypothesis there exists 1 ď

i ď k such that p|mi.

‚ Case II: p - m1 ¨ ¨ ¨mk. Setting a “ m1 ¨ ¨ ¨mk and p “ mk`1, from part

(i) it follows that p|mk`1.

(iii) The argument is similar to (i). We leave it as an exercise. �

Exercises.

(1) Using induction, prove that if m1, . . . ,mk are integers and p is a

prime number such that p|m1 ¨ ¨ ¨mk, then there exists an index 1 ď

i ď k such that p|mi.

(2) Compute gcdp56, 24q. Then compute integers x, y P Z such that

gcdp56, 24q “ 24x` 56y.

(3) Complete the proof of Lemma 1.15(ii).

(4) Write down the proof of Lemma 1.15(iii).

(5) Let m,n, d P Z such that gcdpd, nq “ 1 and d|mn. Prove that d|m.

Hint. First choose x, y P Z such that nx ` dy “ 1. Then write m “

mpnx` dyq, and continue as in the proof of Lemma 1.15(i).

An important theorem in elementary arithmetic, whose proof we skip

(but it appears in the textbook) is the following.

Theorem 1.16. (Fundamental Theorem of Arithmetic) Let n ą 1 be an in-teger. Then n can be written as a product of prime numbers. Furthermore, thefactorization of n as a product of primes is unique: if

n “ p1 ¨ ¨ ¨ pk “ q1 ¨ ¨ ¨ q`

where the pi’s and the qj’s are prime numbers, then k “ ` and the qj’s are just arearrangement of the pi’s.

Theorem 1.17. There exist infinitely many prime numbers.

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Proof. We give Euclid’s original proof. Assume, on the contrary, that there

are only finitely many primes. Label them as p1, . . . , pk. Now consider

N :“ 1` p1 ¨ ¨ ¨ pk.

by Theorem 1.16, we can find a prime number p|N . Thus p “ pi for some

1 ď i ď k. This implies that p|p1 ¨ ¨ ¨ pk, and therefore

p|pN ´ p1 ¨ ¨ ¨ pkq “ 1,

which is a contradiction. �

Lemma 1.18. Let N ą 1 be an integer. Assume that the prime factorization of Nis

N “ pa11 ¨ ¨ ¨ pakk

where p1, . . . , pk are distinct prime numbers and a1, . . . , ak P N. Let p be a primenumber such that pk|N for some k P N. Then p “ pi for a unique 1 ď i ď k.Furthermore k ď ai.

Proof. Step 1. From pk|N it follows that p|N , hence p|pa11 ¨ ¨ ¨ pakk . Now Lemma

1.15(ii) implies that p|pi for some 1 ď i ď k. Since p and pi are prime num-

bers, from p|pi it follows that p “ pi.

Step 2. We prove k ď ai. Becasue of the symmetry of the indices we can

assume that i “ 1. Assume that, on the contrary, k ą a1; thus k ě a1 ` 1.

From pk1|N we have

N “ pk1d for some d P Z,

hence N “ pk1d “ pa1`11 ppk´pa1`1q1 dq and d1 :“ p

k´pa1`1q1 d P Z. Hence by

cancellation of pa11 from both sides we obtain

pa11 ¨ ¨ ¨ pakk “ pa1`11 d1 ñ pa22 ¨ ¨ ¨ p

akk “ p1d

1.

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The latter implies that p1|pa22 ¨ ¨ ¨ pakk . Thus Lemma 1.15(ii) implies that p|pi

for some 2 ď i ď k. This is a contradiction because the pi’s are distinct

primes. �

The prime factorization provides another method to compute the gcd of

integers:

Proposition 1.19. Let a, b P N, with prime factorizations

a “ pa11 ¨ ¨ ¨ pakk and b “ pb11 ¨ ¨ ¨ p

bkk ,

where p1, . . . , pk are distinct primes and ai, bi P Zě0 :“ tx P Z : x ě 0u for all1 ď i ď k. Then

gcdpa, bq “ pminta1,b1u1 ¨ ¨ ¨ p

mintak,bkuk .

Proof. Step 1. First we show that M :“ pminta1,b1u1 ¨ ¨ ¨ p

mintak,bkuk is a common

divisor of a and b. To do so, we prove that M |a, and the proof for M |b is

similar.

Step 2. To proveM |a, note that for every 1 ď i ď k we have mintai, biu ď ai,

hence ai “ mintai, biu ` ci for some ci P Zě0. Thus we have

a “Mpc11 ¨ ¨ ¨ pckk ,

and pc11 ¨ ¨ ¨ pckk P Z because it is a product of primes with non-negative inte-

ger exponents. This completes the proof of M |a.

Step 3. To show that M is indeed equal to gcdpa, bq, we need to show the

following: if d|a and d|b for some d P Z, then d|M . This completes the proof

of the fact that M “ gcdpa, bq.

Step 4. We now proceed towards the proof of the claim of Step 3. Assume

that d P Z satisfies d|a and d|b. If p is a prime such that pk|d for some k P N,

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then

pk|d and d|a implies pk|a,

and therefore by Lemma 1.18 we have p “ pi for some 1 ď i ď k, and

k ď ai. Similarly, from d|b we obtain pk|b, hence by Lemma 1.18 we have

p “ pj for some 1 ď j ď k, and k ď bj. Since the primes p1, . . . , pk are

distinct, it follows that i “ j, and consequently, k ď mintai, biu.

Step 4. From what is shown in Step 3 it follows that in the prime factor-

ization of d, the only primes that can appear are p1, . . . , pk; morevoer, this

prime factorization is of the form

d “ pc11 ¨ ¨ ¨ dekk , where ei ď mintai, biu for all 1 ď i ď k.

Therefore d|M , because

M “ dpminta1,b1u´e11 ¨ ¨ ¨ p

mintak,bku´ekk ,

and the exponents minta1, b1u ´ e1 are non-negative integers. �

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GROUPS

Integers modulo n. Let n ě 1 be an integer. The division algorithm mod-

ulo n yields remainders r such that 0 ď r ă n. We want to define addition

and multiplication operations on the set of these remainders, namely on

Zn :“ t0, 1, . . . , n´ 1u.

For a, b P Zn we define

a ` b :“ pa` bq modulo n,

and

a ˆ b :“ ab modulo n.

Example 2.1. We have Z4 “ t0, 1, 2, 3u. Then we can check that

1 ` 2 “ 3 and 2 ` 3 “ 1.

Also,

2 ˆ 3 “ 2 and 2 ˆ 2 “ 0.

The formal definition of the set Zn is as follows. Consider the binary

relation ”n on Z defined as follows: for a, b P Z,

a ”n b ô n|pa´ bq.

Exercises.

(1) Let n P N.

(i) Prove that ”n is an equivalence relation, i.e., it is reflexive, sym-

metric, and transitive.

(ii) Prove that the integers 0 ď k ď n ´ 1 are representatives of dis-

tinct equivalence classes of ”n. Furthermore, these equivalence

classes are indeed all of the equivalence classes of ”n.

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(iii) Let a, b, a1, b1 P Z. Prove that if a ”n a1 and b ”n b

1, then

(2.1) a` b ”n a1` b1 and ab ”n a

1b1.

(2) Compute the last 2 digits from the right of 7111.

Definition 2.2. Fix an integer n ě 1. For each a P Z, let ras denote the

equivalence class of a, i.e.,

ras :“ tb P Z : b ”n au.

We denote the set of the equivalence classes of ”n by Zn. Then by the

preceding exercise each of these equivalence classes contains a unique el-

ement a P Z such that 0 ď a ď n ´ 1. Therefore we can naturally identify

Zn with t0, 1, . . . , n´ 1u.

Remark 2.3. From (2.1) it follows that the addition and multiplication the

we defined on Zn is indeed “induced” from the corresponding operations

on Z. We will learn more about this phenomenon in the future.

Proposition 2.4. Let n ě 1, and let a, b, c P Zn.

(i) a ` b “ b ` a.(ii) a ˆ b “ b ˆ a.

(iii) a ` pb ` cq “ pa ` bq` c.(iv) a ˆ pb ˆ cq “ pa ˆ bqˆ c.(v) a ` 0 “ 0 ` a “ a.

(vi) a ˆ 1 “ 1 ˆ a “ a.(vii) a ˆ pb ` cq “ a ˆ b ` a ˆ c.

(viii) a ` p´aq “ 0, where by ´a we denote the equivalence class of n´ a.(ix) gcdpa, nq “ 1 if and only if there exists an element a1 P Zn such that

a ˆ a1 “ 1.

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Proof. We only prove (i) and (ix). The remaining parts are left as an exer-

cise.

(i) Set r :“ a ` b. Then a ` b “ qn ` r, hence b ` a “ qn ` r, from which it

follows that b ` a “ r.

(ix) Assume that gcdpa, nq “ 1. By Theorem 1.8 there exist x, y P Z such

that ax`ny “ 1. Thus ax ”n 1. Now set a1 :“ x modulo n. Then a1 “ x`tn

for some t P Z, hence

aa1 “ ax` atn “ 1´ ny ` atn “ 1` np´y ` atq,

so that a ˆ a1 “ 1.

Conversely, assume that there exists a1 P Zn such that a ˆ a1 “ 1. Then

aa1 “ nq`1, so that we have aa1´nq “ 1. Now if d|a and d|n then we have

d|aa1 ´ nq “ 1, hence d “ ˘1. Therefore gcdpa, nq “ 1. �

Solving equations in Zn. From now on we always equip Zn with the de-

fined addition and multiplication operations. Moreover, we will stop us-

ing the fancy notation ` and ˆ for addition and multiplication of Zn. Thus,

henceforth we will use the usual symbols for addition and multiplication

in both Z and Zn.

Definition 2.5. Let n P N. An element a P Zn is called invertible if there

exists some b P Zn such that ab “ 1. If a is invertible, we denote the inverse

of a by a´1.

Proposition 2.6. Let n P N, and a, b P Zn. Assume that gcdpa, nq “ 1. Then theequation ax “ b has a unique solution in Zn, given by x “ a´1b.

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Proof. Assume that ax “ b for some x P Zn. By multiplying both sides

by a´1 (where multiplication takes place in Zn), we obtain x “ a´1b. This

proves uniqueness and existence simultaneously. �

Computation of a´1. Let n P N and let a P Zn. Assume that gcdpa, nq “ 1.

Then a is invertible in Zn. But how can we compute a´1? This can be done

using the proof of Proposition 2.4(ix), as follows. Choose x, y P Z such that

ax` by “ 1. Then set a´1 :“ x mod n.

Exercises.

(1) Verify, using the proof of Proposition 2.4(ix), that the above method

indeed yields a´1.

Example 2.7. Consider 3 P Z7. Note that gcdp3, 7q “ 1, and therefore 3 is

invertible in Z7. To compute 3´1, we note that 3p´2q ` 7p1q “ 1, so that

3´1 “ ´2 mod 7 “ 5.

Example 2.8. We want to solve the equation 8x “ 3 in Z35. First we com-

pute gcdp35, 8q. The Euclidean algorithm yields:

35 8 3 2 1 0 ¨ ¨ ¨

Therefore gcdp35, 8q “ 1. Thus 8 is invertible in Z35. Now to compute a´1

we need to write the gcd as a linear combination of 35 and 8:

1 “ 3´ 2

“ 3´ p8´ 3ˆ 2q “ p3ˆ 3q ´ 8

“ 3ˆ p35´ p4ˆ 8qq ´ 8 “ 3ˆ 35´ 13ˆ 8.

It follows that 8´1 “ ´13 “ 35´ 13 “ 22. Now

x “ p3qp22q “ 66 mod 35 “ 31 in Z35.

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Symmetries of polygons. A symmetry of a polygon is a map on the set

of its vertices which preserves adjacency, distances, and angles. It follows

easily that every symmetric is a bijective map on the set of the nodes of

the polygon. Note that the composition of any two symmetries is again a

symmetry.

Example 2.9. Let us investigate the symmetries of a rectangle:

A B

C D

The rectangle has 4 symmetries:

(i) The identity map (which fixes every vertex), denoted by f0.

(ii) A horizontal reflection:

f1pAq “ C, f1pBq “ D, f1pCq “ A, f1pDq “ B.

(iii) A vertical reflection:

f2pAq “ B, f2pBq “ A, f2pCq “ D, f2pDq “ C.

(iv) The composition of the two maps in (ii) and (iii). There are two ways

to compose, but they both yield the same map:

f3pAq “ D, f3pBq “ C, f3pCq “ B, f3pDq “ A.

The multiplication table of the symmetries of the rectangle is given below.

The entry on row i and column j is fi ˝ fj.

Example 2.10. Let us investigate symmetries of an equilateral triangle.

- 19 -

f0 f1 f2 f3

f0 f0 f1 f2 f3

f1 f1 f0 f3 f2

f2 f2 f3 f0 f1

f3 f3 f2 f1 f0

FIGURE 1. The multiplication of symmetries of a rectangle.

A B

C

There are 6 different symmetries:

(i) The identity map.

(ii) The 60 degree clockwise rotation ρ1.

(iii) The 60 degree counter-clockwise rotation ρ2.

(iv) The reflection µ1 that fixes A.

(v) The reflection µ2 that fixes B.

(vi) The reflection µ3 that fixes C.

The multiplication table of these rotations is given in the textbook.

Permutations. A permutation is just a special kind of bijection, as defined

below.

Definition 2.11. Let n ě 1. A permutation of t1, . . . , nu is a bijection

σ : t1, . . . , nu Ñ t1, . . . , nu.

Since permutations are functions, we can compose them, as shown the

the following example. Clearly the composition of any two permutations

is again a permutation.

- 20 -

Example 2.12. Let n “ 4. The maps σ, τ : t1, 2, 3, 4u Ñ t1, 2, 3, 4u defined

by

σp1q “ 3, σp2q “ 4, σp3q “ 1, σp4q “ 2

and

τp1q “ 1, τp2q “ 3, τp3q “ 2, τp4q “ 4

are permutations of t1, 2, 3, 4u. Note that σ2 “ τ 2 “ id, where id denotes

the identity permutation (i.e., idpkq “ k for all 1 ď k ď n). Furthermore,

ρ :“ σ ˝ τ is given by

ρp1q “ 3, ρp2q “ 1, ρp3q “ 4, ρp4q “ 2.

We sometimes denote the permutation σ by

(2.2)

˜

1 2 ¨ ¨ ¨ n´ 1 n

σp1q σp2q ¨ ¨ ¨ σpn´ 1q σpnq

¸

Definition 2.13. The set of all of the permutations of t1, . . . , nu is denoted

by Sn.

Example 2.14. The permutation σ P S4 defined by

σp1q “ 4, σp2q “ 3, σp3q “ 1, σp4q “ 2

is denoted by˜

1 2 3 4

4 3 1 2

¸

.

Groups: definitions and examples. A binary operation on a set X is a map

‹ : X ˆX Ñ X which maps any pair pa, bq to an element a ‹ b.

Definition 2.15. A group is a set G equipped with a binary operation ‹ :

GˆGÑ G which satisfies the following:

(i) The binary operation ‹ is associative, i.e., pa ‹ bq ‹ c “ a ‹ pb ‹ cq for all

a, b, c P G.

- 21 -

(ii) G has an identity element. That is, there exists an element e P G such

that a ‹ e “ e ‹ a “ a for all a P G.

(iii) Every a P G has an inverse, i.e., for every a P G there exists an element

a´1 P G such that a ‹ a´1 “ a´1 ‹ a “ e.

Example 2.16. Here are some examples of groups:

(i) The set Z together with the binary operation a ‹ b :“ a` b (the usual

addition) is a group. The identity element of Z is 0 and the inverse

of every a P Z is ´a.

(ii) The set R` consisting of positive real numbers, equipped with the

binary operation a‹b :“ ab (the usual multiplication) is a group. The

identity element of R` is 1 P R`, and the inverse of every a P R` is1a .

(iii) The set Zn together with the binary operation a‹ b :“ a` b (the usual

addition of Zn) is a group. The identity element of Z is 0. The inverse

of 0 P Zn is 0, and the inverse of every a P Znzt0u is n´ a.

Remark 2.17. The importance of associativity of a groupG is that the value

of an expression g1 ‹ ¨ ¨ ¨ ‹ gk does not depend on how it is parsed using

brackets. For example, associativity implies that for all choices of elements

g1, g2, g3, g4 P G,

ppg1 ‹ g2q ‹ g3q ‹ g4 “ pg1 ‹ g2q ‹ pg3 ‹ g4q “ g1 ‹ ppg2 ‹ g3q ‹ g4q “ ¨ ¨ ¨

We will prove this statement in Theorem 3.5.

Example 2.18. Let G be the set of symmetries of a rectangle (see Example

2.9), equipped with composition of maps considered as a binary operation.

Then G is a group. Associativity of the binary operation of G is a natural

property of composition of functions. The identity element of G is f0, and

we have f´1i “ fi for 1 ď i ď 3.

- 22 -

Definition 2.19. Let G be a finite group whose elements are g1, . . . , gn in

some order. The Cayley table of G is the n ˆ n table whose pi, jq-entry is

gi ‹ gj. Note that the Cayley table depends on the chosen ordering of the

elements of G.

For example, Figure 2.9 is the Cayley table of the group described in

Example 2.18.

Definition 2.20. A group G is called abelian if a‹ b “ b‹a for every a, b P G.

Exercises.

(1) Verify that the set of symmetries of an equilateral triangle, equipped

with the composition of maps considered as a binary operation, is a

group. Is this group abelian?

(2) Verify that all of the groups in Example 2.16 are abelian.

The symmetric group Sn. Now let n P N. Consider the binary operation

on Sn defined by

σ ‹ τ :“ σ ˝ τ,

where the right hand side means composition of the two maps σ and τ .

Remark 2.21. If

σ “

˜

1 2 3 4

4 3 1 2

¸

and τ “

˜

1 2 3 4

2 3 4 1

¸

then σ ‹ τ “ σ ˝ τ “

˜

1 2 3 4

3 1 2 4

¸

Proposition 2.22. The set Sn equipped with the above binary operation is a group

For the proof of this proposition we note that:

(i) Associativity of ‹ is a natural property of composition of functions.

- 23 -

(ii) The identity element of ‹ is the identity permutation ι P Sn, defined

by ιpkq “ k for all 1 ď k ď n.

(iii) The inverse of a permutation σ is simply the inverse map σ´1, de-

fined by

σ´1pkq “ ` if and only if σp`q “ k.

The group of units of Zn. For n ě 1, set Upnq :“ ta P Zn : gcdpa, nq “ 1u.

By Proposition 2.4(ix), Upnq is precisely the set of invertible elements of Zn.

Thus,

Upnq “ ta P Upnq : au.

Proposition 2.23. The set Upnq, equipped with the multiplication of Zn, is anabelian group.

Proof. Step 1. The multiplication of Zn induces a binary operation on Upnq:

If a, b P Upnq then ab P Upnq because:

a P Upnq ñ Da1PZnaa1“ 1,

and

b P Upnq ñ Db1PZnbb1“ 1,

and now we have pabqpb1a1q “ apbb1qa1 “ ap1qa1 “ aa1 “ 1, hence ab P Upnq.

Thus Upnq is closed under multiplication.

Step 2. Associativity of the multiplication on Upnq is inherited from Zn.

Step 3. The identity element of Upnq is 1 P Zn. Note that 1 P Upnq because

gcdp1, nq “ 1.

Step 4. For a P Upnq, the multiplicative inverse a1 also belongs to Upnq

(because it is also invertible).

- 24 -

Step 5. Finally, Upnq is abelian because of Proposition 2.4(ii). �

Example 2.24. We have Up12q “ t1, 5, 7, 11u. The identity element is 1. We

have 5´1 “ 5, 7´1 ` 7, 11´1 “ 11.

Example 2.25. We have Up11q “ t1, 2, 3, 4, 5, 6, 7, 8, 9, 10u (note that 11 is

prime). We have 2´1 “ 6 and 3´1 “ 4, etc.

The vector space Rn. The standard addition of vectors is a binary opera-

tion on the vector space Rn “ tpx1, . . . , xnq : @1ďiďnxi P Ru:

px1, . . . , xnq ` py1, . . . , ynq “ px1 ` y1, . . . , xn ` ynq.

Proposition 2.26. Rn equipped with the standard addition of vectors is an abeliangroup.

Proof. We leave the proof as an exercise. We only remark that the iden-

tity element of Rn is the vector p0, . . . , 0q, and the inverse of px1, . . . , xnq is

p´x1, . . . ,´xnq. �

The matrix group GL2pRq. Let GL2pRq be the set of invertible 2ˆ2 matrices

with real entries. Then GL2pRq is a group (its binary operation is the usual

matrix multiplication). It is not an abelian group because for example«

1 1

0 1

ff«

1 0

1 1

ff

‰

«

1 0

1 1

ff«

1 1

0 1

ff

2.1 Digression on complex numbers*. Numbers of the form a`bi, where

a, b P R and i “?´1 (so that i2 “ ´1), are called complex numbers. We add

complex numbers componentwise:

pa` biq ` pc` diq “ pa` cq ` pb` dqi.

We multiply complex numbers just like ordinary numbers, taking into ac-

count that i2 “ ´1:

pa` biqpc` diq “ ac` bdi2 ` bci` adi “ pac´ bdq ` pad` bcqi.

- 25 -

For example,

p1´ 3iq ` p2` 4iq “ 3` i and p1´ iqp1` 2iq “ 3` i.

The mutiplicative inverse of a nonzero complex number a` bi is

1

a` bi“

a

a2 ` b2´

b

a2 ` b2i.

The quaternion group. Consider the matrices

I “

«

1 0

0 1

ff

, I :“

«

0 1

´1 0

ff

, J :“

«

0 i

i 0

ff

, K :“

«

i 0

0 ´i

ff

,

where i “?´1. Then we have

I2 “ J2“ J2

“ K2“ ´I,

and

IJ “ K “ ´JI, JK “ I “ ´KJ, KI “ J “ ´IJ.

The following proposition is now starighforward:

Proposition 2.27. The set of matrices Q :“ t˘I,˘I,˘J,˘Ku is a group.

Clearly this groups is not abelian.

- 26 -

ELEMENTARY PROPERTIES OF GROUPS

In this section we study some fundamental properties of general groups.

Proposition 3.1. Let G be any group.

(i) The identity element of G is unique.(ii) For every a P G, the inverse of a is unique.

Proof. (i) Let e, e1 P G be two identity elements in G. Then

ee1 “ e because e1 is an identity element

and

ee1 “ e1 because e is an identity element.

From these relations it follows that e “ e1.

(ii) Let g P G, and assume that h1, h2 P G are two inverses of g. As usual

let e P G be the identity element. Then

h1 “ h1e “ h1pgh2q “ ph1gqh2 “ eh2 “ h2.

�

Proposition 3.2. Let G be any group.

(i) For all a, b P G we have pabq´1 “ b´1a´1.(ii) For every a P G we have pa´1q´1 “ a.

Proof. (i) We have pabqpb´1a´1q “ apbb´1qa´ “ apeqa´1 “ aa´1 “ e. Simi-

larly, pb´1a´1qpabq “ e. Thus the inverse of ab is b´1a´1.

(ii) We have pa´1qa “ apa´1q “ e, hence by uniqueness of inverse, the

inverse of a´1 is equal to a. �

Proposition 3.3. LetG be any group, and let a, b P G. Then the equations ax “ b

and xa “ b have unique solutions in G.

- 27 -

Proof. Assume that ax “ b. Multiplying on the left by a´1, we obtain

a´1paxq “ a´1b.

Thus x “ ex “ pa´1aqx “ a´1paxq “ a´1b. Furthermore, x “ a´1b satis-

fies the equation ax “ b. Thus this equation has a unique solution. The

converse is proves similarly. �

Proposition 3.4. (Left and Right Cancellation Laws) Let G be a group and leta, b, c P G.

(i) If ab “ ac then b “ c.(ii) If ba “ ca, then b “ c.

Proof. (i) Assume that ab “ ac. Multiplying both sides by a´1, we have

a´1pabq “ a´1pacq.

By associativity of G, we can write the latter equality as pa´1aqb “ pa´1aqc.

Thus eb “ ec, hence b “ c.

(ii) The argument is similar and left as an exercise. �

By a group’s defining properties we know that g1pg2g3q “ g1pg2g3q for all

elements g1, g2, g3 P G. Is it also true that g1pg2pg3g4qq “ ppg1g2qg3qg4? The

next theorem answers this question.

Theorem 3.5. (General Associativity Theorem) Let G be a group and letg1, . . . , gn P G. Then the value of the product g1 ‹ ¨ ¨ ¨ ‹ gn is independent ofthe way brackets are placed.

Proof. We prove the statement by strong induction on n.

Step 1. The statement is trivial for n ď 3.

Step 2. Assume the statement holds for all n ď n0. Now assume n “ n0`1

and consider a way of parsing g1 ‹ ¨ ¨ ¨ ‹ gn by brackets. There exists an

- 28 -

outermost part of brackets, i.e., some 1 ď k ď n´1 such that the expression

looks like

pg1 ‹ ¨ ¨ ¨ ‹ gkq ‹ pgk`1 ‹ ¨ ¨ ¨ ‹ gnq.

By induction hypothesis we can assume that the left bracket is equal to

ppg1 ‹ g2q ‹ ¨ ¨ ¨ q ‹ gkq

and the right bracket is equal to

gk`1 ‹ pgk`1 ‹ ¨ ¨ ¨ ‹ pgn´1 ‹ gnqq.

Thus we have

pg1 ‹ ¨ ¨ ¨ ‹ gkq ‹ pgk`1 ‹ ¨ ¨ ¨ ‹ gnq

“ ppg1 ‹ ¨ ¨ ¨ ‹ gk´1loooooomoooooon

a

q ‹ gkloomoon

b

q ‹ pgk`1 ‹ pgk`1 ‹ ¨ ¨ ¨ ‹ gnlooooooomooooooon

loooooooooooomoooooooooooon

c

qq

“ pg1 ‹ ¨ ¨ ¨ ‹ gk´1loooooomoooooon

a

q ‹ p gkloomoon

b

‹pgk`1 ‹ ¨ ¨ ¨ ‹ gnlooooooomooooooon

c

qq,

where we are using pabqc “ apbcq. Continuing in this fashion, we obtain

pg1 ‹ ¨ ¨ ¨ ‹ gkq ‹ pgk`1 ‹ ¨ ¨ ¨ ‹ gnq “ g1 ‹ pg2 ‹ pg3 ‹ ¨ ¨ ¨ ‹ gnqq.

Thus the value of any bracketed expression pg1 ‹ ¨ ¨ ¨ ‹ gkq ‹ pgk`1 ‹ ¨ ¨ ¨ ‹ gnq is

equal to the right-aligned bracketing g1 ‹ pg2 ‹ pg3 ‹ ¨ ¨ ¨ ‹ gnqq. In particular,

we have shown using the induction hypothesis that the value of bracketed

expressions in any g1, ¨ ¨ ¨ , gn P G for n “ k`1 is independent of the choice

of the bracketing. This completes the proof by induciton. �

Powers of an element g P G. Let G be a group. For every g P G and n P Zwe define gn as follows:

- 29 -

gn :“

$

’

’

’

’

’

&

’

’

’

’

’

%

g ¨ ¨ ¨ gloomoon

n times

if n ą 0,

e if n “ 0,

pg´1qn if n ă 0.

Example 3.6. Let G “ pZ,`q; this notation means the binary operation of

Z is the usual addition. Then in this notation gm “ m ¨ g for g P G “ Z and

m P Z. For instance, 35 “ 15 and p´2q4 “ ´8. This notation should not be

confused with the usual exponentiation of numbers.

Example 3.7. Let G “ Z24. Then for g P G and n P Z we have

gn :“ ng mod 24.

Again, this notation should not be confused with the usual exponentiation

on Z24, which pertains to the binary operator of multiplication on Z24. For

example, 32 :“ 3` 3 “ 6 and 23 :“ 2` 2` 2 “ 6.

Example 3.8. Let G “ Up24q. This time the binary operation of G is the

restriction of the multiplication of Z24. Therefore gn is consistent with the

usual notion of exponentiation that is obtained from the multiplication of

Z24. Thus, for instance 52 “ 25 modulo 24 “ 1.

- 30 -

The concept of the power of an element can be a bit con-

fusing when working with arithmetic modulo n:

‚ Suppose that G “ Zn “ t0, 1, . . . , n ´ 1u. Then the

identity element is 0. In this group, the binary oper-

ation is “`”. Therefore ak for a P G and k P N means

a ‹ ¨ ¨ ¨ ‹ aloooomoooon

k times

“ a` ¨ ¨ ¨ ` alooooomooooon

k times

.

For example, if n “ 8 and a “ 3 and k “ 2, then ak

means a` a “ 3` 3 “ 6.

‚ Now suppose that G “ Upnq. Note that Upnq Ă Zn,

but the group operations of Zn are Upnq are different.

The identity element of G is 1.

The binary operation of G is induced by multipli-

cation modulo n, which does not make Zn a group.

For a P G and k P N, the element ak in G means

a ‹ ¨ ¨ ¨ ‹ aloooomoooon

k times

“ aˆ ¨ ¨ ¨ ˆ alooooomooooon

k times

.

For example, if n “ 8 and a “ 3 and k “ 2, then ak

means aˆ a “ 3ˆ 3 “ 1.

Theorem 3.9. Let G be a group. Then the following statements hold:

(i) gmgn “ gm`n for all g P G and m,n P Z.(ii) pgmqn “ gmn for all g P G and m,n P Z.

(iii) pghq´n “ ph´1g´1qn for all , h P G and n P Z.

Proof. (i) We argue by considering various cases. Let us first assume that

m,n ą 0. Then

gmgn “ pg ¨ ¨ ¨ gloomoon

m times

qpg ¨ ¨ ¨ gloomoon

n times

q “ g ¨ ¨ ¨ gloomoon

m` n times

“ gm`n.

- 31 -

Next assume that m ą 0 and n “ 0. Then gm`n “ gm “ gme “ gmgn. The

argument for the case m “ 0 and n ą 0 is similar.

Next assume that m ą 0 and n ă 0. Then n “ ´k for some k ą 0. Thus

gmgn “ pg ¨ ¨ ¨ gloomoon

m times

qpg´1 ¨ ¨ ¨ g´1loooomoooon

k times

q “

$

&

%

gm´k “ gm`n if m ě k,

pg´1qk´m “ gm´k if m ă k.

We leave the remaining cases as an exercise. �

Cartesian product of groups. Let G1 and G2 be two groups. We equip the

Cartesisn product

G1 ˆG2 :“ tpg1, g2q : g1 P G1, g2 P G2u

with the binary operation defined by

pg1, g2qph1, h2q :“ pg1h1, g2h2q for g1, h1 P G1, g2, h2 P G2.

Proposition 3.10. G1 ˆG2 is a group.

Proof. We leave the details as an exercise. We only mention that the iden-

tity element of this group is pe1, e2q, where ei is the identity element of Gi

for i “ 1, 2, and the inverse of an element pg1, g2q is pg´11 , g´12 q. �

Example 3.11. The set Z6 ˆ Z4 is a group. We have

p2, 3q´1 “ p4, 1q and p3, 3qp3, 3q “ p3` 3 modulo 6, 3` 3 modulo 4q “ p0, 2q.

Isomorphisms of groups. Of course changing the labelings/names of el-

ements of a group does not change its intrinsic structure. Therefore it is

natural to only consider groups up to isomorphisms:

Definition 3.12. LetG1 andG2 be two groups. An isomorphism f : G1 Ñ G2

is a bijection such that fpg1g2q “ fpg1qfpg2q for all g1, g2 P G1. If an isomor-

phism exists between G1 and G2, then we say G1 and G2 are isomorphic,

and we write G1 – G2.

- 32 -

Remark 3.13. For an isomorphism, we have fpe1q “ e2 where ei P Gi de-

notes the identity element. This is because we ave

fpe1q “ fpe21q “ fpe1q2, or equivalently, fpe1qe2 “ fpe1qfpe1q.

Now the left cancellation law in G2 implies e2 “ fpe1q.

Furthermore, fpg´1q “ fpgq´1. This is because we have

fpgqfpg´1q “ fpgg´1q “ fpe1q “ e2 and similarly fpg´1qfpgq “ e1.

By uniqueness of inverses in G2, we obtain fpgq´1 “ fpg´1q.

Example 3.14. Consider the setH : t´1,`1uwith the usual multiplication,

i.e.,

p1qp1q “ 1, p1qp´1q “ p´1qp1q “ ´1, p´1qp´1q “ 1.

Then H is a group. Furthermore, the map

φ : Z2 Ñ H,

given by φp0q “ 1 and φp1q “ ´1 is an isomorphism.

Example 3.15. There is an isomorphism φ : Z2 Ñ Up3q: Recall that

Z2 “ t0, 1u and Up3q “ t1, 2u.

The isomorphism φ is given by

φp0q “ 1 and φp1q “ 2.

Example 3.16. The map φ : Z2 ˆ Z3 Ñ Z6 given by

φp0, 0q “ 0, φp0, 1q “ 4, φp0, 2q “ 2, φp1, 0q “ 3, φp1, 1q “ 1, φp1, 2q “ 5

is an isomorphism. To check this, you need to verify that

φpg1g2q “ φpg1qφpg2q

- 33 -

for every pair of elements g1, g2 chosen from Z2ˆZ3. Since this group has 6

elements, there are 6ˆ6 “ 36 pairs g1, g2 that one can choose, and therefore

there are 36 relations to check. This is a tedious task. Later, we will study

this isomorphism more conceptually.

Example 3.17. Let k P Z5 such that k ‰ 0. Then the map φ : Z5 Ñ Z5,

defined by

φpiq “ ki,

is an isomorphism.

Exercises.

(1) Verify that the map in Examples 3.15, 3.16 and 3.17 are indeed iso-

morphisms.

(2) Is the map in Example 3.16 the only possible isomorphism?

- 34 -

GROUPS OF SMALL CARDINALITY

As shown below, it is not difficult to classify all groups of small cardinal-

ity up to isomorphism. Clearly up to isomorphism there is only one group

with one element.

Proposition 4.1. Let G be a group such that |G| “ 2. Then G – Z2.

Proof. Step 1. Since G has an identity element, it is of the form G “ te, au

where a ‰ e. Now if a2 “ a then by cancellation laws we obtain a “ e,

which is impossible. Thus we should have a2 “ e. Step 2. Now consider

the map

φ : Z2 Ñ H, φp0q “ e and φp1q “ a.

It is straightforward to verify that φ is an isomorphism. �

Proposition 4.2. Let G be a group such that |G| “ 3. Then G – Z3.

Proof. As before we can express G as G “ te, a, bu where e is the identity

element.

Step 1. We show that ab “ e. Assume the contrary. Then we should be in

one of the following two cases:

(i) ab “ a. Then cancellation implies b “ e, which is a contradiction.

(ii) ab “ b. Then cancellation implies a “ e, which is a contradiction.

Step 2. We show that a2 “ b. Assume the contrary. Then we should be in

one of the following two cases:

(i) a2 “ a. By cancellation this implies a “ e which is a contradiction.

(ii) a2 “ e. Then by Step 1 we have a2 “ ab, hence by cancellation we

obtain a “ b which is a contradiction.

Step 3. Similar to Step 2, we can show that b2 “ a. Thus the distinct

element of G are G “ te, a, a2u, and we have a3 “ apa2q “ pa2qa “ e and

pa2qpa2q “ a.

- 35 -

Step 4. We can now easily verify that the map

φ : Z3 Ñ G

defined by

φp0q “ e, φp1q “ a, φp2q “ a2

is an isomorphism. �

Before we describe all groups of order 4, we prove a general proposition:

Proposition 4.3. Let G be any finite group. Then every row and every columnin the Cayley table of G contains a permutation of the elements of G.

Proof. The column of the Cayley table corresponding to a P G contains

elements of the form xa for x P G. Now from Proposition 3.3 it follows

that every g P G appears precisely once in this column. The argument for

the rows in analogous. �

Proposition 4.4. Let G be a group such that |G| “ 4. Then either G – Z4 orG – Z2 ˆ Z2.

Proof. We can write the elements of G as G “ te, a, b, cu where e is the

identity element.

Step 1. Similar to the proof of Proposition 4.2, we know that ab ‰ a and

ab ‰ b. Thus we have ab P te, cu. We consider two separate cases:

Case I: ab “ e.

Step I.1. We show ac “ b. Assume the contrary. Then we are in one of the

following the following cases:

‚ ac “ e. Then ac “ ab, hence c “ b which is a contradiction.

‚ ac “ a. Then by cancellation c “ e which is a contradiction.

‚ ac “ c. Then by cancellation a “ e which is a contradiction.

- 36 -

Step I.2. We show a2 “ c. Assume the contrary. Then we are in one of the

following cases:

‚ a2 “ e: Then a2 “ ab, hence a “ b, which is a contradiction.

‚ a2 “ a: Then a “ e which is a contradiction.

‚ a2 “ b: Then a2 “ ac, hence a “ c which is a contradiction.

Step I.3. We show bc “ a. Assume the contrary. Then we are in one of the

following cases:

‚ bc “ b. Then by cancellation c “ e which is a contradiction.

‚ bc “ e. Then bc “ ba (since ab “ e implies ba “ e), hence by cancella-

tion c “ a which is a contradiction.

‚ bc “ c. Then by cancellation b “ e, which is a contradiction.

Step I.4. So far the Cayley table of G looks like

e a b c

e e a b c

a a c e b

b b e a

c c

By Proposition 4.3 we can complete the rest of the Cayley table to the fol-

lowing:

e a b c

e e a b c

a a c e b

b b e c a

c c b a e

We can now check that that the map φ : Z4 Ñ G defined by

φp0q “ e, φp1q “ a, φp2q “ c, φp3q “ b

- 37 -

is an isomorphism.

Case II: ab ‰ e.

Step II.1. By symmtry, we can assume that ba ‰ e, bc ‰ e, cb ‰ a, ca ‰ e,

and ac ‰ e (otherwise after relabelling the variables we will be back in

Case I).

Step II.2. It follows (since ab P te, cu, etc.) that

ab “ c “ ba, bc “ a “ ab, ca “ b “ ac.

Step II.3. So far the Cayley table of G looks like

e a b c

e e a b c

a a c b

b b c a

c c b a

By Proposition 4.3 we can complete the rest of the Cayley table to the fol-

lowing:e a b c

e e a b c

a a e c b

b b c a a

c c b a e

It is now a straightforward calculation that the map

φ : Z2 ˆ Z2 Ñ G

defined by

φp0, 0q “ e, φp1, 0q “ a, φp0, 1q “ b, φp1, 1q “ c

is an isomorphism. �

- 38 -

Remark 4.5. The groups Z4 and Z2ˆZ2 are not isomorphic. In fact suppose

that there exists an isomorphism

φ : Z4 Ñ Z2 ˆ Z2.

Set pa, bq “ φp1q. Now on the one hand 1` 1 ‰ 0, hence

φp1` 1q ‰ φp0q,

since φ is a bijection. On the other hand for any a P Z2 we have a ` a “ 0,

hence

φp1` 1q “ φp1qφp1q “ pa, bqpa, bq “ pa` a, b` bq “ p0, 0q “ φp0q,

which is a contradiction.

- 39 -

SUBGROUPS

Definition 5.1. Let G be any group. A subset H Ď G is called a subgroup

of G if the restriction of the binary operation of G to H makes H a group.

If H is a subgroup of G, then we write H ď G.

Example 5.2. (i) Z is a subgroup of Q (the binary operation on both is

the usual addition).

(ii) For any group G, the subset teu where e “ eG is a subgroup. It is

called the trivial subgroup of G.

(iii) Up5q is not a subgroup of Z5 even though Up5q Ď Z5.

(iv) The set t˘I,˘Iu is a subgroup of the quaternion groupQ of Example

2.1.

(v) Zˆ t0u is a subgroup of Zˆ Z.

(vi) Zˆ t1u is not a subgroup of Zˆ Z.

(vii) Let n ě 2. The set tσ P Sn : σpnq “ nu is a subgroup of Sn.

Remark 5.3. (i) Note that if H Ď G is a subgroup, then the identity

elements eH of H and eG of G coincide: this is because

peHq2“ eH “ eHeG,

and the cancellation law (in G) implies eH “ eG.

(ii) It is also straightforward to verify that the inverse inH of an element

g P H is the same as its inverse in G. We leave the verification of this

statement as an exercise.

Theorem 5.4. (Subgroup Test) LetG be any group and letH Ď G be any subsetof G. Then H is a subgroup of G if and only if the following are satisfied.

(i) e P G.(ii) For all h, h1 P H we have hh1 P H .

(iii) For all h P H we have h´1 P H , where h´1 denotes the inverse of h in G.

- 40 -

Proof. Assume that conditions (i)–(iii) are satisfied. We show that H is a

subgroup of G.

Step 1. From (ii) it follows that the binary operation of G induces a binary

operation on H . Associativity of this operation follows from the fact that

G is a group.

Step 2. From (i) it follows that H has an identity element (the identity of G

is also an identity of H).

Step 3. From (iii) it follows that every h P H has an inverse in H .

Thus all of the conditions of Definition 2.15 are satisfied for H .

Conversely, if H is a subgroup of G, then it is straightforward to verify

the conditions (i)–(iii). We leave this verification as an exercise.

�

Exercises.

(1) Using Theorem 5.4 show that the following sets of elements is a sub-

group of the given G.

(i) tX P GLnpRq : detpXq “ 1u, G “ GLnpRq.(ii) dZ :“ tkd : k P Zu “ t0,˘d,˘2d,˘3d, . . .u where d is a fixed

integer, G “ Z.

(2) Determine all of the subgroups of Z2 ˆ Z2.

Theorem 5.5. (Subgroup Test - Concise Version) (Subgroup Test - Finite

Groups Version) Let G be any group and let H Ď G be a nonempty subset of G.Then H is a subgroup of G if and only if H satisfies the following property:

for all g, h P H we have gh´1 P H .

- 41 -

Proof. If H is a subgroup of G, then it satisfis the property stated in the

theorem. We leave this straightforward argument as an exercise.

Conversely, assume that H is a nonempty subset that satisfies the given

condition, i.e., for all g, h P H we have gh´1 P H . To prove that H is a sub-

group, we verify that the conditions (i)–(iii) of Theorem 5.4 are satisfied:

Step 1. Since H is nonempty, we can find h0 P H . Then e “ h0h´10 P H ,

thus assumption (i) of Theorem 5.4 holds.

Step 2. We show that for every a P H we have a´1 P H . Since e P H , setting

g :“ e and h :“ a, we obtain

gh´1 “ ea´1 “ a´1 P H.

Note that this verifies assumption (iii) of Theorem 5.4.

Step 3. Finally, we show that for all a1, a2 P H we have a1a2 P H . This

verifies assumption (ii) of Theorem 5.4.

Let a1, a2 P H . Then by Step 2 we have pa2q´1 P H . Now setting g “ a1

and a2 “ pa2q´1 we obtain gh´1 “ a1a2 P H . �

Theorem 5.6. (Subgroup Test - Finite Groups Version) LetG be a finite groupand let H Ď G be a nonempty subset of G. Then H is a subgroup of G if and onlyif H is closed under the binary operation of G, i.e.,

for all h, h1 P H we have hh1 P H .

Proof. We show that assumptions (i) and (iii) of Theorem 5.4 are satisfied.

Step 1. Since H ‰ ∅, we can choose some h0 P H . Now consider the set

Sph0q :“ th0, h20, h

30, . . .u “ th

k0 : k P Nu.

From the assumption of the theorem it follows that Sph0q Ď H .

- 42 -

Step 2. Since H is finite, Sph0q should also be finite. Thus, there exist

integers k ą ` ě 1 such that hk0 “ h`0. Multiplying by hk`0 , we obtain that

hm0 “ e where m :“ k ´ ` P N.

Thus e P S. Consequently, e P H .

Step 3. Finally, we show that for every h P H , we have h´1 P H . To this

end, note that as in Step 2, we have hm for some m P N. If m “ 1, then

h “ e, and clearly h´1 “ e P H . If m ą 1, then

h´1 “ eh´1 “ hmh´1 “ hm´1 P S,

and therefore h´1 P H . �

Example 5.7. Theorem 5.6 is not correct if G is not finite. For example, if

G “ Z (equipped with addition) and H “ N, then H satisfies the condition

of Theorem 5.6, but H is not a subgroup of Z because 1 P H but ´1 R H .

Proposition 5.8. LetG be a group and letH ď G andK ď G. ThenHXK ď G.

Proof. We verify the assumptions of Theorem 5.4 for H XK:

Step 1. Since e P H and e P K, we have e P H XK.

Step 2. For a, b P H XK, we have ab P H because a and b are in H , and we

have ab P K because a and b are in K. Thus ab P H XK.

Step 3. By a similar argument as in Step 2, it follows that for h P H XK we

have h´1 P H XK.

�

- 43 -

Example 5.9. If H,K are subgroups of a group G, then H YK is not nec-

essarily a subgroup of G: Set G “ Z6 and

H :“ t0, 3u and K :“ t0, 2, 4u.

Then H and K are subgroups of Z6 but H Y K “ t0, 2, 3, 4u is not a sub-

group because 2` 3 “ 5 R H YK.

The centre of a group. The centre and centralizer of an element of a group

are some of the most ubiquitous examples subgroups of groups.

Definition 5.10. Let G be any group. The centre of G is the subset of G

defined by

ZpGq :“ tg P G : gh “ hg for every h P G.u .

In other words, the centre of G is the set of elements of G which commute

with every element of G.

Theorem 5.11. Let G be any group. Then the centre of G is an abelian subgroupof G.

Proof. We use the concise subgroup test (Theorem 5.5):

Step 1. Clearly eg “ ge “ g for all g P G, hence e P ZpGq. In particular,

ZpGq ‰ ∅.

Step 2. Assume that a, b P ZpGq. We want to show that ab´1 P G. Then for

every g P G we have

bg “ gb.

Multiplying on both sides by b´1, we obtain

gb´1 “ b´1g for all g P G.

- 44 -

It follows that for all g P G we hvve

pab´1qg “ apb´1gq “ apgb´1q “ pagqb´1 “ pgaqb´1 “ gpab´1.

Thus ab´1 P ZpGq.

Step 3. Finally, ZpGq is abelian, because the definition of ZpGq implies that

the relation ab “ ba holds for all a, b P ZpGq. �

Example 5.12. (i) IF G is an abelian group, then ZpGq “ G.

(ii) Let Q “ t˘I,˘I,˘J,˘Ku be the quaternion group (see Proposition

2.27). Then ZpQq “ t˘Iu.

(iii) The centre of S3 is the trivial subgroup of S3. That is, ZpS3q “ teu.

Exercises.

(1) Verify the claim of Example 5.12(iii).

(2) Let G and H be two groups, and let f ;G Ñ H be an isomorphism

between G and H . Prove that for every subgroup G1 of G, the set

fpG1q :“ tfpgq : g P G1u

is a subgroup of H .

Definition 5.13. Let G be any group, and let a P G be any element. The

centralizer of a, denoted by CGpaq, is defined as

CGpaq :“ tg P G : ag “ gau.

Proposition 5.14. Let G be any group. Then CGpaq is a subgroup of G.

Proof. We use the Concise Subgroup Test. Note that CGpaq ‰ ∅ since

e P CGpaq.

- 45 -

Step 1. Let x, y P CGpaq. We need to show that xy´1 P CGpaq. We have

ya “ ay ñ y´1pyaqy´1 “ y´1payqy´1 ñ py´1yqay´1 “ y´1apyy´1q

ñ eay´1 “ y´1ae ñ ay´1 “ y´1a.

Step 2. Now we have

pxy´1qa “ xpy´1aq “ xpay´1q “ pxaqy´1 “ paxqy´1 “ apxy´1q.

This implies that xy´1 P CGpaq. Thus the assumption of Theorem 5.5 holds.

�

Example 5.15. (i) Let Q “ t˘I,˘I,˘J,˘Ku be the quaternion group

(see Proposition 2.27). Then CQpJq “ t˘I,˘Ju.

(ii) Let G “ GL2pRq, and set A :“

«

1 1

0 1

ff

. Then

CGpAq “

#«

a b

0 a

ff

: a P C˚ and b P C.

+

.

To prove the latter claim, take B “

«

x y

z t

ff

P CGpAq. Then

AB “ BA implies

«

x` z y ` t

z t

ff

“

«

x x` y

z z ` t

ff

.

It follows that x` y “ x, hence z “ 0, and y ` t “ x` y, hence t “ x.

Thus B is of the form«

x y

0 x

ff

.

Since B is invertible, we have x ‰ 0. Furthermore, by a direct cal-

culation we can verify that all matrices of the latter form belong to

CGpAq.

- 46 -

CYCLIC GROUPS

Definition 6.1. (i) We say a group G is a cyclic group if there exists an

element g P G such that

G “ tgk : k P Zu.

(ii) Let G be any group. A subgroup H of G which is of the form

H “ thk : k P Zu

is called a cyclic subgroup of G generated by h. Such a cyclic sub-

group is denoted by xhy.

Example 6.2. (i) The group Zn is cyclic because Zn “ x1y.(ii) Note that the generator of a cyclic group is not necessarily unique.

For instance, we have

Z6 “ x1y “ x5y.

(iii) The group Z is cyclic. Note that Z “ x1y “ x´1y.(iv) The set S :“ tpa, 0q : a P Znu is a cyclic subgroup of Zn ˆ Zn. Note

that S – Zn, where the isomorphism is the natural map pa, aq ÞÑ a.

(v) The group Up9q “ t1, 2, 4, 5, 7, 8u is cyclic. Indeed Up9q “ x2y. How-

ever, note that we have x4y “ t1, 4, 8u Ĺ Up9q.

Our next goal is to classify all cyclic groups. We begin with the following

definition.

Definition 6.3. Let G be any group, and let g P G.

(i) If there exists some k P N such that gk “ e, then we define the orderof g to be opgq :“ mintk P N : gk “ eu.

(ii) If gk ‰ e for all k P N, then we set opgq “ 8.

Example 6.4. (i) The order of 3 in Z12 is 4. That is, op3q “ 4.

- 47 -

(ii) The order of 3 in Z is infinity. That is, op3q “ 8.

(iii) opeq “ 1 for every group G.

Exercises.

(1) let G be any group. Show that opgq “ opg´1q for all g P G.

(2) Assume that G is an abelian group and g, h P G such that opgq “ 2

and ophq “ 3. Prove that opghq “ 6.

Lemma 6.5. Let G be a finite group. Then opgq ă 8 for every g P G.

Proof. Fix g P G. The set Spgq :“ tg, g2, g3, . . .u “ tgk : k P Nu is a subset of

G, hence in particular it is finite. In particular, there exist integers k ą ` ě 1

such that gk “ g`. It follows that gk´` “ gkg´` “ e. Note that k ´ ` P N.

Now from Definition 6.3(i) it follows that opgq ă 8.

�

Theorem 6.6. Let G be any group, and let g P G such that opgq ă 8. Setn :“ opgq.

(i) For k P Z, we have gk “ e if and only if n|k.(ii) For k, ` P Z, we have gk “ g` if and only if n|pk ´ `q.

(iii) xgy “

e, g, ¨ ¨ ¨ , gn´1(

.

Proof. (i) For k P Z, by the division algorithm we can write k “ nq ` r,

where 0 ď r ď n´ 1. Note that

n|k if and only if r “ 0.

Now

gk “ gnq`r “ gnqgr “ pgnqqgr “ eqgr “ gr.

If r “ 0 then gk “ e, and if 1 ď r ă n then gk ‰ e because k ă n “ opgq.

This completes the proof of (i).

- 48 -

(ii) Note that gk “ g` if and only if gk´` “ e. Now we write pk´ `q “ nq` r

by the division algorithm, where 0 ď r ď n ´ 1. From (i) it follows that

gk´` “ e if and only if n|pk ´ `q.

(iii) From (i) it follows that for any k P Z, gk “ gr for some 0 ď r ď n ´ 1.

From (ii) it follows that the elements e “ g0, g1, . . . , gn´1 are distinct. Thus

xgy “ te, g, . . . , gn´1u.

�

Theorem 6.7. Let G be a cyclic group.

(i) If G is finite, then G – Zn for some n P N.(ii) If G is not finite, then G – Z.

Proof. By definition of a cyclic group, we know thatG “ xgy for some g P G.

If G is finite, then by Lemma 6.5 we know that opgq P N. Set n :“ opgq.

We define a map

φ : Zn Ñ G : φpkq :“ gk.

We show that this map is an isomorphism.

Step 1. Clearly φpZnq Ď xgy. From Theorem 6.6(iii) it follows that indeed φ

is a bijection onto xgy.

Step 2. We show that φpa` bq “ φpaqφpbq. For further clarity, we represent

the addition of Zn by `, and the usual addition of Z by `. We have

φpa ` bq “ ga`b and φpaqφpbq “ gagb “ ga`b.

From the definition of ` it follows that n| rpa` bq ´ pa ` bqs, hence by The-

orem 6.6(ii) it follows that ga`b “ ga`b.

- 49 -

Next assume that G is not finite. Since G “ xgy, we must have opgq “ 8.

Now we consider the map

φ : ZÑ G , φpkq “ gk.

We will show that φ is an isomorphism.

Step 1. Clearly φ is a surjection onto xgy. We show that it is also an injec-

tion. Assume, on the contrary, that φpkq “ φp`q for integers k ą `. Then

gk´` “ gkg´` “ e. Since k ´ ` P N, from Definition 6.3(i) it follows that we

should have opgq ă 8. This is a contradition.

Step 2. To complete the proof, it is enough to verify that φpa`bq “ φpaqφpbq

for all a, b P Z. To this end, note that

φpa` bq “ ga`b “ gagb “ φpaqφpbq,

which verifies the latter claim. �

Computing opgdq from opgq and d. Suppose we know opgq for some g P G.

How can we compute opgdq? This is answered by the next theorem.

Theorem 6.8. Let G be any group and let g P G such that opgq ă 8. Then forevery d P Z we have

opgdq “opgq

gcdpopgq, dq.

Proof. Set m :“ opgqgcdpopgq,dq and m1 “ d

gcdpopgq,dq . Note that m,m1 P Z. Indeed

m P N.

Step 1. We show that pgdqm “ 1. We have

pgdqm “ gdm,

- 50 -

and

dm “ dopgq

gcdpopgq, dq“

d

gcdpopgq, dqopgq “ m1opgq.

From the latter equation it follows that opgq|dm. Thus Theorem 6.6(i) im-

plies that gdm “ e.

Step 2. Let k P N be such that pgdqk “ e. We claim that m|k. To prove this

claim, note that From gdk “ e and Theorem 6.6(i) it follows that opgq|dk.

Thus

dk “ opgqt for some t P Z.

Now

dk “ opgqt ñ m1k “d

gcdpopgq, dqk “

opgq

gcdpopgq, dqt “ mt ñ m|m1k.

From Corollary 1.11 it follows that

gcd pm,m1q “ 1.

Therefore from m|m1k, gcdpm,m1q “ 1, and Lemma 1.15(iii) it follows that

m|k. In particular, m ď k (since m, k ě 1).

Step 3. From Step 2 it follows that if pgdqk “ e for some k P N, then m ď k,

and therefore m “ min

k P N : pgdqk “ e(

, hence m “ opgq. �

Exercises.

Let G be a group and let g, h P G such that opgq “ m and ophq “ n. Assume

that gh “ hg. Prove that opghq “ mn.

Subgroups of cyclic groups. Our next goal is to classify subgroups of

cyclic groups. We start with a general statement about the order of powers

of an element.

- 51 -

Remark 6.9. For any d, n P N such that d|n, set

dZn :“ tk P Zn : d|ku “ t0, d, 2d, . . . , n´ du.

Note that dZn is the set of elements of Zn that are divisible by d. In partic-

ular,

nZn “ t0u.

It is straightforward to check using the subgroup test that dZn is a sub-

group of Zn.

Example 6.10. Here are some examples of the groups dZn:

(i) 4Z12 “ t0, 4, 8u.

(ii) 3Z18 “ t0, 3, 6, 9, 12, 15u.

Theorem 6.11. Let n ě 1. The subgroups of Zn are precisely the sets dZn forintegers d P N such that d|n. Here

dZn :“ tk P Zn : d|ku “ t0, d, 2d, . . . , n´ du.

Proof. Fix a subgroup H of Zn. If H “ t0u then we can set d “ n. Next

assume that H ‰ t0u.

Step 1. Set d :“ mintk P H : k ą 0u. Our goal is to prove that indeed

H “ dZn.

Step 2. We prove that d|n. In order to prove this claim, write n “ qd` r for

0 ď r ă d by the division algorithm. (Note that q ě 0 because n ě d ě 1.)

To verify that d|n, it is enough to show that r “ 0.

Now assume, on the contrary, that r ą 0. Then from r` qd “ n it follows

that r is the inverse of qd in Zn. However, qd “ d` ¨ ¨ ¨ ` dlooooomooooon

q times

P H . It follows

- 52 -

that the inverse of qd also belongs to H . Namely, r P H . But this contra-

dicts the fact that d “ mintk P H : k ą 0u.

Step 3. Since d P H and H is closed under the addition of Zn, it is clear that

dZn Ď H . Next we prove the reverse inclusion: H Ď dZ.

Take any element k P H . Then by the division algorithm we can write

k “ qd` r for some 0 ď r ă d.

As usual, let ´qd denote the inverse of qd in Zn. As in Step 2, we have

qd P H and therefore ´qd P H . By adding ´qd to both sides of k “ qd ` r,

we obtain

k ` p´qdq “ r, hence r “ k ` p´qdq P H.

If r ‰ 0, then by the definition of d we should have d ď r, but we know

that r ă d, and this is a contradiction. Thus we should have r “ 0, and

therefore k “ qd. This implies that k P dZn. �

Example 6.12. The subgroups of Z12 are

Z12, 2Z12, 3Z12, 4Z12, 6Z12, 12Z12.

Exercises.

(1) Verify that for n, d P N such that d|n, we have p´dqZn “ dZn.

(2) Verify that the subgroup of Z18 generated by 10 is the same as the

subgroup of Z18 generated by 2. Note that 2|18 but 10 - 18. Why

does this not contradict Theorem 6.11?

(3) Let n, d P N such that d|n. Prove that the subgroup dZn of Zn is

isomorphic Znd. This implies that every subgroup of Zn is a cyclic

group.

- 53 -

Remark 6.13. For d P Z, set dZ :“ tk P Z : d|ku. Note that dZ is precisely

the set of integers that are divisible by d (indeed for d “ 0 we have 0Z “t0u). Using the subgroup test, it is easy to verify that dZ is a subgroup of

Z.

Example 6.14. For example, 3Z “ t0,˘3,˘6,˘9, . . .u.

Theorem 6.15. The subgroups of Z are precisely the sets dZ :“ tk P Z : d|ku,where d P Z.

Proof. The proof is similar to the one given for Theorem 6.11. Fix a sub-

group H of Z. If H “ t0u then H “ 0Z and we can set d “ 0. From now

on, we assume that H ‰ t0u.

Step 1. Set d :“ mintx P H : x ą 0u. Our final goal is to show thatH “ dZ.

Let us verify that dZ Ď H . Indeed from d P H it follows that

@kPN : kd “ d` ¨ ¨ ¨ ` dlooooomooooon

k times

P H

and ´kd is the inverse of kd in G, so ´kd P H as well (since H is a group).

This completes the proof of dZ Ď H .

Step 2. We prove that H Ď dZ. Assume, on the contrary, that there

existsm P H such that m R dZ. Now by the division algorithm we can

write

m “ qd` r where 0 ď r ă d.

Indeed r ‰ 0 because m R dZ. Now m, qd P H , and therefore

r “ m´ qd P H.

Since 0 ă r ă d, this contradicts the fact that d is the smallest positive

element of H . �

- 54 -

THE CIRCLE AND DIHEDRAL GROUPS

7.1 The circle group T. Set T :“ tz P C : |z| “ 1u. Since we have

|z1z2| “ |z1| ¨ |z2|

for any two z1, z2 P C, the set of complex numbers T is closed under mul-

tiplication.

Proposition 7.1. The set T, equipped with multiplication, is an abelian group.

Proof. T is a subset of the group C˚ of nonzero complex numbers. Using

the Concise Subgroup Test it is straightforward to verify that T is indeed a

subgroup of C˚. We leave the details as an exercise. �

A geometric way to think about T is as follows. Every element of T is

a point on the unit circle. This point forms an angle 0 ď θ ă 2π with the

x-axis (we measure the angle counterclockwise). We denote this point by

eiθ. Recall that

i “?´1.

The standard Cartesian coordinates of eiθ are pcos θ, sin θq. Therefore we

are defining

eiθ :“ cospθq ` i sinpθq.

θ

eiθ

- 55 -

For example

eiπ “ cospπq ` i sinpπq “ ´1, and e2πi3 “ cosp2π3 q ` i sinp

2π3 q “

12 `

?32 i.

Using trigonometry, one can show that for any two numbers θ1, θ2 P R,

(7.1) eipθ1`θ2q “ eiθ1eiθ2.

Here the multiplication on the right hand side is taking place in complex

numbers.

Exercises.

(1) Using the formulas for cospa ` bq and sinpa ` bq in terms of the val-

ues of cospaq, sinpaq, cospbq, and sinpbq from trigonometry, prove the

relation (7.1).

(2) Verify that eπ “ ´1 and e2πi “ 1 and eπ2 i “ i and e

π3 i “ 1

2 “?32 i.

(3) Prove that every nonzero complex number z can be written uniquely

in the form z “ reiθ, where r ą 0 and 0 ď θ ă 2π. What happens for

z “ 0?

Hint: Set r :“ |z|.

Remark 7.2. You are most likely familiar with et for a real number t. You

know that

(7.2) et “ 1` t`t2

2!`t3

3!` ¨ ¨ ¨

In this section, we defined et for imaginary values of t. Interestingly enough,

the formula (7.2) still remains valid. We will not need this formula. But it

is good to know that everything remains consistent!

Let us now look at the points e2πkn i for k “ 0, . . . , n´1. These points form

the angles 2πkn with the x-axis. Therefore they are the vertices of a regular

n-gon. The following picture illustrates these points for n “ 5:

- 56 -

‚

‚

‚

‚

‚

e2π5 i

e4π5 i

e6π5 i

e8π5 i

1

Proposition 7.3. Let n P N, and set

Rn :“!

1, e2πn i, . . . , e

2πkn i, . . . , e

2πpn´1qn i

)

“

!

e2πkn i : k P Z and 0 ď k ď n´ 1

)

.

Then Rn is the set of complex roots of the equation zn “ 1.

Proof. (i) First note that from (7.1) it follows that for every k P Z,´

e2πkn i¯n

“ en2πkn i“ e2πki “ 1.

Thus every element of Rn is a root of zn “ 1.

Conversely, assume that z P C satisfies zn “ 1. Then

1 “ |1| “ |zn| “ |z|n,

and since |z| is a non-negative real number, it follows that |z| “ 1. This

means that z P T. Therefore z “ eiθ for some 0 ď θ ă 2π. Now

zn “ peiθqn “ eniθ “ cospnθq ` i sinpnθq.

Since zn “ 1, we find that

cospnθq “ 1 and sinpnθq “ 0.

- 57 -

These relations (together with the well-known properties of sinx and cosx)

imply that

nθ “ 2πk for some k P Z.

But we know that 0 ď θ ă 2π, and therefore 0 ď nθ ă 2πn. Consequently,

we should have

nθ “ 2πk for some k P t0, 1, . . . , n´ 1u.

Equivalently,

θ “2πk

nfor some k P t0, 1, . . . , n´ 1u.

It follows that the complex number z “ eiθ belongs to Rn. �

Example 7.4. The set of complex roots of the equation z5 “ 1 is

R5 “

!

1, e2π5 i, e

4π5 i, e

6π5 i, e

8π5 i)

.

The set of complex roots of the equation z4 “ 1 is

R4 “

!

1, e2π4 i, e

4π4 i, e

6π4 i,

)

“ t1, i,´1,´iu.

Remark 7.5. Let n P N Let a P Z, and let b be the remainder of a modulo n.

(Thus b P t0, . . . , n´ 1u.) Then

e2πan i“ e

2πbn i.

The reason is that by the division algorithm we can write a “ nq` b where

q P Z. Then,

e2πan i“ e

2πpnq`bqn i

“ e2πnqn ie

2πbn i“ e2πqie

2πbn i“ pe2πiqqe

2πbn i“ p1qqe

2πbn i“ e

2πbn i.

Proposition 7.6. Let n P N, and let Rn be the subset of T defined in Proposition7.3. Then Rn is a subgroup of T and it is isomorphic to Zn.

- 58 -

Proof. Step 1. We show that Rn is a subgroup of T. This can be done by

the Concise Subgroup Test (note that 1 P Rn, so that Rn ‰ ∅). We need to

show that for any two elements z1, z2 P Rn, we have z1z´12 P Rn. Recall that

by Proposition 7.3, the set Rn is the set of all roots of zn “ 1 in C. Thus it

is sufficient to check that z1z´12 is a root of the equation zn “ 1. This can be

verified as follows:

pz1z´12 q

n“ zn1 z

´n2 “ p1qp1q “ 1.

Step 2. We define an isomorphism φ : Zn Ñ Rn. We set

φ : Zn Ñ T , φpkq :“ e2πikn for k P t0, 1, . . . , n´ 1u.

This map is clearly a bijection. It is an isomorphism because

φpk1 ` k2q “ e2πpk1`k2 mod nq

n i“ e

2πpk1`k2qn i

“ e2πk1n ie

2πk2n i“ φpk1qφpk2q.

Note that in the last calculation, we are using Remark 7.5. �

The dihedral groups Dn, n ě 1.

Definition 7.7. Let n ě 3 be an integer. A regular n-gon is a convex poly-

gon with n sides of equal length such that the angles between its adjacent

sides are all equal.

Example 7.8. An equilateral triangle is a regular 3-gon and a square is a

regular 4-gon.

Remark 7.9. The points e2πkn i P C with k P t0, . . . , n ´ 1u are the vertices of

a regular n-gon.

Definition 7.10. Let n ě 3 be an integer. The group of symmetries of a

regular n-gon is called the n-th dihedral group.

- 59 -

The reflections of a regular pentagon.

Let us describe an explicit realization of the symmetries of the n-gon of

Remark 7.9. Consider the following symmetries:

(i) n reflections about the lines passing through the origin and the point

eπkn i for k “ 0, . . . , n´ 1.

(ii) n counterclockwise rotations with angles 2kπn , for k “ 0, 1, . . . , n ´ 1,

around the origin.

Consider the following symmetries of the aforementioned n-gon:

(i) r : the counterclockwise rotation about the origin with angle 2πn .

(ii) s : the reflection about the x-axis.

Clearly r, s P Dn. Furthermore, by geometric investigation we observe that

oprq “ n and opsq “ 2 and srs “ srs´1 “ r´1.

Thus the elements of Dn are precisely

e, r, r2, . . . , rn´1, s, rs, r2s, . . . , rn´1s(

.

Remark 7.11. (i) rks is reflection about the line passing throught the ori-

gin and the point eπkn i.

(ii) srks is a clockwise rotation with angle 2πkn , which is the same as a

counterclockwise rotation with angle 2πpn´kqn .

- 60 -

Example 7.12. Consider the element rsr3sr´1ssr P D5. We can simplify

this element as follows:

rsr3sr´1ssr “ rsr3sr4pssqr “ rsr3spr4rq

“ rsr3s “ rsr3s´1 “ rpsrs´1q3 “ rpr´1q3 “ r´2 “ r3.

Exercises.

(1) Prove or disprove: D4 – Q.

(2) Assume n ě 3. Prove that trk : 0 ď k ď n ´ 1u is a cyclic subgroup

of Dn.

(3) Determine all of the subgroups of D6.

- 61 -

SYMMETRIC GROUPS

Recall that Sn denotes the group of permutations of t1, . . . , nu. The num-

ber of elements of Sn is n! “ npn´ 1qpn´ 2q ¨ ¨ ¨ 2 ¨ 1.

Remark 8.1. From now on, we denote the identity permutation by ε. Thus

εpkq “ k for all integers 1 ď k ď n.

We have already represented permutations as an array in (2.2). We are

going to introduce a more effective representation of permutations. The

idea is the following: given a permutation σ, we can start from some k

and compute the values

σpkq , σpσpkq , σpσpσpkqqq , . . . .

Since the values are always integers between 1 and n, some numbers should

reoccur. Since σ is a bijection, we can “back up” from any repetition. This

implies thatthere exists a j P N such that σpjqpkq “ k. Thus, the numbers

k, σpkq, σpσpkqq, . . . , σpj´1qpkq

form a “cycle”.

Definition 8.2. Let d, n P N such that 2 ď d ď n. A d-cycle (or a cycle of

length d) in Sn is a permutation σ P Sn with the following property: there

exist d distinct numbers a1, . . . , ad such that

(i) σpaiq “ ai`1 for 1 ď i ď d´ 1 and σpadq “ a1,

(ii) σpkq “ k for all k P t1, . . . , nu such that k R ta1, . . . , adu.

We denote this d-cycle by

σ “ pa1 a2 ¨ ¨ ¨ ad´1 adq.

- 62 -

Example 8.3. The permutation

σ :“

˜

1 2 3 4 5

1 4 2 3 5

¸

is a cycle of length 3 in S5 because

σp2q “ 4 , σp4q “ 3 , σp3q “ 2

and σpkq “ k for k R t1, 5u.

Proposition 8.4. Let n ě 1 and let σ “ pa1 a2 ¨ ¨ ¨ ad´1 adq P Sn be a d-cycle.Then σ´1 “ pad ad´1 ¨ ¨ ¨ a2 a1q. In paricular, σ´1 is also a d-cycle.

Proof. This is straightforward. �

Proposition 8.5. Let n P N and let σ :“ pa1 a2 ¨ ¨ ¨ ad´1 adq be a d-cycle inSn. Then

σ “ pai ai`1 ¨ ¨ ¨ ad´1 ad a1 a2 ¨ ¨ ¨ ai´2 ai´1q for every 2 ď i ď d.

Proof. It is easy to verify that both cycles map ai to ai`1 for 1 ď i ď d ´ 1,

map ar to a1, and fix all x P t1, . . . , nuzta1, . . . , adu. �

Remark 8.6. Let σ P Sn and assume that σ is a d-cycle, i.e., σ “ pa1 ¨ ¨ ¨ adq.

Obviously we have n ě maxta1, . . . , adu. Now, for any m ą n, we can also

think of σ as a d-cycle in Sm.

Products of cycles. Since cycles are permutations, products of cycles is

simply the composition of their corresponding permutations. To compute

the product of cycles in Sn, we should act ona letter x P t1, . . . , nu from the

rightmost cycle, and move towards the left.

- 63 -

Example 8.7. Let σ1 “ p3 1 5q and σ2 “ p5 4 1q and σ3 “ p5 2q be

permutations in S6. Then

p3 1 5qp5 4 1qp5 2q “ σ1σ2σ3 “

˜

1 2 3 4 5 6

3 4 1 5 2 6

¸

.

Disjoint cycles and cycle decomposition.

Definition 8.8. Two cycles

σ “ pa1 ¨ ¨ ¨ arq and τ “ pb1 ¨ ¨ ¨ bsq

in Sn are called disjoint if the numbers a1, . . . , ar, b1, . . . , bs are all distinct.

Example 8.9. The cycles σ, τ P S10 given by

σ “ p2 4 1 6q and τ “ p5 8 3q

are disjoint.

We now give a characterisation of disjoint cycles in terms of their fixed

point sets.

Definition 8.10. Let n ě 1 and let σ P Sn. The set of points moved by σ,

denoted by Mσ, is defined by

Mσ :“ tx P N : 1 ď x ď n and σpxq ‰ xu.

Example 8.11. Let τ :“ pa1 ¨ ¨ ¨ adq be a d-cycle in Sn. Then τ only moves

a1, . . . , ad. In other words, Mτ “ ta1, . . . , adu.

Remark 8.12. let n P N and let σ, τ be two cycles in Sn. From the definition

of disjoint cycles it follows that σ and τ are disjoint if and only ifMσXMτ “

∅. In plain terms, σ and τ commute if and only if the set of elements of

t1, . . . , nu that are moved by σ and by τ are disjoint sets.

An important property of disjoint cycles is that they commute.

- 64 -

Proposition 8.13. Let n ě 1 and let σ, τ P Sn be disjoint cycles. Then στ “ τσ.

Proof. Assume that

σ “ pa1 ¨ ¨ ¨ arq and τ “ pb1 ¨ ¨ ¨ bsq.

Fix x P t1, . . . , nu. We need to show that σpτpxqq “ τpσpxqq. One of the

following 3 cases happens:

Case I. x P ta1, . . . , aru. Then τpxq “ x because x R tb1, . . . , bsu. Hence

(8.1) σpτpxqq “ σpxq.

Also, τpσpxqq “ σpxq since σpxq P ta1, . . . , aru and therefore σpxq R tb1, . . . , bsu.

Consequently,

(8.2) τpσpxqq “ σpxq.

From (8.1) and (8.2) it follows that σpτpxqq “ τpσpxqq.

Case II. x P tb1, . . . , bsu. This case is very similar to Case I. We leave the

details of the proof as an exercise.

Case III. x R ta1 . . . , aru Y tb1, . . . , bsu. In this case σpxq “ τpxq “ x and

therefore

σpτpxqq “ σpxq “ x and τpσpxqq “ τpxq “ x. �

Definition 8.14. Let σ P Sn be a permutation. Starting from an element

x P t1, . . . , nu, we consider the elements

x, σpxq, σpσpxqq, . . . , σpkqpxq, . . .

We call the set of these elements the orbit of x under σ and denote it by

Oσpxq. Thus

Oσpxq “!

σpiqpxq : i P NY t0u)

.

- 65 -

The number of distinct elements of Oσpxq is called the length of the orbit of

x.

Example 8.15. For the permutation σ P S10 given by

σ “

˜

1 2 3 4 5 6 7 8 9 10

4 10 1 7 6 5 3 8 2 9

¸

,

the orbits are as follows:

Oσp1q “ t1, 4, 7, 3u “ Oσp3q “ Oσp4q “ Oσp7q.

Oσp2q “ t2, 10, 9u “ Oσp9q “ Oσp10q.

Oσp5q “ t5, 6u “ Oσp6q

Oσp8q “ t8u.

Note that several orbits are identical. Also, non-identical orbits are disjoint

sets. These properties indeed hold in general. Our next goal is to prove

these two properties in Proposition 8.16 and Proposition 8.17.

Proposition 8.16. Let σ P Sn.

(i) Let x P t1, . . . , nu. Then there exists an integer k P N such that σpkqpxq “x. If

` :“ mintk P N : σpkqpxq “ xu,

then the elements x, σpxq, . . . , σp`´1qpxq are all distinct and

Oσpxq “ tx, σpxq, . . . , σp`´1q

pxqu “!

σpiqpxq : 0 ď i ď `´ 1)

.

(ii) For any y P Oσpxq, we have Oσpyq “ Oσpxq.

Proof. (i) Assume, on the contrary, that σpjqpxq “ σpiqpxq for some i, j such

that 0 ď i ă j ď `´ 1. Since σ is a bijection, we have

σpj´1qpxq “ σpi´1qpxq and σpj´2qpxq “ σpi´2qpxq and . . .

- 66 -

Continuing in this fashion, we obtain σpj´iqpxq “ x. But then note that

1 ď j ´ i ď `´ 1, hence

` “ mintk P N : σpkqpxq “ xu ď j ´ i ď `´ 1,

which is a contradiction.

Clearly, tx, σpxq, . . . , σp`´1qpxqu Ď Oσpxq. Furthermore,

σpσp`´1qpxqq “ σp`qpxq “ x, hence σpkqpxq “ σpk modulo `qpxq.

Therefore there are no other elements in Oσpxq.

(ii) Observe that y “ σpiqpxq for some 1 ď i ď `´ 1. Thus clearly,

(8.3) Oσpyq Ď Oσpxq.

Furthermore,

x “ σp`qpxq “ σp`´iqpσpiqpxqq “ σp`´iqpyq P Oσpyq.

That is, x P Oσpyq, hence

(8.4) Oσpxq Ď Oσpyq.

From (8.3) and (8.4) it follows that Oσpxq “ Oσpyq. �

Proposition 8.17. Let σ P Sn, and let x, y P t1, . . . , nu. Then either

Oσpxq “ Oσpyq,

or Oσpxq XOσpyq “ ∅.

Proof. Assume that Oσpxq XOσpyq ‰ ∅. We prove that Oσpxq “ Oσpyq.

Let `x and `y denote the orbit lengths of x and y, respectively. Step 1.

Take a P Oσpxq X Oσpyq. Then there exist 0 ď i ď `x ´ 1 and 0 ď j ď `y ´ 1

such that

a “ σpiqpxq “ σpjqpyq.

- 67 -

Step 2. First assume that i ě j. (Because of the symmetry between x and

y, the argument for the case i ă j is similar.) Since σ is a bijection, by

precomposing with σ´1 we have

σpi´1qpxq “ σpj´1qpyq and σpi´2qpxq “ σpj´2qpyq and . . .

By continuing in this fashion, we obtain σpi´jqpxq “ y. This means that

y P Oσpxq, hence Proposition 8.16(ii) implies that Oσpxq “ Oσpyq. �

In the next proposition, we explain how to obtain the cycle decomposi-

tion of a permutation from its orbits.

Proposition 8.18. Let σ P Sn, where n P N. Suppose that x1, . . . , xr are chosensuch that the collection of subsets

tOσpx1q, . . . , Oσpxrqu

contains each of the orbits of σ of length ě 2 precisely once. For each 1 ď i ď r,let `i be the orbit length of xi, and let τi be the `i-cycle defined by

τi “`

xi, σpxiq, . . . , σp`i´1qpxiq

˘

.

Then the cycles τ1, . . . , τr are mutually disjoint and σ “ τ1 ¨ ¨ ¨ τr.

Proof. Step 1. We prove that the given cycles are disjoint. For each i such

that 1 ď i ď r, the cycle τi only moves the elements of Oσpxiq. In other

words, Mτi “ Oσpxiq. Now by Proposition 8.17, for i ‰ j we have

Oσpxiq XOσpxjq “ ∅.

Therefore τi and τj are disjoint cycles (see Remark 8.12).

Step 2. We prove that σpxq “ τ1 ¨ ¨ ¨ τrpxq for every x P t1, . . . , nu. Fix

x P t1, . . . , nu. One of the following two cases can occur:

- 68 -

Case I: x P Oσpxiq for some 1 ď i ď r. Since the cycles τ1, . . . , τr are disjoint,

τi is the only cycle among τ1, . . . , τr that moves the elements of Oσpxiq. But

we know that x, σpxq P Oσpxiq. Therefore

τjpxq “ x and τjpσpxqq “ σpxq for all j ‰ i.

It follows that

τ1 ¨ ¨ ¨ τrpxq “ τ1 ¨ ¨ ¨ τipxq “ τ1 ¨ ¨ ¨ τi´1pσpxqq “ σpxq.

Case II. x does not belong to any of Oσpx1q, . . . , Oσpxrq. Then the orbit of

x should have length 1, and therefore σpxq “ x. In addition, each τi only

moves the elements of Oσpxiq, and therefore τipxq “ x for all 1 ď i ď r. It

follows that

τ1 ¨ ¨ ¨ τrpxq “ x “ σpxq. �

Example 8.19. Let us apply 8.18 to the permutation of Example 8.15. The

collection of distinct orbits of length ě 2 is:

tOσp1q, Oσp2q, Oσp5qu.

Therefore

σ “ p1 4 7 3qp2 10 9qp5 6q.

Theorem 8.20. (Cycle Decomposition Theorem) Every permutation σ P Sn

can be written in a unique way as a product of mutually disjoint cycles. In otherwords:

(i) There exist mutually disjoint cycles τ1, . . . , τr such that σ “ τ1 ¨ ¨ ¨ τr.(ii) If τ 11, . . . , τ 1s is another list of mutually disjoint cycles such that

σ “ τ 11 . . . τ1s,

then r “ s and the τ 1j’s are identical to the τi’s up to a rearragement.

- 69 -

Proof. Existence follows immediately from Proposition 8.18.

Uniqueness is proved as follows. Suppose that σ “ τ 11 ¨ ¨ ¨ τ1s, for disjoint

cycles τ 11, . . . , τ 1s. Then the elements moved by each cycle (i.e., the sets Mτ 1i)

must be the orbits of σ with length ě 2. Therefore the decomposition

σ “ τ 11 ¨ ¨ ¨ τ1s is uniquely determined by the orbits of σ. �

Exercises.

(1) Using the method of proof of Proposition 8.13, prove that the fol-

lowing more general statement is true: for any two permutations

σ, τ P Sn such that Mσ XMτ “ ∅, we have στ “ τσ.

(2) Compute the cycle decompostion of

σ “

˜

1 2 3 4 5 6 7 8

7 3 4 2 6 8 5 1

¸

(3) Let σ be a permutation such that σ “ σ´1. Prove that in the cycle

decomposition of σ, all of the cycles are 2-cycles.

(4) Let σ be a permutation such that the cycles in its cycles decomposi-

tion are either 2-cycles or 3-cycles. Prove that σ6 “ ε.

Transpositions and parity. Our next goal is to describe another useful

way of expressing permutations as products of (not necessarily disjoint)

cycles.

Definition 8.21. A 2-cycle is called a transposition.

Lemma 8.22. Let σ :“ pa1 ¨ ¨ ¨ adq be a d-cycle in Sn, where n ě d ě 2. Then

σ “ pa1 a2qpa2 a3q ¨ ¨ ¨ pai ai`1q ¨ ¨ ¨ pad´1 adq

“ pad´1 adqpad´2 adq ¨ ¨ ¨ pai adq ¨ ¨ ¨ pa1 adq.

- 70 -

Proof. It is not difficult to trace the “journey” of an ai under the product.

For the first product, we have

ai Ñ ¨ ¨ ¨ Ñ ai Ñloomoon

pai ai`1q

ai`1 Ñ ai`1 Ñ ¨ ¨ ¨ Ñ ai`1 for i ă d,

and

ad Ñloomoon

pad´1 adq

ad´1 Ñloomoon

pad´2 ad´2q

ad´2 Ñ ¨ ¨ ¨ Ñ a1.

For the second product, we leave the argument as an exercise. �

Exercises.

Prove the remaining part of Lemma 8.22.

Theorem 8.23. (Parity Theorem) Let n P N. Then every permutation σ P Sncan be written as a product of transpositions. Furthermore, if

σ “ τ1 ¨ ¨ ¨ tτr “ τ 11 ¨ ¨ ¨ τ1s

where the τi’s and τ 1j’s are transpositions, then r and s have the same parity (thatis, they are either both even or both odd).

Proof. From the Cycle Decomposition Theorem and Lemma 8.22 it follows

that every permutation can be expressed as a product of transpositions.

Next we prove the statement about parity of r and s. Note that pτ 1jq´1 “ τ 1j,

and therefore from the equality τ1 ¨ ¨ ¨ τr “ τ 11 ¨ ¨ ¨ τ1s it follows that

τ1 ¨ ¨ ¨ τrτ1s ¨ ¨ ¨ τ

11 “ ε.

Thus, we have reduced the statement to the following claim:

if ε “ η1 ¨ ¨ ¨ ηk where the ηi’s are transpositions, then k is even.

We give a tricky proof for this claim, which is simple and beautiful, but it

looks like its idea came out of the blue!

- 71 -

Let I denote the nˆn identity matrix. Clearly detpIq “ 1. We know from

linear algebra that if we swap two columns of any matrix, the determinant

changes sign. Now consider the effect of the column swaps by the ηi’s on

the columns of a matrix: if ηi “ pa bq, then we swap the a-th column and

the b-th column of the matrix.

By applying ηk, ηr´1, . . . , η1 successively on the columns of I , each time

the sign of the determinant of the result matrix changes. Therefore the

final matrix should have determinant p´1qk. However, the final matrix

is the identity matrix (because η1 ¨ ¨ ¨ ηk “ ε). Therefore its determinant

should be 1. It follows that

p´1qk “ 1, hence k is even. �

Theorem 8.23 allows us to make the following definition.

Definition 8.24. Let n P N. A permutation Sn is called even if it can be ex-

pressed as a product of an even number of transpositions. A permutation

in Sn is called odd if it can be expressed as a product of an odd number of

transpositions.

Example 8.25. (i) Let σ “

˜

1 2 3 4 5

3 2 4 1 5

¸

. Then σ is even because

σ “ p1 3qp3 4q.

(ii) For every n P N, the identity permutation ε P Sn is even.

Proposition 8.26. Let n P N, and let σ, τ P Sn be arbitrary permutations.

(i) If σ and τ are both even, then στ is even.(ii) If σ is even and τ is odd, then στ is odd.

(iii) If σ is odd and τ is even, then στ is odd.(iv) If σ and τ are both odd, then στ is even.

- 72 -

Proof. These statements are consequences of obvious properties of even

and odd numbers. For example, we can prove (i) as follows: assume that

σ and τ are both even. Then

σ “ η1 ¨ ¨ ¨ η2r and τ1 “ η11 ¨ ¨ ¨ η12s for r, s P NY t0u,

where the ηi’s and the η1j’s are transpositions. Now

στ “ η1 ¨ ¨ ¨ η2rη11 ¨ ¨ ¨ η

12s,

which is a product of 2pr ` sq transposition. Hence στ is even. The proofs

of (ii)–(iv) are similar. �

Lemma 8.27. Let n P N and let σ P Sn.

(i) If σ is even, then σ´1 is also even.(ii) If σ is odd, then σ´1 is also odd.

Proof. (i) Assume that σ is even. Then we can express σ as σ “ η1 ¨ ¨ ¨ η2r,

where the ηi’s are transpositions. Since pηiq´1 “ ηi for every 1 ď i ď 2r, we

have

σ´1 “ η2r ¨ ¨ ¨ η1.

Therefore σ´1 is an even permutation.

(ii) The argument is similar. �

Proposition 8.28. Let n P N. Then the subset An of Sn, consisting of the evenpermutations in Sn, is a subgroup of Sn.

Proof. From 8.26(i) and Lemma 8.27(i) it follows that for all σ, τ P An we

have στ´1 P An. Thus by the Concise Subgroup Test, An is a subgroup of

Sn. �

Definition 8.29. The subgroup An of the symmetric group Sn is called the

alternating group on n letters.

- 73 -

Exercises.

(1) Write down all of the elements of A3.

(2) Prove that for every integer n ě 2, the group An has preciselyn!

2elements.

Conjugate permutations. So far we have not determined the centre of the

group Sn. If σ is in the center of Sn, then for any τ P Sn we have στ “

τσ, or equivalently, τ´1στ “ σ. Therefore understanding the centre of Snnaturally leads to investigating permutations of the form τ´1στ .

Definition 8.30. Two permutations σ, σ1 P Sn are called conjugate if there

exists a permutation τ P Sn such that σ1 “ τ´1στ .

Remark 8.31. Note that the relation σ1 “ τ´1στ can be written as

σ “ τσ1τ´1 “ η´1σ1η where η “ τ´1.

This implies that being conjugate is a symmetric relation on permutations

(i.e, if σ is conjugate to σ1, the σ1 is also conjugate to σ).

Example 8.32. The permutations

σ “

˜

1 2 3 4

3 2 1 4

¸

and τ “

˜

1 2 3 4

1 2 4 3

¸

are conjugate. Indeed σ “ η´1τη where η “

˜

1 2 3 4

3 1 4 2

¸

. But how do we

find such an η? This is explained in the rest of this section.

Let n P N, and let σ P Sn be an arbitrary permutation. Suppose that the

cycle decomposition of σ is as follows:

σ “ pa1 ¨ ¨ ¨ ad1qpb1 ¨ ¨ ¨ bd2q ¨ ¨ ¨

- 74 -

Now let τ P Sn be another permutation. What is the cycle decomposition

of τ´1στ? We will answer this question in the next proposition.

Proposition 8.33. Let σ and τ be as above. Then the cycle decomposition ofτ´1στ is as follows:

τ´1στ “`

τ´1pa1q ¨ ¨ ¨ τ´1pad1q

˘`

τ´1pb1q ¨ ¨ ¨ τ´1pbd2q

˘

¨ ¨ ¨

Proof. We can trace the “journey” of an element x P t1, . . . , nu under the

composition τ´1στ . For example, if x “ τ´1paiq for some 1 ď i ă d1 then it

is moved as follows:

τ´1paiq Ñloomoon

by τ

ai Ñloomoon

by σ

ai`1 Ñloomoon

by τ´1

τ´1pai`1q,

and if x “ τ´1pad1q then a similar reasoning yields that under τ´1στ the

element x is mapped to τ´1pa1q. For the other cycles the argument is the

same. �

Example 8.34. Let σ “ p2 4 6 3 5qp7 1q, and let τ “

˜

1 2 3 4 5 6 7 8

5 3 1 2 8 4 6 7

¸

.

Then the cycle decomposition of τ´1στ is

p4 6 7 2 1qp8 3q.

Note that the cycles that appear in the decompositions of σ and τ´1στ are

different, but the lengths of the cycles that appear in the two decomposi-

tions are the same.

Example 8.35. Are the permutations σ1, σ2 P S8 defined by

σ1 “

˜

1 2 3 4 5 6 7 8

7 1 5 2 3 6 4 8

¸

and σ2 “

˜

1 2 3 4 5 6 7 8

4 3 6 5 1 2 7 8

¸

- 75 -

conjugate? To answer this question, we write the cycle decompositions of

σ1 and σ2:

σ1 “ p1 7 4 2qp3 5q and σ2 “ p1 4 5qp2 6 3q.

Now one consequence of Proposition 8.33 is the following: if two permu-

tations σ1 and σ2 in Sn are conjugate, then for any 1 ď ` ď n, the number of

`-cycles that appear in the cycle decompositions of σ1 and σ2 should be the

same. However, there is one 2-cycle in the cycle decomposition of σ1, and

there are no 2-cycles in the cycle decomposition of σ2. Therefore σ1 and σ2

are not conjugate.

Theorem 8.36. Let n ě 3. Then ZpSnq is equal to the trivial subgroup tεu.

Proof. Let σ ‰ ε. We will prove that σ R ZpSnq. We can express σ as a

product of disjoint cycles. Then one of the following cases occurs:

Case I: There is at least one cycle in the cycle decomposition of σ of length

at least 3. Thus, we can write the cycle decomposition as:

σ “ pa1 ¨ ¨ ¨ ad1qpb1 ¨ ¨ ¨ bd2q ¨ ¨ ¨ ,

where d1 ě 3. Now let consider the permutation τ P Sn defined as follows:

τpa1q “ a2, τpa2q “ a1,

and τpxq “ x for all x R ta1, a2u. Then

τ´1στ “ pa2 a1 a3 a4 a5 ¨ ¨ ¨ ad1qpb1 ¨ ¨ ¨ bd2q ¨ ¨ ¨

Clearly τ´1στpa1q “ a3 whereas σpa1q “ a2. Thus τ´1στ ‰ σ, and therefore

σ R ZpGq.

Case II: The cycle decomposition of σ has at least two 2-cycles. Thus we

- 76 -

can express this cycle decomposition as

σ “ pa bqpc dqτ3 ¨ ¨ ¨ τr,

where a, b, c, d are distinct. Now consider the permutation τ P Sn defined

by

τpaq “ c, τpcq “ b, τpbq “ a, τpdq “ d,

and τpxq “ x for all x R ta, b, c, du. Then the cycle decomposition of τ´1στ

is

pb cqpa dqτ1 ¨ ¨ ¨ τr.

In particular, σpaq “ b, whereas τ´1στpaq “ d. Therefore σ ‰ τ´1στ , and

hence σ R ZpSnq.

Case III: The cycle decomposition of τ consists of a single 2-cycle. Thus

σ “ pa bq

for distinct a, b. Since n ě 3, we can choose an element c P t1, . . . , nu such

that c ‰ a and c ‰ b. Now let τ “ pa b cq. Then

σpaq “ b and τ´1στpaq “ c,

therefore σ ‰ τ´1στ , and in particular σ R ZpSnq. �

- 77 -

COSETS AND LAGRANGE’S THEOREM

Definition 9.1. Let G be any group and let H be a subgroup of G. For an

element a P G, the left coset ofH generated by a is the set aH :“ tah : h P Hu.

The right coset of H generated by a is the set Ha :“ tha : h P Hu.

Example 9.2. (i) Let G “ Z and let

H “ 3Z “ tx P Z : x “ 3k for some k P Zu “ t0,˘3,˘6,˘9, . . .u.

Then 0`H “ H “ 3Z and

1`H “ tx P Z : x “ 3k ` 1 for some k P Zu

and

2`H “ tx P Z : x “ 3k ` 2 for some k P Zu.

Note that

0`H “ H ` 0 , 1`H “ H ` 1 , 2`H “ H ` 2.

(ii) Let G “ Z2 ˆ Z2 and let H “ Z2 ˆ t0u “ tp0, 0q, p1, 0qu. Then the

distinct left cosets of H are

p0, 0q `H “ H and p0, 1q `H “ tp0, 1q, p1, 1qu.

(iii) Let G “ S3 and let H “ tε, p1 2qu. Then we have

p1 3qH “ tp1 3q, p1 2 3qu but Hp1 3q “ tp1 3q, p1 3 2qu.

In particular, the left coset p1 3qH and the right coset Hp1 3q are notidentical.

Theorem 9.3. Let G be a group, let H Ď G be a subgroup of G, and let a, b P G.

(i) H “ He “ eH where e P G denotes the identity of G.(ii) Ha “ H if and only if a P H .

(iii) Ha “ Hb if and only if ab´1 P H .

- 78 -

(iv) a P Hb if and only if Ha “ Hb.(v) Either Ha “ Hb or HaXHb “ ∅.

(vi) The right cosets of H form a partition of G.

Proof. (i) is obvious.

(ii) If Ha “ H then a “ ea P Ha, hence a P H .

Conversely, if a P H , then for any x P Hawe have x “ ha for some x P H ,

hence ha P H . This implies Ha Ď H . Also, for any h P H we have

h “ ha´1a,

and since a´1 P H , we have h P Ha. This implies H Ď Ha.

(iii) Assume that Ha “ Hb. Then a “ ea P Ha, hence a “ hb for some

h P H . It follows that ab´1 “ hbb´1 “ h P H .

Conversely, assume that ab´1 P H . Then ab´1 “ h for some h P H , that

is, a “ hb. Now:

‚ For every x P Ha we have x “ h1a for some h1 P H . Thus

x “ h1a “ h1phbq “ ph1qhb and h1h P H ñ x P Hb.

This proves that Ha Ď Hb.

‚ Similarly, for every x P Hb we have x “ h2b for some h2 P H . Thus

x “ h2b “ h2ph´1aq “ ph2h

´1qa and h2h

´1P H ñ x P Ha.

This proves that Hb Ď Ha.

(iv) If a P Hb then a “ hb for some h P H , hence ab´1 “ hbb´1 “ h P H .

Now (iii) implies that Ha “ Hb. Conversely, if Ha “ Hb, then a “ ea P

Ha “ Hb, so a P Hb.

- 79 -

(v) Assume that HaXHb ‰ ∅. Choose x P HaXHb. Then x “ h1a “ h2b

for h1, h2 P H . Thus

a “ ph1q´1h1a “ ph1q

´1h2b P Hb.

Now (iv) implies that Ha “ Hb.

(vi) Note that a P Ha for any a P G, hence G “Ť

aPGHa. From (v) we

know that the right cosets are either identical or disjoint. This implies that

the right cosets induce a partition of the elements of G. �

Remark 9.4. There is a result analogous to Theorem 9.3 for left cosets. The

only difference is that Theorem 9.3(iii) should be formulated as

aH “ bH if and only if a´1b P H.

Example 9.5. (i) Let G “ Z6 and let H “ 2Z6 “ t0, 2, 4u. Then the

distinct left cosets of H in G are

0`H “ t0, 2, 4u “ 2`H “ 4`H and 1`H “ t1, 3, 5u “ 3`H “ 5`H.

(ii) Let G “ R˚ “ Rzt0u be the group of nonzero real numbers (where

the binary operation is the usual multiplication). Let

H “ R` :“ tx P R : x ą 0u.

Then the distinct left cosets of H in G are

1 ¨H “ R` and p´1qH “ R´ “ tx P R : x ă 0u.

(iii) Let G “ R2, the 2-dimensional real vector space equipped with the

usual vector addition. Also let H “ R ˆ t0u “ tpx, 0q : x P Ru.Then H is a subspace of G, and therefore also a subgroup of G. The

distinct left cosets of H are the sets

p0, bq `H “ tpa, bq : a P Ru for every b P R.

- 80 -

Example 9.6. More generally, let G “ Zn for some n P N, and let H “ dZnfor some d P N which satisfies d|n. We want to determine the distinct left

cosets of H .

Step 1. Observe that dZ “ t0, d, 2d, . . . , n ´ du is the set of elements of

t1, . . . , nu that are divisible by d, i.e., they have remainder 0 module d.

Step 2. There are d possible remainders modulo d:

0, 1, . . . , d´ 1.

For any r P t0, 1, . . . , d´1u, the elements of the left coset r`H “ r`dZ are

precisely those elements in Zn which have remainder r modulo d. Thus,

every element of Zn belongs to precisely one of the sets r ` H . Therefore

there are in total d distinct left cosets of H , namely:

0`H, 1`H, . . . , pd´ 1q `H.

Example 9.7. Let G “ Z ˆ Z and let H “ 2Z ˆ 3Z “ tp2k, 3sq : k, s P Zu.It is straightforward to verify that H is a subgroup of G. We would like to

determine the distinct left cosets of H . First note that for any pa, bq P ZˆZ,

we have

pa, bq `H “ tp2k ` a, 3s` bq : k, s P Zu

“ tpm,nq P Zˆ Z : m ” a mod 2 and n ” b mod 3u.

Thefore the left coset pa, bq ` H is determined by the remainders of the

components modulo 2 and 3.

The possible remainders modulo 2 are 0,1 and the possible remainders

modulo 3 are 0,1,2. Therefore we have a total of 2ˆ 3 “ 6 pairs of possible

remainders:

p0, 0q, p0, 1q, p0, 2q, p1, 0q, p1, 1q, p1, 2q.

For each of the above six pairs pa, bq we can form the corresponding left

coset pa, bq`H , and every element pm,nq P ZˆZ appears in precisely one of

- 81 -

these cosets (according to the remainder ofmmodulo 2, and the remainder

of n modulo 3). Therefore in total there are six distinct left cosets of H :

p0, 0q `H, p0, 1q `H, p0, 2q `H, p1, 0q `H, p1, 1q `H, p1, 2q `H.

Example 9.8. Let G “ S5 and let H “ tσ P G : σp5q “ 5u. Note that H

is a subgroup of G, and indeed H – S4. Let us determine the distinct left

cosets of H in G.

For any permutation σ P S5 we have σp5q P t1, 2, 3, 4, 5u. Now consider

the transpositions

τi “ p5 iq.

Then the following statement holds for any σ P S5:

Let 1 ď i ď 5. If σp5q “ i, then τiσ P H. Conversely, if τiσ P H, then σp5q “ i.

For the “conversely” part, note that τ 2i “ ε, so that

σp5q “ τ 2i σp5q “ τiτipσp5qq “ τip5q “ i.

It follows that σiH “ tσ P S5 : σp5q “ iu. Thus, the left cosets

σ1H, σ2H, σ3H, σ4H, σ5H

are distinct and their union is equal to G “ S5. Consequently, the left

cosets

σ1H, σ2H, σ3H, σ4H, σ5H

are all of the distinct left cosets of H .

Remark 9.9. If G is an abelian group, then for any subgroup H of G and

any a P G we have aH “ Ha.

Exercises.

(1) Determine all of the left and right cosets of H “ tε, p1 2qu in S3.

- 82 -

(2) Let G “ R ˆ R. Let H be the set of elements of G of the form px, xq,

i.e.,

H “ tpx, yq P G : x “ yu .

Prove that H is a subgroup of G. Also, show that the sets

px, 0q `H for all x P R.

are all of the distinct cosets of H in G.

Lemma 9.10. LetG be a group, and letH be a subgroup ofG. Then the followingstatements are true:

(i) For every a P G, the maps

`a : H Ñ aH , h ÞÑ ah

and

ra : H Ñ Ha , h ÞÑ ha

are bijections.(ii) If G is finite, then for every a P H we have |H| “ |Ha| “ |aH|.

Proof. (i) We prove the statement for the first map. (For the second map

the argument is similar.)

To show that the map `a is an injection:

`apxq “ `apyq ñ ax “ ay ñ x “ y,

where the last step follows from the Left Cancellation Law.

To show `a is surjective: given any x P aH we have x “ ah for some

h P H . Therefore `aphq “ ah “ x. Hence `a is a surjection onto aH .

(ii) Follows from (i). �

Theorem 9.11. (Lagrange’s Theorem) Let G be a finite group, and let H be asubgroup of G. Then

- 83 -

(i) |H| divides |G|.(ii) The number of distinct right cosets of H in G is equal to |G|

|H| .

(iii) The number of distinct left cosets of H in G is equal to |G||H| .

Proof. By Theorem 9.3(vi), G is partitioned into the right cosets of H . By

Lemma 9.10(ii), all of the right cosets of H have |H| elements. It follows

that

|G| “ |H| ˆ number of distinct right cosets of H.

This completes the proof of (i) and (ii). The proof of (iii) is similar. �

Definition 9.12. Let G be a group and let H be a subgroup of G. The indexof H in G, denoted by rG : Hs, is defined as the number of distinct left (or

right) cosets of H in G.

Example 9.13. LetG “ Z9 and letH “ 3Z9 “ t0, 3, 6u. Then the distinct left

cosets ofH inG are 0`H “ t0, 3, 6u, 1`H “ t1, 4, 7u, and 2`H “ t2, 5, 8u.

Thus rG : Hs “ 3.

Corollary 9.14. Let G be a finite group and let g P G. Then opgq divides |G|.

Proof. Note that opgq ă 8 because G is finite (see Lemma 6.5). Set H “ xgy,

the cyclic subgroup generated by G. Then Theorem 6.6(iii) implies that

|H| “ opgq. Furthermore, Theorem 9.11(i) implies that |H| divides |G|.

Therefore opGq divides |G|. �

Example 9.15. Does S5 have an element of order 48? The answer is no,

because |S5| “ 5! “ 120 and 48 - 120.

Corollary 9.16. LetG be a finite group and let g P G. Set n :“ |G|. Then gn “ e.

Proof. From Corollary 9.14 it follows that opgq|n. Thus

gn “ pgopgqqnopgq “ e

nopgq “ e.

�

- 84 -

Let n P N. Recall that Upnq is the (multiplicative) group of invertible

elements of Zn. We define

φpnq :“ |Upnq|.

For example φp1q “ φp2q “ 1, φp3q “ φp4q “ 2, etc. The function φpnq is

usually called the Euler function.

Corollary 9.17. (Fermat’s Little Theorem/Euler’s Theorem) Let n P N. Forevery a P Z such that gcdpa, nq “ 1, the residue of aφpnq modulo n is equal to 1.

Proof. The statement is equivalent to showing that for any k P Upnq, we

have kφpnq “ 1 in Upnq. In this form, the statement follows from Corollary

9.16 for G “ Upnq. �

Example 9.18. Suppose we want to compute the remainder of the division

of 21000 modulo 37. Then we note that 37 is a prime number, hence φp37q “

36. Note that Corollary 9.17 implies that the remainder of 236 modulo 37 is

1. Now

1000 “ 36ˆ 27` 28,

and therefore

21000 “ p236q27 ˆ 228 “ 228 modulo 37.

To further simplify this calculation, note that 27 “ 128 “ 3ˆ 37` 17, hence

228 “ p27q4 “ 174 “ 83521 “ 2257ˆ 37` 12.

Consequently, 21000 modulo 37 is equal to 12.

Example 9.19. Let G “ Dn, the n-th dihedral group (see Definition 7.10).

Then |Dn| “ 2n and therefore by Lagrange’s theorem for every g P Dn we

have opgq|2n. In fact there are no elements of order 2n. Below we determine

the orders of elements of D6. We have

D6 “ te, r, r2, r3, r4, r5, s, sr, sr2, sr3, sr4, sr5u,

- 85 -

where s2 “ e “ r6 and srs´1 “ r´1. The orders of the elements of D6 are

given as follows:

g e r r2 r3 r4 r5 s sr sr2 sr3 sr4 sr5

opgq 1 6 3 2 3 6 2 2 2 2 2 2

All of these numbers of divisors of 12, and this is consistent with La-

grange’s theorem.

Theorem 9.20. Let G be a finite group such that p :“ |G| is a prime number.Then G is isomorphic to Zp.

Proof. Let g P G such that g ‰ e. Then opgq ‰ 1, and by Corollary 9.14

we know that opgq divides p. However, p is prime and its only positive

divisors are 1 and p. It follows that opgq “ p.

Now consider the cyclic subgroup of G generated by g, i.e., H “ xgy. By

Theorem 6.6(iii), H has opgq “ p elements. Thus H “ G. It follows that G

is a cyclic group with p elements. From the classification of cyclic groups,

it follows that G – Zp. �

Exercises.

(1) Show that if n is a prime number, then φpnq “ n´ 1.

(2) Show that if n “ pq where p and q are prime numbers, then

φpnq “ pp´ 1qpq ´ 1q.

(3) Calculate the remainder of 31000 modulo 55.

- 86 -

CAYLEY’S THEOREM

The symmetric groups Sn have so far been the most interesting family of

examples of finite non-abelian groups. These groups have the following

remarkable property:

Theorem 10.1. (Cayley’s Theorem) Let G be a finite group, and set n :“ |G|.Then the group Sn has a subgroup isomorphic to G.

Proof. Since |G| “ n, we can think of Sn as the group of permutations of

the elements of G. We denote the group of permutations of elements of G

by SG. Thus, SG – Sn. Every element of SG is a bijection from G onto G.

Step 1. For any g P G, define the map σg : G Ñ G by σgpxq :“ gx. By the

Left Cancellation Law, the maps σg are injective. By Proposition 3.3, these

maps are also surjective. Therefore the maps σg are permutations of the

elements of G.

Step 2. Let g1, g2 P G. We show that σg1g2 “ σg1 ˝ σg2. To verify this state-

ment, observe that for every x P G,

σg1 ˝ σg2pxq “ σg1pg2xq “ g1pg2xq “ pg1g2qx “ σg1g2pxq.

Step 3. From Step 2 it follows in particular that

σg ˝ σg´1 “ σe “ ε and σg´1 ˝ σg “ ε.

Therefore pσgq´1 “ σg´1 for every g P G. (Here we interpret the left hand

side of the last equation as the inverse of σg in SG.)

Step 4. Set H :“ tσg : g P Gu. Thus H is a subset of SG. We prove that H

is a subgroup of G. To this end, we use the Concise Subgroup test (clearly

- 87 -

H ‰ ∅). For any two elements σg, σh P H , where g, h P G, by Steps 2 and 3

we have

σgσ´1h “ σgσh´1 “ σgh´1 P H.

Thus by the Concise Subgroup Test, H is a subgroup of SG.

Step 5. We now construct an isomorphism φ : GÑ H . Set

φpgq :“ σg for all g P G.

Clearly φ is a surjection. Furthermore, φ is an injection because

φpg1q “ φpg2q ñ σg1 “ σg2 ñ σg1peq “ σg2peq ñ g1 “ g2.

Finally, for every g1, g2 P G,

φpg1g2q “ σg1g2 “ σg1 ˝ σg2 “ φpg1qφpg2q.

Thus φ : GÑ H is an isomorphism.

�

Example 10.2. Let G “ Z3. Then Theorem 10.1 states that G is isomorphic

to a subgroup of S3. The embedding is as follows: we have

G “ Z3 “ t0, 1, 2u,

and therefore

σ0pxq “ x, σ1pxq “ x` 1, σ2pxq “ x` 2 for x “ 0, 1, 2.

Thus as permutations of t0, 1, 2uwe have

σ0 “

˜

0 1 2

0 1 2

¸

, σ1 “

˜

0 1 2

1 2 0

¸

, σ2 “

˜

0 1 2

2 0 1

¸

.

- 88 -

Remark 10.3. Theorem 10.1 is a “no-go” theorem: it basically says that a

complete understanding of all of the subgroups of Sn will include a com-

plete understanding of all finite groups, and therefore it is an extremely

difficult problem.

However, it is still useful to know that every finite group appears as a

subgroup of a permutation group. Indeed in the first half of 20th century,

qualitative results about subgroups of permutation groups played a major

role in the development of group theory.

In the rest of this section we give an application of Cayley’s theorem. We

begin by two lemmas.

Lemma 10.4. LetG be a group such that |G| is even. Then there exists an elementg P G such that opgq “ 2.

Proof. We need to show that there is element g P G such that g “ g´1 and

g ‰ e.

Step 1. Suppose G “ tg1, . . . , g2nu where n P N. Connect each g P G to g´1

by an arc. If g “ g´1, then this arc will become a loop at g.

Step 2. There is at least one loop (it is the arc connecting e to itself). Ob-

serve that the claim of the lemma is equivalent to showing that there is at

least one more loop.

Step 3. Now suppose that, on the contrary, there are no loops other than

the one connecting e to e. If there are k other arcs (which are not loops),

then the end points of each of these arcs identify two distinct elements of

G (which are also different from e). Therefore the total number of elements

of G sohuld be p2 ˆ kq ` 1, which is an odd number. This contradicts the

assumption of the lemma. �

- 89 -

Remark 10.5. The following generalization of Lemma 10.4 is also true: if

G is a finite group, then for every prime number p which divides |G| there

exists an element g P G such that opgq “ p. However, the proof of this

statement requires techniques that are beyond this course.

Lemma 10.6. Let G be a subgroup of Sn, where n P N. Assume that G Ć An,where An Ď Sn is the alternating subgroup of Sn. Then G X An is a subgroup ofG that satisfies rG : GX Ans “ 2. In particular, |G| is an even number.

Proof. Since G X An is the intersection of two subgroups of Sn, it is also a

subgroup of Sn (or G).

Next we show that rG : G X Ans “ 2. Since G Ć An, at least one element

ofG is an odd permutation. Moreover, ε P G is an even permutation. Now

let m “ |G|, and write the elements of G as

σ1, . . . , σm,

such that σ1, . . . , σk are odd permutations, whereas σk`1, . . . , σm are even

permutations. By the above discussion, we have 1 ă k ă m.

Step 1. Consider the following elements of G

σ1σ1, . . . , σ1σi, . . . , σ1σm.

By the Left Cancellation Law, these m elements are distinct. Since G has m

elements, every element of G occurs in the above list. Thus, the above list

contains every element of G precisely once.

Step 2. By Proposition 8.26, the permutations σ1σi are even for 1 ď i ď k,

and odd for k` 1 ď i ď m. It follows that the elements of G should consist

of k even permutations and m´ k odd permutations.

- 90 -

Step 3. Comparing the conclusion of Step 2 with the definition of k prior

to Step 1, we obtain that k “ m ´ k. Thus the number of elements of G

is m “ 2k (hence it is an even number), and the number of elements of

GX An is k. It follows that

rG : GX Ans “|G|

|GX An|“ 2. �

Theorem 10.7. Let G be a group such that |G| “ 2n, where n is an odd integer.Then G contains a subgroup of order n.

Proof. Step 1. By Cayley’s Theorem, we can assume that G is a subgroup

of SG. Recall from the proof of Cayley’s Theorem that the element of SGcorresponding to any g P G is the permutation

σg : GÑ G , σgpxq “ gx.

Step 2. By Lemma 10.4, there exists an element g˝ P G such that opg˝q “ 2.

Note that in particular g˝ ‰ e. Then for any x P G we have

σg˝pxq “ g˝x ‰ x and σg˝pσg˝pxqq “ g2˝x “ x.

It follows that σg˝ is a product of n disjoint transpositions of the form

px, σg˝pxqq.

Step 3. Since n is odd, Step 2 implies that the permutation σg˝ is an odd

permutation. Now let AG denote the alternating subgroup of SG. (In other

words AG is the subgroup of SG consisting of even permutations.) Since

σg˝ P G, it follows that G Ć AG. Therefore Lemma 10.6 implies that rG :

G X AGs “ 2. Consequently, G X AG is a subgroup of G that satisfies

|GX AG| “ n. �

- 91 -

NORMAL SUBGROUPS AND QUOTIENTS

This section introduces an important construction in algebra: the quo-

tient construction.

Definition 11.1. Let G be any group and let H be a subgroup of G. We call

H a normal subgroup of G if gH “ Hg for every g P H .

Example 11.2. (i) Let G “ Z and let H “ 2Z “ t0,˘2,˘4,˘6, . . .u. The

left cosets of H are of the form k`H for k P Z, and the distincts right

cosets of H are of the form H ` k for k P Z. Since Z is abelian, we

have

k `H “ H ` k for all k P Z.

Therefore H is a normal subgroup of G.

(ii) Let G “ S3, and let H “ tε, p1 2qu. In Example 9.2(iii) we noted that

p1 3qH ‰ Hp1 3q.

Therefore H is not a normal subgroup of G.

(iii) Let G be an abelian group. Then every subgroup of G is normal.

Let G be a group and let H be a subgroup of G. Then for every g P G we

define

gHg´1 “ tghg´1 : g P Gu.

Theorem 11.3. Let G be any group and let H be a subgroup of G. Then thefollowing statements are equivalent:

(i) H is a normal subgroup of G.(ii) gHg´1 “ H for every g P G.

(iii) gHg´1 Ď H for every g P G.

Proof. (i)ñ(ii): Let g P G. Then gH “ Hg, and therefore

gHg´1 “ pgHqg´1 “ pHgqg´1 “ Hpgg´1q “ He “ H.

- 92 -

(ii)ñ(iii): This is trivial.

(iii)ñ(i): Let g P G. Then gHg´1 Ď H , and replacing g by g´1 we have

g´1Hg Ď H . Multiplying the latter relation on the left by g and on the

right by g´1, we obtain :

gpg´1Hgqg´1 Ď gHg´1 and consequently, H Ď gHg´1.

Now we have

gHg´1 Ď H and H Ď gHg´1,

hence H “ gHg´1. Multiplying the latter relation on the right by g, and

using associativity of G, we obtain

Hg “ pgHg´1qg “ pgHqpg´1gq “ gHpeq “ gH. �

Proposition 11.4. The alternating group An is a normal subgroup of the sym-metric group Sn for every n P N.

Proof. Step 1. If σ is an even permutation then σAn “ An, because σ P An

and therefore the map η ÞÑ ση is a permutation of the elements ofAn. (This

follows from Proposition 3.3.)

Step 2. A similar argument also implies that If σ is an even permutation

then Anσ “ An.

Step 3. Now if σ is an odd permutation, then the elements of σAn are a

product of an even permutation and an odd permutation. By Proposition

8.26, such products are always odd permutations. Therefore σAn Ď SnzAn.

In addition, every odd permutation τ P Sn belongs to σAn. Indeed we have

τ “ ετ “ pσσ´1qτ “ σpσ´1τq,

and σ´1τ P An because σ´1 and τ are both odd permutations (see Proposi-

tion 8.26 and Lemma 8.27).

- 93 -

Step 4. By Step 3, for every odd permutation we have σAn “ SnzAn. A

small modification of the argument of Step 3 shows that we also have

Anσ “ SnzAn.

Step 5. From Steps 1, 2, and 4 it follows that

$

&

%

σAn “ Anσ “ An if σ is even,

σAn “ Anσ “ SnzAn if σ is odd.

Thus in particular An is a normal subgroup of Sn. �

Proposition 11.5. LetG be any group. Then the centre ofG is a normal subgroupof G.

Proof. Let g P G. Then for any h P ZpGqwe have ghg´1 “ h. Therefore,

gZpGqg´1 “ tghg´1 : h P ZpGqu “ ZpGq.

Consequently, ZpGq is a normal subgroup of G. �

The following statement extends Proposition 11.4.

Proposition 11.6. Let G be a group and let H be a subgroup of G such thatrG : Hs “ 2. Then H is a normal subgroup of G.

Proof. Step 1. Since rG : Hs “ 2, the subgroup H has only two distinct

right cosets. These cosets can be represented as H “ He and Hg, where

g R H . Since H Y Hg “ G and H X Hg “ ∅, it follows that Hg “ GzH .

Consequently, we have

Hg “

$

&

%

H if h P H,

GzH if h R H.

- 94 -

Step 2. The argument of Step 1 also works for left cosets, and therefore we

obtain that

gH “

$

&

%

H if h P H,

GzH if h R H.

Step 3. From Step 1 and Step 2 it follows that Hg “ gH for every g P G.

Hence H is a normal subgroup of G. �

Exercises.

(1) Let G be any group. Prove that every subgroup of ZpGq is a normal

subgroup of G.

(2) Let D5 be the 5-th dihedral group (which is the group of symmetries

of a regular pentagon). Does D5 have a normal subgroup with 5

elements?

(3)

Lemma 11.7. Let G be any group, and let H and K be two normal subgroups ofG. Then H XK is also a normal subgroup of G.

Proof. In Proposition 5.8 we proved that the intersection of any two sub-

groups is a subgroup. Hence H X K is a subgroup of G. Next we show

that it is normal.

Since H and K are normal, for every g P G we have

gHg´1 “ H and gKg´1 “ K.

Thus

gpH XKqg´1 “ tgxg´1 P G : x P H XKu Ď gHg´1 X gKg´1 “ H XK.

Now Theorem 11.3 implies that H XK is a normal subgroup of G. �

- 95 -

Remark 11.8. Let G be a group and let H,K be subgroups of G. We define

HK :“

x P G : DhPHDkPKpx “ hkq(

.

Note that HK is not necessarily a subgroup of G. For example, G “ S3,

H “ tε, p1 2qu, and K “ tε, p1 3qu. Then

HK “ tε, p1 2q, p1 3q, p1 3 2qu

and it is easy to verify that HK is not a subgroup of S3.

In the next proposition, we provide some conditions that determine when

HK is a subgroup of G.

Proposition 11.9. Let G be a group and let H,K be subgroups of G. The follow-ing statements are equivalent:

(i) HK is a subgroup of G.(ii) HK “ KH .

(iii) KH is a subgroup of G.

Proof. (i)ñ (ii): We need to show that HK Ď KH and KH Ď HK.

KH Ď HK: Let x P KH . Then x “ hk for some elements h P H and

k P K. We need to show that kh P HK. We have

h´1 P H and k´1 P K,

since H and K are subgroups. Thus h´1k´1 P HK. Set g “ h´1k´1. Since

HK is a subgroup, we have g´1 P HK. But g´1 “ ph´1k´1q´1 “ kh. Thus

kh P HK, as desired.

HK Ď KH : Let x P HK. Then x´1 P HK because HK is a subgroup of

G. It follows that

x´1 “ hk for some elements h P H and k P K.

- 96 -

Now

x “ px´1q´1phkq´1 “ k´1h´1 P KH,

because k´1 P K and h´1 P H (since H and K are subgroups of G).

(ii)ñ (iii): As usual let e denote the identity element of G. Since e P H and

e P K, we have e P KH . We now use the Concise Subgroup Test. To this

end, we need to verify that if k1h1 P KH and k2h2 P KH , then

pk1h1qpk2h2q´1P KH.

We have pk1h1qpk2h2q´1 “ k1h1h´12 k´12 . Now

k1h1h´12 P KH “ HK,

hence there there exist elements h3 P H and k3 P K such that k1h1h´12 “

h3k3. It follows that

pk1h1qpk2h2q´1“ pk1h1h

´12 qk

´12 “ ph3k3qk

´12 “ h3pk2k3q

´1P HK,

as desired.

The proof of (iii)ñ(ii) is similar to (i)ñ(ii), and the proof of (ii)ñ(iii) is

similar to (ii)ñ(i). �

Proposition 11.10. Let G be a group and let H,K be subgroups of G.

(i) If at least one of H and K is a normal subgroup of G, then HK is a sub-group of G.

(ii) If both H and K are normal subgroups of G, then HK is also a normalsubgroup of G.

Proof. (i) First assume that H is normal.

Step 1. By Proposition 11.9, in order to show that HK is a subgroup of

G it is enough to verify that HK “ KH . To this end, we will verify that

HK Ď KH and KH Ď HK.

- 97 -

Step 2. We verify that HK Ď KH . Let g P HK. Then there exist elements

h P H and k P K such that g “ hk. Now

g “ hk “ ehk “ pkk´1qhk “ kpk´1hkq.

Since H is normal, we have k´1Hk Ď H and therefore k´1hk P H . There-

fore g P KH .

Step 3. The proof of the reverse inclusion KH Ď HK is similar to to Step

2:

g P KH ñ g “ k1h1 where k1 P K,h1 P H

ñ g “ k1h1 “ pk1hk´11 qk1 P HK ñ g P HK,

since k1Hk´11 Ď H .

Now assume that K is normal. Then by the argument given above, it

follows that KH is a subgroup of G. In particular, Proposition 11.9 implies

that KH “ HK. Thus HK is a subgroup of G.

(ii) By part (i), we know that HK is a subgroup of G. Thus we only need

to show that HK is normal. For any g P G,

gHKg´1 “ gHeKg´1 “ gHg´1gKg´1 “ pgHg´1qpkHk´1q “ pHqpKq “ HK.

Therefore for every g P G we have gHKg´1 “ HK. Consequently, HK is

a normal subgroup of G. �

Quotient groups. Let G be a group and let N be a normal subgroup of G.

Our next goal is to define a group structure on the set of left cosets of G.

We define

G{N “ tgN : g P Gu.

- 98 -

Example 11.11. Let G “ Z4 “ t0, 1, 2, 3u and let N “ 2Z4 “ t0, 2u. Then N

is a normal subgroup of G, and it has two distinct left cosets: 0 ` N and

1`N . Therefore G{N is the two-element set

G{N “ t0`N, 1`Nu .

We now define the following binary operation on G{N :

(11.1) pg1Nq ‹ pg2Nq :“ g1g2N.

Lemma 11.12. Let G be a group and let N be a normal subgroup of G. Thenthe binary operation defined in (11.1) is well-defined, i.e., if g1N “ h1N andg2N “ h2N for g1, g2, h1, h2 P G, then

g1N ‹ g2N “ h1N ‹ h2N.

Proof. From the definition of the binary operation ‹ on the cosets, given in

Equation (11.1) above, it follows that we need to verify that

g1g2N “ h1h2N.

By Theorem 9.3, this is equivalent to the relation

ph1h2q´1pg1g2q P N.

In the following steps we prove this relation.

Step 1. From g1N “ h1N and Theorem 9.3 it follows that h´11 g1 P N . Simi-

larly, from g2N “ h2N it follows that h´12 g2 P N .

Step 2. We can now write

(11.2) ph1h2q´1g1g2 “ h´12 h´11 g1g2 “ h´12 ph

´11 g1qg2.

Since h´11 g1 P N , we have ph´11 g1qg2 P Ng2. Since N is normal, we have

Ng2 “ g2N . Therefore ph´11 g1qg2 P g2N . In particular, there exists an ele-

ment h P N such that

ph´11 g1qg2 “ g2h.

- 99 -

Step 3. Now using Equation (11.2) we have

ph1h2q´1g1g2 “ h´12 pg2hq “ ph

´12 g2qh P N,

because h´12 g2 P N by Step 1 and h P N . �

Example 11.13. Let G and N be as in Example 11.11. Then the binary

operation on G{N has the following Cayley table:

‹ 0`N 1`N

0`N 0`N 1`N

1`N 1`N 0`N

Example 11.14. Let G “ Z and let N “ 3Z. Then

G{N “ t0` Z, 1` Z, 2` Zu ,

and the binary operation on G{N has the following Cayley table:

‹ 0`N 1`N 2`N

0`N 0`N 1`N 2`N

1`N 1`N 2`N 0`N

2`N 2`N 0`N 1`N

Theorem 11.15. Let G be a group and let H be a normal subgroup of G. Thenthe set G{N equipped with the binary operation (11.1) is a group.

Proof. We verify the three properties of a group:

Associativity of ‹: we have

pg1N ‹ g2Nq ‹ g3N “ pg1g2qN ‹ g3N “ ppg1g2qg3qN

and similarly,

g1 ‹ pg2N ‹ g3Nq “ pg1pg2g3qqN.

Now associativity of G{N follows from pg1g2qg3 “ g1pg2g3q, i.e., from asso-

ciativity of G.

- 100 -

Identity element: as usual let e P G denote the identity element. We have

eN ‹ gN “ egN “ gN and gN ‹ eN “ geN “ gN.

Thus eN is the identity element of G{N .

Inverse: for every gN P G{N , we have

gN ‹ g´1N “ gg´1N “ eN and g´1 ‹ gN “ g´1gN “ eN

�

Definition 11.16. Let G be a group and let N be a normal subgroup of G.

Then the group G{N whose binary operation is defined in (11.1) is called

the quotient of G by N .

Example 11.17. Let G “ Z and let N “ 3Z. Then

G{N “ t0`N, 1`N, 2`Nu .

By Theorem 11.15,G{N is a group. Since |G{N | “ 3, it follows from Propo-

sition 4.2 that G{N is isomorphic to the group Z3.

Example 11.18. More generally, let G “ Z and let N “ dZ where d P N.

Then

G{N “ t0`N, 1`N, . . . , pd´ 1q `N.

The identity element of G{N is 0 ` N . Furthermore if we set x “ 1 ` N ,

then for every 1 ď k ď d´ 1 we have:

x ‹ ¨ ¨ ¨ ‹ xloooomoooon

k times

“ p1`Nq ‹ ¨ ¨ ¨ ‹ p1`Nqlooooooooooooomooooooooooooon

k times

“ p1` ¨ ¨ ¨ ` 1looooomooooon

k times

q `N “ k `N,

and

x ‹ ¨ ¨ ¨ ‹ xloooomoooon

d times

“ 0`N.

It follows that G{N is generated by x “ 1 ` N . Therefore G{N is a cyclic

group of order d. Hence G{N – Zd.

- 101 -

Example 11.19. Let n ě 1 and let G “ Zn. Also let N “ dZn where d|n.

Then N is a normal subgroup of G (because G is abelian). As we showed

in Example 9.6, the elements of G{N are

t0`N, 1`N, . . . , pd´ 1q `Nu.

The identity element of G{N is 0`N , and furthermore

k `N “ p1`Nq ` ¨ ¨ ¨ ` p1`Nqloooooooooooooomoooooooooooooon

k times

for every 1 ď k ď d´ 1.

Therefore G{N is a cyclic group of order d. It follows that G{N – Zd.

Example 11.20. Let G “ R2 “ R ˆ R and let H “ R ˆ t0u. In Example

9.5(iii) we showed that the distinct left cosets of H are of the form

p0, yq `H for all y P R.

Thus there is a natural bijective correspondence between elements of Rand the distinct left cosets of H . This bijection is given by the map

φ : RÑ G{H , φpyq “ p0, yq `H.

Moreover, from the definition of the operation ‹ on G{H it follows that for

every y1, y2 P R, we have

pp0, y1q `Hq ‹ pp0, y2q `Hq “ pp0, y1q ` p0, y2qq `H “ p0, y1 ` y2q `H.

The left hand side of the above relation is φpy1q ‹ φpy2q, and its right hand

side is φpy1 ` y2q. It follows that φpy1 ` y2q “ φpy1q ‹ φpy2q. Therefore the

bijection φ is indeed an isomorphism of groups. Thus G{H – R.

Example 11.21. Let G “ R ˆ R and let H “ tpx, xq : x P Ru. In one of the

exercises of Section 9 you have shown that H is a subgroup of G, and the

sets

px, 0q `H

- 102 -

are all of the distinct left cosets of G. Thus there is a natural bijection φ

between elements of R and G{H :

φ : RÑ G{H , φptq “ pt, 0q `H.

Moreover, from the definition of the operation ‹ on G{H we have

ppx1, 0q `Hq ‹ pH ` px2, 0q `Hq “ ppx1, 0q ` px2, 0qq `H “ px1 ` x2q `H.

The left hand side of the above relation is φpx1q ‹ φpx2q, and its right hand

side is φpx1 ` x2q. It follows that φpx1 ` x2q “ φpx1q ‹ φpx2q. Consequently,

φ is an isomorphism. This implies that G{H – R.

Exercises.

(1) Let G be a group and let N be a subgroup of G. Prove that if the

binary operation defined on G{N in (11.1) is well-defined, then N

should be normal.

(2) Let G be any group and let H Ď GˆG be defined by:

H “ tpx, xq : x P Gu.

Prove that H is a subgroup of GˆG. Further, prove that H is normal

if and only if G is abelian.

(3) Let G be a group and let H be the subset of the Cartesian product

GˆG defined as follows:

H “ tpx, yq P GˆG : y “ x´1u.

Prove that H is a subgroup of GˆG if and only if G is abelian.

- 103 -

GROUP HOMOMORPHISMS

We will now generalize the notion of group isomorphisms.

Definition 12.1. Let G and H be two groups. A map φ : G Ñ H which

satisfies

φpg1g2q “ φpg1qφpg2q for all g1, g2 P G

is called a group homomorphism from G to H .

Lemma 12.2. Let G and H be two groups and let φ : G Ñ H be a group homo-morphism. Then the following statements hold:

(i) φpeGq “ eH , where eG and eH denote the identity elements of G and G.(ii) φpg´1q “ φpgq´1 for every g P G.

Proof. The proof of this lemma is exactly the same as the argument given

for Remark 3.13. We leave the adaptation of the proof as an exercise. �

Example 12.3. (i) Let G “ H “ Z, and let φ : G Ñ H be defined by

φpkq “ 2k for every k P G “ Z. Then φ is a group homomorphism.

However, φ is not surjective and therefore it is not an isomorphism.

(ii) Let G “ Z and let H “ Z3. Consider the map

φ : ZÑ Z3 , φpkq “ the remainder of k modulo 3.

Then φ is a group homomorphism because for every k1, k2 P Z,

φpk1 ` k2q “ the remainder of k1 ` k2 modulo 3

“ p the remainder of k1 modulo 3q ` p the remainder of k2 modulo 3q .

However, φ is not an isomorphism because it is not a bijection. In-

deed

φ´1p0q “ t0,˘3,˘6,˘9,˘12, . . .u.

- 104 -

(iii) Let n P N and let G “ Sn. Also, let H “ Z2. Then consider the map

φ : Sn Ñ Z2 , φpσq “

$

&

%

0 if σ is an even permutation,

1 if σ is an odd permutation.

From Proposition 8.26 it follows that φ is a homomorphism of groups.

Furthermore, φ´1p0q “ An, whereAn is as usual the subgroup of even

permutations of Sn.

Definition 12.4. Let φ : G Ñ H be a group homomorphism. The image of

φ, denoted by imφ, is defined as

imφ “ tφpgq : g P Gu.

The kernel of φ, denoted by kerφ, is defined by

kerφ “ tg P G : φpgq “ eHu.

Example 12.5. Here are some examples of image and kernel of homomor-

phisms.

(i) Let G “ Z, H “ Z, and φ : GÑ H be the map defined by φpkq :“ 2k

for every k P Z. Then imφ “ 2Z Ă Z, and kerφ “ t0u.

(ii) Let G “ Z, H “ Z3, and φ : Z Ñ Z3 be the homomorphism defined

in Example 12.3(ii). Then

imφ “ Z3 and kerφ “ 3Z “ t0,˘3,˘6,˘9, . . .u.

(iii) More generally, let d P N and let φ : Z Ñ Zd be the map defined as

follows: for every k P Z we set

φpkq “ the remainder of k modulo d.

Then φ is a group homomorphism: to prove this claim, take any

k1, k2 P Z and set r1 “ φpk1q and r2 “ φpk2q. Then by the division

- 105 -

algorithm we have

k1 “ dq1 ` r1 and k2 “ dq2 ` r2 where q1, q2 P Z.

Thus

pk1 ` k2q ´ pr1 ` r2q “ dpq1 ` d2q,

and it follows that the remainder of k1 ` k2 modulo d is equal to the

remainder of r1 ` r2 “ φpk1q ` φpk2qmodulo d. Consequently,

φpk1 ` k2q “ φpk1q ` φpk2q in Zd.

The kernel of φ is the set of all elements of Z which have remainder

0 modulo d. Thus

kerφ “ dZ.

(iv) Let G “ Sn, let H “ Z2, and let φ : G Ñ H be the homomorphism

defined in Example 12.3(iii). Then

imφ “ Z2 and kerφ “ An.

(v) Let G be any group and let H “ teu be the one-element group. Then

the map φ : G Ñ H which maps every element of g to e P H is a

group homomorphism. Moreover, kerφ “ G.

Example 12.6. Let G “ H “ R2 “ R ˆ R, and let φ : G Ñ H be the map

defined by

φ

˜«

x

y

ff¸

“

«

1 ´3

2 ´6

ff«

x

y

ff

.

Here we are representing the elements of G “ R ˆ R as a vertical vector.

Thus we have

φpx, yq “ px´ 3y, 2x´ 6yq.

- 106 -

Then kerφ is equal to kerpAq, i.e., the solution set of the homogeneous lin-

ear system$

&

%

x´ 3y “ 0

2x´ 6y “ 0

In this linear system x is a basis variable and y is a free variable. Therefore:

kerφ “

#«

x

y

ff

: x “ 3y

+

“

#«

3t

t

ff

: t P R

+

.

Proposition 12.7. Let G and H be two groups and let φ : G Ñ H be a grouphomomorphism. Then the following statements hold:

(i) imφ is a subgroup of H .(ii) kerφ is a normal subgroup of G.

(iii) φ is an injection if and only if kerφ “ teu, where e denotes the identityelement of G.

Proof. (i) Clearly imφ is nonempty. We now use the Concise Subgroup Test.

For all x, y P imφ,

x “ φpaq and y “ φpbq for some a, b P G.

Since φpbq´1 “ φpb´1q, we have

xy´1 “ φpaqφpbq´1 “ φpaqφpb´1q “ φpab´1q,

hence ab´1 P im pφq.

(ii) First we show that kerφ is a subgroup of G. Note that e P kerφ so

kerφ ‰ ∅. Now we use the Concise Subgroup Test. Let a, b P kerφ. Then

φpaq “ e and φpbq “ e ñ φpab´1q “ φpaqφpb´1q “ φpaqφpbq´1 “ e,

and it follows that ab´1 P kerφ.

- 107 -

Next we show that kerφ is a normal subgroup of G. To this end, it is

enough to show that gpkerφqg´1 Ď kerφ for every g P G. To prove the

latter statement, note that for any x P kerφ we have

φpgxg´1q “ φpgqφpxqφpg´1q “ φpgqeφpg´1q

“ φpgqφpg´1q “ φpgg´1q “ φpeq “ e,

hence gxg´1 P kerφ.

(iii) Assume that φ is an injection. If φpxq “ eH , then φpxq “ eH “ φpeGq,

and injectivity of φ implies that x “ eG.

Conversely, assume that φpxq “ eH implies that x “ eG. Now assume

φpaq “ φpbq for some a, b P G. Then

φpab´1q “ φpaqφpb´1q “ φpaqφpbq´1 “ eH ,

hence ab´1 “ eG. It follows immediately that a “ b. �

Example 12.8. Let G “ Z ˆ Z, let H “ Z, and let φ “ G Ñ H be given by

φpx, yq “ 3x´ 2y. Then

kerpφq “ tpx, yq P Zˆ Z : 3x´ 2y “ 0u.

We have

3x “ 2y ñ 3|2y, but gcdp2, 3q “ 1, therefore by we have 3|y.

A similar reasoning implies that 2|x. Thus

x “ 2x1 and y “ 3y1 for x1, y1 P Z.

Now

3x “ 2y ñ 3p2x1q “ 2p3y1q ñ 6x1 “ 6y1 ñ x1 “ y1.

Denoting x1 “ y1 by k, we obtain that

H “ tp2k, 3kq : k P Zu.

- 108 -

Example 12.9. Let m ą n ě 1 be integers. Now let φ : Rm Ñ Rn be the

map defined by

φpx1, . . . , xmq “ px1, . . . , xnq for every px1, . . . , xmq P Rm.

It is straightforward to verify that this map is a homomorphism of groups.

We have

px1, . . . , xnq P kerφ if and only if xn`1 “ ¨ ¨ ¨ “ xm “ 0.

Thus kerφ “ tpx1, . . . , xmq P Rm : x1 “ ¨ ¨ ¨ “ xn “ 0u.

Example 12.10. LetA be an nˆmmatrix with entries in R, wherem,n P N.

We denote the elements of the groups/vectorspaces Rm and Rn by vertical

vectors. Now let φ : Rm Ñ Rn be the map defined by φpvq “ Av for every

v P Rm. Then φ is a group homomorphism because

φpv ` wq “ Apv ` wq “ Av ` Aw “ φpvq ` φpwq.

The image of A is the set of all vectors in w P Rn such that w “ phipvq for

some v P Rm. Equivalently, it is the set of all vectors w P Rn of the form

w “ Av for some v P Rm.

From linear algebra, we know that:

w P imφ if and only if w P ColA.

Thus

imφ “ ColA.

Now let us determine kerφ. We have v P kerφ if and only if φpvq “ 0, or

equivalently if and only if Av “ 0. Thus

kerφ “ NulA.

- 109 -

Example 12.11. Let T be the circle group, and let n P N. Now let φn : TÑ Tbe the map defined by φnpzq “ zn for every z P T. Then φ is a group

homomorphism because

φnpz1z2q “ pz1z2qn“ zn1 z

n2 “ φnpz1qφnpz2q for every z1, z2 P T.

This homomorphism is surjective because every z˝ P T is of the form

z˝ “ eiθ for some 0 ď θ ă 2π,

and we have φnpeθn iq “ z˝. The kernel of φn is the set of complex numbers

z P T such that zn “ 1. Thus

kerφn “ Rn “

!

e2πkn i : 0 ď k ď n´ 1

)

.

For example, if n “ 4 then kerφn “ t1,´1, i,´iu.

Recall that if φ : G Ñ H is an isomorphism of groups, then for every

g P G we have opgq “ opφpgqq. This property might not hold for group

homomorphisms, but a weaker version of that is true:

Lemma 12.12. Let φ : G Ñ H be a group homomorphism, and let g P G. Thenopφpgqq

ˇ

ˇopgq.

Proof. Set m “ opgq. Then φpgqm “ φpgmq “ eH Therefore from Theorem

6.6(i) it follows that opφpgqq|m.

�

Image and inverse image of subgroups. Let φ : G Ñ H be a group ho-

momorphism. For any subset S Ď G, we define

fpSq :“ tfpgq : g P Su.

Thus fpSq Ď H .

Similarly, for any subset T Ď H , we define

f´1pT q :“ tx P G : fpgq P T u.

- 110 -

Thus f´1pT q Ď G.

Proposition 12.13. Let φ : GÑ H be a group homomorphism. Then:

(i) If K Ď G is a subgroup of G, then φpKq is a subgroup of H .

(ii) If L Ď H is a subgroup of H , then φ´1pLq is a subgroup of G.

Proof. (i) Recall that φpKq “ tφpkq : k P Ku. Obviously φpKq ‰ ∅. To

prove that φpKq is a subgroup of H we use the Concise Subgroup Test:

take any x, y P φpKq. Then x “ φpaq and y “ φpbqwhere a, b P K. Now

xy´1 “ φpaqφpbq´1 “ φpaqφpb´1q “ φpab´1q P φpKq,

because ab´1 P K.

(ii) Recall that φ´1pLq “ tg P G : φpgq P Lu. Obviously φ´1pLq ‰ ∅. To

prove that φ´1pLq is a subgroup of G we use the Concise Subgroup Test:

take any x, y P φ´1pLq and set a “ φpxq and b “ φpyq. Then a, b P L. Now

φpxy´1q “ φpxqφpy´1q “ φpxqφpyq´1 “ ab´1 P L,

because L is a subgroup of H . �

Proposition 12.14. Let φ : GÑ H be a surjective group homomorphism. Then:

(i) If K Ď G is a normal subgroup of G, then φpKq is a normal subgroup ofH .

(ii) If L Ď H is a normal subgroup of H , then φ´1pLq is a normal subgroup ofG.

Proof. Proposition 12.13 implies that φpKq and φ´1pLq are subgroups of the

corresponding groups. Next we show that they are normal:

(i) Take any h P H . Since φ is surjective, we have h “ φpgq for some

g P G. Now

hφpKqh´1 “ φpgqφpKqφpgq´1 “ φpgKg´1q Ď φpKq,

- 111 -

since K is normal. Now Proposition 11.3 implies that φpKq is a normal

subgroup of H .

(ii) Take any g P G. We need to prove that gφ´1pLqg´1 Ď φ´1pLq. Now

take any x P φ´1pLq. We have φpxq P L, and therefore

φpgxg´1q “ φpgqφpxqφpg´1q P φpgqLφpgq´1 Ď L.

It follows that

gxg´1 P φ´1pLq.

This completes the proof of the inclusion gφ´1pLqg´1 Ď φ´1pLq. �

Proposition 12.15. Let f : GÑ H be a surjective group homomorphism. Then:

(i) For every subgroup K Ď G such that ker f Ď K, we have f´1pfpKqq “K.

(ii) For every subgroup L Ď H , we have fpf´1pLqq “ L.

Proof. (i) We need to prove two inclusions:

S Ď f´1pfpKqq and f´1pfpKqq Ď K.

K Ď f´1pfpKqq: Let x P K and set A “ fpKq. Then fpxq P fpKq “ A, and

therefore by definition of f´1pAq we have x P f´1pAq “ f´1pfpKqq. This

argument does not need surjectivity of f or the assumption ker f Ď K.

f´1pfpKqq Ď K: Let x P f´1pfpKqq. Then fpxq P fpKq, and hence

fpxq “ fpkq for some k P K.

Since f is a homomorphism, it follows that

fpxk´q “ fpxqfpkq´1 “ eH , that is, xk´1 P ker f.

but ker f Ď K and therefore x “ pxk´1qk is a product of two elements of

K. It follows that x P K. This argument uses ker f Ď K but does not need

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surjectivity of f .

(ii) Again we need to show that

fpf´1pLqq Ď L and L Ď fpf´1pLqq.

fpf´1pLqq Ď L: Let x P fpf´1pLqq. Then x “ fpgq for some g P f´1pLq.

Thus,

x “ fpxq and fpgq P L.

It follows that x P L. This argument does not need surjectivity of f .

L Ď fpf´1pLqq: Let x P L. By surjectivity of L, we have

x “ fpgq for some g P G.

Now

fpgq “ x and x P L imply that g P f´1pLq.

Consequently,

x “ fpgq P fpf´1pLqq. �

Exercises.

(1) Let α : G Ñ H and β : H Ñ K be group homomorphisms. Prove

that β ˝ α : GÑ K is also a group homomorphism.

(2) Let φ1 : G1 Ñ H1 and φ2 : G2 Ñ H2 be group homomorphisms.

Prove that the map φ : G1 ˆG2 Ñ H1 ˆH2 defined by

φpg1, g2q “ pφ1pg1q, φ2pg2qq for all g1 P G1, g2 P G2

is also a group homomorphism. (Typically, the map φ is denoted by

φ1 ˆ φ2.)

(3) Let G be a group and let K be a subgroup of G such that K Ď ZpGq.

Assume that G{K is a cyclic group. Prove that G is abelian.

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(4) Let G “ R and let H “ C˚, where R is considered as a group with

respect to addition, whereas H is a group with respect to multipli-

cation. As usual we set i “?´1. Prove that the map φ : G Ñ H ,

φptq “ eit is a group homomorphism. Determine and kerφ explicitly.

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ISOMORPHISM THEOREMS

We now come back to the quotient group G{N , and we explore the rela-

tion between its structure and the group structure of G more thoroughly.

Theorem 13.1. Let G be a group and let N be a normal subgroup of G. Then:

(i) The map

φ : GÑ G{N , φpgq :“ gN

is a group homomorphism.

(ii) imφ “ G{N and kerφ “ N .

(iii) Set

A “ tH Ď G : H is a subgroup of G and N Ď Hu

and

B “ tK Ď G{N : K is a subgroup of G{Nu.

Then the assignments

AÑ B , H ÞÑ φpHq

and

B Ñ A , K ÞÑ φ´1pKq

are bijections between A and B. These bijections are inverse to each other.

Proof. (i) We have φpg1g2q “ g1g2N “ g1N ‹ g2N “ φpg1q ‹ φpg2q.

(ii) Every element of G{N is of the form gN for some g P G, and we have

φpgq “ gN . Thus φ is surjective.

Next we determine the kernel of φ. If g P kerφ, then φpgq “ eN , hence

gN “ eN , and therefore e´1g P N , which means g P N . This proves that

kerφ Ď N . Furthermore, for g P N we have φpgq “ gN “ N “ eN , hence

g P kerφ. Thus N Ď kerφ.

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(iii) LetH be a subgroup ofG such thatN Ď H . Then by Proposition 12.14,

φpHq is a subgroup of G{N . Similarly, if K is a subgroup of G{N , then by

Proposition 12.14, φ´1pKq is a subgroup of G, and indeed N Ď φ´1pKq

because for any x P N , from N “ kerφ (see part (ii) above) it follows that

x P kerφ, hence φpxq “ eH P K, and consequently x P φ´1pKq.

Thus the assignments H ÞÑ φpHq and K ÞÑ φ´1pKq are maps from A to Band from B to A respectively.

Finally, Proposition 12.15 implies that these assignments are inverse to

each other (and in particular, they are bijections). �

Proposition 13.2. In the bijective correspondence of Theorem 13.1(iii), normalsubgroups of G{N correspond to normal subgroups of G that contain N .

Proof. This follows from Proposition 12.14. �

Example 13.3. Let G “ Z4 ˆ Z4 and let

N “ Z4 ˆ t0u “ tpx, 0q : x P Z4u.

Then N is a normal subgroup of G because G is abelian. The distinct left

cosets of N in G are:

p0, 0q `N, p1, 0q `N, p2, 0q `N, p3, 0q `N.

By writing down the Cayley table of G{N we can see that G{N – Z4.

Indeed there is a natural isomorphism f : Z4 Ñ G{N , given by

fpkq “ k ` Z4 for every k P Z4.

Now Theorem 13.1(iii) states that there is a bijection between subgroups

of G{N and subgroups of G – which contain N . This correspondence can

be described explicitly as follows. Subgroups of G{N – Z4 are:

‚ K1 “ tp0, 0q `Nu.

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‚ K2 “ tp0, 0q `N, p2, 0q `Nu.

‚ K3 “ tp0, 0q `N, p1, 0q `N, p2, 0q `N, p3, 0q `Nu.

Subgroups of G “ Z4 ˆ Z4 that contain N are:

‚ H1 “ N “ tpx, 0q : x P Z4u.

‚ H2 “ tpx, yq : x P Z4 and y P 2Z4u.

‚ H3 “ G “ Z4 ˆ Z4.

In the bijective correspondence of Theorem 13.1, the subgroup Hi corre-

sponds to the subgroup Ki for i “ 1, 2, 3. More precisely,

φpHiq “ Ki and φ´1pKiq “ Hi for i “ 1, 2, 3.

Example 13.4. Let G “ Z and let N “ dZ where d P N. Then N is a

normal subgroup of G because G is abelian. In Example 11.18 we saw that

G{N – Zd, and the map yielding this isomoprhism is

f : Zd Ñ Z{dZ , fpkq “ k ` dZ for all k P t0, 1, . . . , d´ 1u.

Theorem 13.1(iii) states that there is a bijective correspondence between

the following:

‚ subgroups of Zd,‚ subgroups of Z which contain dZ.

This correspondence can be described explicitly as follows: subgroups of

Zd are of the form mZd where m P N and m|d. The subgroup of Z that

corresponds to mZd via Theorem 13.1 is mZ. More precisely, we have

φpmZq “ mZd and φ´1pmZdq “ mZ.

Note that from m|d it follows that dZ Ď mZ.

Theorem 13.5. (First Isomorphism Theorem) Let ψ : G Ñ H be a grouphomomorphism.

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(i) Set N “ kerψ. Then there exists a unique group homomorphism ψ :

G{N Ñ H such that ψ ˝ φ “ ψ, where φ : G Ñ G{N is the canonicalgroup homomorphism defined by φpgq :“ gN (see Theorem 13.1).

G

φ��

ψ

$$G{ kerφ

ψ

// H

(ii) The map ψ of part (i) provides an isomorphism G{ kerψ – imψ.

Proof. (i) The relation ψ ˝ φ “ ψ uniquely defines ψ: we must define

ψpgNq “ ψpgq for all g P G.

Step 1. We show that ψ is well defined. Let g1, g2 P N such that g1N “ g2N .

Then g´11 g2 P N “ kerψ, hence ψpg´11 g2q “ eH . It follows that

ψpg1q´1φpg2q “ ψpg´11 qψpg2q “ ψpg´11 g2q “ eH ,

and consequently ψpg1q “ ψpg2q. Thus ψpg1Nq “ ψpg2Nq.

Step 2. We prove that ψ is a group homomorphism. Let g1, g2 P G. Then

ψpg1N ‹ g2Nq “ ψpg1g2Nq “ ψpg1g2q “ ψpg1qψpg2q “ ψpg1Nqψpg2Nq.

Consequently, ψ is a group homomorphism.

(ii) The image of ψ is equal to ψpGq, and therefore we can consider ψ as

a map from G{ kerψ onto imψ. Thus the only thing left to show is that ψ is

an injection. By Proposition 12.7(iii), it suffices to show that kerψ “ teG{Nu,

where eG{N “ eGN is the identity element of G{N “ G{ kerψ.

Take any element gN P kerψ. Then from the definition of ψ we have

ψpgq “ ψpgNq “ eH ,

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hence g P kerψ, that is, g P N . This implies that

gN “ eN “ eG{N .

Consequently, kerψ “ teG{Nu. �

Example 13.6. Let d P N, and let ψ : ZˆZÑ ZˆZd be the map defined by

ψpx, yq “ p2x, y modulo dq for all px, yq P Zˆ Z.

kerψ: Assume that ψpx, yq “ p0, 0q. Then p2x, y modulo dq “ p0, 0q, hence

x “ 0 and d|y. Therefore kerψ “ t0u ˆ dZ.

imψ: Clearly p2x, y modulo dq P 2ZˆZd, and ψ is a surjection onto 2ZˆZd.Therefore

imψ “ 2Zˆ Zd.

Now by the First Isomorphism Theorem we have

pZˆ Zq{pt0u ˆ dZq – 2Zˆ dZ.

Example 13.7. Let ψ : Zˆ ZÑ Z be the map defined by

ψpx, yq “ 5x` 3y for all px, yq P Zˆ Z.

imψ: We need to determine which integers n P Z can be written in the

form n “ 5x`3y for x, y P Z. By Bezout’s Theorem (see Theorem 1.8), since

gcdp5, 3q “ 1, every n P Z can be written of this form. Hence im, ψ “ Z.

kerψ: Assume that ψpx, yq “ 0. Then we have 5x`3y “ 0, hence 5x “ ´3y.

We have:

5|5x ñ 5| ´ 3y but gcdp5, 3q “ 1, hence 5|y because of Lemma 1.15(iii).

For a similar reason, 3|x. Now we can write

x “ 3x1 and y “ 5y1 for some x1, y1 P Z,

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and thus

5x` 3y “ 0 implies that 15x1 ` 15y1 “ 0, hence x1 “ ´y1.

Denoting x1 “ ´y1 by k P Z, it follows that

px, yq “ p3k,´5kq.

Conversely, any element of Zˆ Z of the form p3k,´5kq is in kerφ because:

φp3k,´5kq “ 5p3kq ` 3p´5kq “ 15k ´ 15k “ 0.

Therefore kerψ “ tp3k,´5kq : k P Zu.Now the First Isomorphism Theorem implies that pZˆ Zq{ kerψ – Z.

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FINITELY GENERATED ABELIAN GROUPS

The problem of classifying all finite groups up to isomorphisms is an

extremely difficult one, and there is no solution to it at the moment. How-

ever, the problem of classifying all finite abelian groups (and even more

generally, all finitely generated abelian groups) is much simpler, and its

solution is a fundamental theorem in group theory.The goal of this section

is describe this classification.

Definition 14.1. Let G be a group. We call G finitely generated if there exists

a finite subset S “ ts1, . . . , sru Ď G such that every element of g P G can be

written as

g “ s˘1i1 ¨ ¨ ¨ s˘1iN,

where 1 ď i1, . . . , iN ď r. If this condition holds, then we say G is generatedby S, and we call S a generating set for G.

Example 14.2. (i) The group Z is finitely generated because we can set

S “ t1u or S “ t´1u or S “ t1, 3u (there are many other choices for

S).

(ii) The group Zˆ Z is finitely generated because we can set

S “ tp1, 0q, p0, 1qu.

Then for every element px, yq P Z we have:

– If x, y ě 0 then

px, yq “ p1, 0q ` ¨ ¨ ¨ ` p1, 0qloooooooooomoooooooooon

x times

`p0, 1q ` ¨ ¨ ¨ ` p0, 1qloooooooooomoooooooooon

y times

.

– If x ě 0 and y ă 0 then

px, yq “ p1, 0q ` ¨ ¨ ¨ ` p1, 0qloooooooooomoooooooooon

x times

`p0,´1q ` ¨ ¨ ¨ ` p0,´1qlooooooooooooomooooooooooooon

´y times

.

- 121 -

– If x ă 0 and y ě 0 then

px, yq “ p´1, 0q ` ¨ ¨ ¨ ` p´1, 0qlooooooooooooomooooooooooooon

´x times

`p0, 1q ` ¨ ¨ ¨ ` p0, 1qloooooooooomoooooooooon

y times

.

– If x, y ă 0 then

px, yq “ p´1, 0q ` ¨ ¨ ¨ ` p´1, 0qlooooooooooooomooooooooooooon

´x times

`p0,´1q ` ¨ ¨ ¨ ` p0,´1qlooooooooooooomooooooooooooon

´y times

.

Note that Zˆ Z is also generated by any subset S 1 satisfying S 1 Ě S.

(iii) Every cyclic group G “ xay is finitely generated because we cans set

S “ tau.

(iv) Every finite group G is finitely generated because we can set S “ G.

(v) For every n P N, the group Zn is finitely generated because we can

take S “ te1, . . . , enu, where

ei “ p0, . . . , 0, 1, 0, . . . , 0q, for every 1 ď i ď n,

is the vector containing only one 1 in the i-th coordinate and 0’s else-

where. Indeed every g “ pg1, . . . , gnq P Zn can be expressed as

g “ g1e1 ` ¨ ¨ ¨ ` gnen.

Remark 14.3. From now on, we focus on abelian groups, and we write

the operations of abelian groups additively. That is, instead of gh for two

elements g, h in an abelian group G, we write g ` h. As a result, instead

of gn for an integer n, we will write ng. This is consistent with the explicit

description of the group Zn. The identity element of an abelian group G

will also be denoted by “0”.

With this notation, the definition of a finitely generated abelian group

can be expressed as follows:

An abelian groupG is finitely generated if and only if there exists a finite

- 122 -

set S “ ts1, . . . , sNu Ď G such that every element g P G can be written as

g “ m1g1 ` ¨ ¨ ¨ `mNgN .

Generating subsets of Zn. Clearly the set of vectors S “ te1, . . . , enu that

was defined in Example 14.2(v) is a generating set of Zn. Now starting

from any generating subset S “ ts1, . . . , snu of Zn we can construct new

generating subsets with the following operations:

I) Replacing one si by ´si (the inverse of si). If g P Zn can be expressed

as

g “ k1s1 ` ¨ ¨ ¨ ` knsn,

then we can write g as

g “ k1s1 ` ¨ ¨ ¨ ` ki´1si´1 ` p´kiqp´siq ` ki`1si`1 ` ¨ ¨ ¨ ` knsn.

Note that the coefficients of the above linear combination are still in

Z.

II) Replacing some si with si ` sj with j ‰ i. if g P Zn can be expressed

as

g “ k1s1 ` ¨ ¨ ¨ ` knsn,

then we can write g as

g “ k1s1 ` ¨ ¨ ¨ ` kipsi ` sjq ` ¨ ¨ ¨ ` pkj ´ kiqsj ` ¨ ¨ ¨ ` knsn.

Note that the coefficients of the above linear combination are still in

Z.

Example 14.4. Let S “ te1, e2, e3u “ tp1, 0, 0q, p0, 1, 0q, p0, 0, 1qu be the stan-

dard generating set of Z3. Then using the operations (I) and (II) above, we

- 123 -

can form new generating sets of S, as follows:

te1, e2, e3upIqÝÑ te1,´e2, e3u

pIIqÝÝÑ te1 ´ e2,´e2, e3u

pIIqÝÝÑ te1 ´ 2e2,´e2, e3u

pIqÝÑ te1 ´ 2e2, e2, e3u

pIqÝÑ te1 ´ 2e2, e2,´e3u

pIIqÝÝÑ te1 ´ 2e2, e2, e2 ´ e3u ÝÑ ¨ ¨ ¨

Definition 14.5. A generating set S “ tv1, . . . , vnu Ď Zn is called a basis if it

satisfies the following property: if m1v1`¨ ¨ ¨`mnvn “ 0, for some integers

m1, . . . ,mn, then m1 “ . . . “ mn “ 0.

Proposition 14.6. Let S :“ tv1, . . . , vnu be a basis of Zn. Then the generatingset obtained from S by each of the operations (I) and (II) is also a basis of Zn.

Proof. We leave the proof as an exercise. �

Remark 14.7. The operations (I) and (II) above can also be applied to gen-

erating sets of any finitely generated abelian group. Thus, if ts1, . . . , snu is a

generating set of a finitely generated abelian group G, then the operations

(I) and (II) above will produce new generating sets of G.

Proposition 14.8. Let G be an abelian group. Then G is finitely generated if andonly if there exists an integer n P N and a surjective homomorphism

φ : Zn Ñ G.

In particular, every finite abelian group G is isomorphic to a quotient group of theform Zn{K, where K is a subgroup of Zn.

Proof. Assume that G is generated by a subset S “ ts1, . . . , snu Ď G. Then

we define the map Zn Ñ G as follows:

φ : Zn Ñ G , φpk1, . . . , knq :“ k1s1 ` ¨ ¨ ¨ ` knsn.

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Since S generates G, the map φ is a surjection. Furthermore, φ is a group

homomorphism because

φpk1 ` k11, . . . , kn ` k

1nq “ pk1 ` k

11qs1 ` ¨ ¨ ¨ ` pkn ` k

1nqsn

“ pk1s1 ` k11s1q ` ¨ ¨ ¨ ` pknsn ` k

1nsnq

“ pk1s1 ` ¨ ¨ ¨ ` knsnq ` pk11s1 ` ¨ ¨ ¨ ` k

1nsnq

“ φpk1, . . . , knq ` φpk11, . . . , k

1nq.

Next assume that a surjective group homomorphism φ : Zn Ñ G exists.

Set

si “ φpeiqwhere ei “ p0, . . . , 0, 1, 0, . . . , 0q,

that is, ei is the unit vector in Zn with a 1 in the i-th position and 0’s else-

where. We show that G is generated by S. Indeed for every g P G we have

g “ φpk1, . . . , knq for some element pk1, . . . , knq P Zn. Then

g “ φpk1, . . . , knq “ k1s1 ` ¨ ¨ ¨ ` knsn. �

Corollary 14.9. Every finite abelian group is isomorphic to a quotient Zn{Kwhere K is a subgroup of Zn.

Proposition 14.10. Let n P N. Then every subgroup of Zn is a finitely generatedabelian group that is generated by a subset S Ď G such that |S| ď n.

Proof. We prove the statement by induction on n. For n “ 1, the statement

follows from the classification of subgroups of Z (see Theorem 6.15).

Next assume that the statement holds for subgroups of Zn´1. We prove

that the statement also holds for all subgroups of Zn. From now on, we

consider a subgroup G of Zn.

Step 1. Write every element g P G as g “ pg1, . . . , gnq, and let F be the

subset of Z consisting of the integers g1 for all g P G. We show that F is

a subgroup of Z. Clearly 0 P F because p0, . . . , 0q P G. Next we use the

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Concise Subgroup Test: take a, b P Z; then there exist elements g, h P G

such that

g “ pg1, . . . , gnqwith g1 “ a, and h “ ph1, . . . , hnqwith h1 “ b.

We need to verify that a´b P F . But a´b is the first component of g´h P G,

because

g ´ h “ pg1 ´ h1, . . . , gn, hnq.

Step 2. Set G0 :“ G X t0u ˆ Zn´1. This is the subset of G consisting of

elements whose first coordinate is equal to 0. Then G0 is a subgroup of Zn

because it is the intersection of two subgroups of Zn.

Step 3. Note that G0 Ď t0u ˆ Zn´1 – Zn´1, and therefore by the induction

hypothesis G0 is generated by a subset S0 Ď G0 such that |S0| ď n´ 1.

Step 4. By the classification of subgroups of Z (see Theorem 6.15), we know

that F “ dZ for some d P Z. If d “ 0, then it follows that

G “ G0,

and the proof is complete by Step 3. Next, we consider the case d ‰ 0.

Step 5. Recall that by Step 4 we can assume that F “ dZ for some d P Zzt0u.Now take an element x P G such that

x “ px1, . . . , xnqwith x1 “ d.

We claim that G is generated by S “ S0 Y txu. Indeed for any g P G we

have

g “ pg1, . . . , gnqwhere g1 “ dk for some k P Z.

Thus we can write

g “ kx` pg ´ kxq,

and

g ´ kx “ pg1 ´ kx1, . . . , gn ´ kxnq “ p0, g2 ´ kx2, . . . , gn ´ kxnq P G0.

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Now Step 3 implies that g ´ kx can be expressed as a linear combination

with integer coefficients of the elements of S0. Consequently, g can be

expressed as a linear combination of the elements of S0 Y txu with integer

coefficients.

�

The Smith normal form of an integral matrix. Let A be an m ˆ n matrix

with integer entries. We are interested in the following six row and column

operations:

RI: Multiplying the i-th row by p´1q, where 1 ď i ď m.

RII: Adding the i-th row to the j-th row, where 1 ď i, j ď m and i ‰ j.

RIII: Switching the i-th row with the j-th row, where 1 ď i, j ď m and

i ‰ j.

CI: Multiplying the i-th column by p´1q, where 1 ď i ď n.

CII: Adding the i-th column to the j-th column, where 1 ď i, j ď n and

i ‰ j.

CIII: Switching the i-th column with the j-th column, where 1 ď i, j ď n

and i ‰ j.

Lemma 14.11. Let Y “ ryijs be an m ˆ n matrix with integer entries. Assumethat y11 P N. Then by operations RI, RII, CI, and CII, we can replace the entriesin the first row and first column of Y , other than the top-left corner of Y , by theirremainders modulo y11.

Proof. This can be done as follows. Say for some 1 ă a ď m we have

y1a “ qy11 ` r,

with q, r P Z and 0 ď r ă y11. If q ă 0, then

r “ y1a ´ qy11,

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and by adding the first column of Y to the a-th column q times (this is CII),

the entry y1a will be substituted by r. No other entry in the first row and

first column of Y is affected.

If q ą 0, then we first multiply the first column of Y by p´1q; this is CI.

Then the top-left corner of the resulting matrix contains ´y11. Now note

that

r “ y1a ` qp´y11q,

and again by adding the first column to the a-th column q-times, the entry

y1a will be replaced by r. Finally we multiply the first column of the result-

ing matrix by p´1q for a second time, so that the entries in the first column

are not affected by the whole process.

Finally, if q “ 0 then y1a “ r and there is nothing to be done. �

Proposition 14.12. (Smith Normal Form) Let A be an m ˆ n matrix withentries in Z. Then using the operations RI–RIII and CI–CIII we can obtain amatrix in the form

(14.1)

«

Dk 0kˆpn´kq

0pm´kqˆk 0kˆk

ff

where D “

»

—

—

—

—

–

d1 0 ¨ ¨ ¨ 0

0 d2 ¨ ¨ ¨ 0... ... ¨ ¨ ¨

...0 ¨ ¨ ¨ 0 dk

fi

ffi

ffi

ffi

ffi

fl

where d1, . . . , dk P N and di|di`1 for 1 ď i ď k ´ 1.

Proof. If A “ 0mˆn then there is nothing to prove. Thus from now on we

assume that A ‰ 0mˆn. We claim that by the above six operations we can

transform A to a matrix of the form

(14.2)

«

d 01ˆpn´1q

0pm´1qˆ1 B

ff

where d P N and B “ rbijs is an pm ´ 1q ˆ pn ´ 1q matrix with entries in Zsuch that d|bi,j for every entry bi,j ofB. The claim of the proposition follows

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from this statement, by continuing the same process in the submatrix B,

and noting that if all of the entries of a matrix are divisible by d, then after

any of the operations RI–RIII and CI–CIII all of the entries of the resulting

matrix will also be divisible by d.

We now prove the above claim. For every nonzero m ˆ n matrix X , we

set

mpXq “ mint|xij| : xij ‰ 0u.

We repeatedly execute the following process on A:

‚ Input: an mˆ n matrix X “ rxijs.

Step 1. Choose i0, j0 such that xi0j0 “ mpXq. If either x11 “ 0 or |xi0j0| ă |x11|,

then using the operations RIII and CIII, move the entry xi0j0 to the

top-left corner position.

Step 2. Using RI if necessary, we can assume that the topleft corner entry

is non-negative. Call the resulting matrix X “ rx1ijs. Thus, x111 “

mpXq “ |xi0j0|.

Step 3. Using RI, RII, CI, and CII, obtain a matrix Y “ ryijs from X 1 such

that

y1a “ x11a modulo x111

and

yb1 “ x1b1 modulo x111

for all 1 ă a ď n and 1 ă b ď m. This is possible by Lemma 14.11.

‚ Output: The mˆ n matrix Y .

If some of the entries in the first row or first column of X 1 other than the

top-left corner are nonzero, then 0 ď mpX 1q ă mpXq. However, by execut-

ing the above process repeatedly, the value ofmpXq cannot decrease indef-

initely (because it is in N). This means that after finitely many iterations

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of the above process, we obtain a a matrix A1 from A in the block-diagonal

form (14.2) such that mpA1q ď mpAq.

If the top-left corner entry d of A1 divides all of the entries of B, then we

are done. However, if there exists an entry bi0j0 of B such that d - bi0j0, then

we do the following:

Step 4. Add the first row to the i0-th row (this is RII), and then by a tech-

nique similar to Lemma 14.11, use the operations RI, RII, CI, and CII to

obtain a a matrix A2 in which bi0j0 is replaced by bi0j0 modulo d. Clearly

mpA2q ă d “ mpA1q. We now execute Steps 1–4 as many times as possi-

ble. Since the value of mpAq keeps decreasing throughout the process, the

process should come to an end. This happens precisely when the resulting

matrix is of the form (14.2) where d divides every entry of B.

�

Using the Smith normal form, we can classify subgroups of Zn explicitly.

Informally, the next theorem states that every subgroup of Zn is naturally

embeds as a product of cyclic groups, after a suitable change of coordi-

nates.

Theorem 14.13. Let G be a subgroup of Zn. Then there exist d1, . . . , dk P N,where k ď n, and an isomorphism

φ : Zn Ñ Zn

such that

φpGq “ d1Zˆ ¨ ¨ ¨ ˆ dkZˆ t0u ˆ ¨ ¨ ¨ ˆ t0u

“

pd1m1, . . . , dkmk, 0, . . . , 0q : m1, . . . ,mk P Z(

,

and di|di`1 for 1 ď i ď k ´ 1.

Proof. Recall that E “ te1, . . . , enu is the standard generating set of Zn. By

Proposition 14.10 we can also choose a generating set S “ ts1, . . . , smu for

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G (which satisfies m ď n).

Step 1. Each si P S can be expressed as

si “nÿ

j“1

cijej.

We can now form the mˆ n matrix

A “

»

—

—

–

c11 ¨ ¨ ¨ c1n... ...

cm1 ¨ ¨ ¨ cmn

fi

ffi

ffi

fl

Thus, each row of A represents the “coordinates” of one of the si’s when it

is expressed in terms of e1, . . . , en.

Step 2. We now interpret the operations RI–RIII and CI–CIII in terms of

operations on the generating sets E and S.

Operations RI–RIII mean that we are changing the generating set of G.

For example, if we multiply the i-th row of A by p´1q, this means that we

are replacing the set S by the set

ts1, . . . , si´1,´si, si`1 . . . , smu.

Similarly, if we add the i-th row of A to the j-th row, this means that we

are replacing the set S by the set

ts1, . . . , sj´1, sj ` si, sj`1 . . . , smu.

Operations CI—CIII on A mean that we are changing the geneating set

E of Zn, and therefore the rows of the “new” A give the coordinates of

s1, . . . , sm when they are expressed in terms of this new generating set of

Zn.

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Step 3. Consider a sequence of the operations RI–RIII and CI—CIII on a

matrix A. Each “R” operation is equivalent to multiplication on the left by

an invertiblemˆmmatrix P . Each “C” operation is equivalent to multipli-

cation on the right by an invertible n ˆ n matrix Q. Thus, the sequence of

operations of “R” and “C” type turns A into B “ P1 ¨ ¨ ¨PrAQ1 ¨ ¨ ¨Qs. Since

matrix multiplication is associative, the ”R” operations commute with the

”C” operations. Consequently, in the process of obtaining B from A, we

can assume that first all the “R” operations are implemented, then all the

“C” operations are implemented.

Step 4. By Proposition 14.12 we can find a sequence of “R” and “C” oper-

ations such that implementing them on A yields a matrix

B “ P1 . . . PrAQ1 . . . Qs

in Smith normal form. We now need the following observations:

‚ The rows of M :“ P1 . . . PrA give coordinates of elements of a gen-

erating set S 1 “ ts11, . . . , s1mu of G in terms of the standard generating

set E of Zn.

‚ The rows of B “ MQ1 . . . Qs give coordinates of the elements of S 1

in terms of a generating set E 1 “ te11, . . . , e1nu of Zn.

Now consider the map φ : Zn Ñ Zn defined by

φpřni“1 xie

1iq :“

řni“1 xiei for all x1, . . . , xn P Z.

Proposition 14.6 implies that this map is well-defined and injective. Since

E “ te1, . . . , enu is a generating set for Z6n, this map is also surjective.

Therefore φ is indeed an isomorphism Zn Ñ Zn.

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Step 5. Finally, we investigate the structure of φpGq under the map φ of

Step 4. Note that

s1i “ die1i for 1 ď i ď k and s1i “ 0 for i ą k.

Therefore we have

φps1iq “ φpdie1iq “ diei for 1 ď i ď k and φps1iq “ 0 for i ą k.

Since the group φpGq is generated by φps11q, . . . , φps1kq, it follows that

φpGq “

pd1m1, . . . , dkmk, 0, . . . , 0q : m1, . . . ,mk P Z(

. �

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