section 6 -- cyclic groupsjupiter.math.nctu.edu.tw/~weng/courses/alg_2007/algebra...

Elementary properties of cyclic groupsStructure of cyclic groups

Subgroups of finite cyclic groups

Section 6 – Cyclic Groups

Instructor: Yifan Yang

Fall 2006

Instructor: Yifan Yang Section 6 – Cyclic Groups

Outline

1 Elementary properties of cyclic groups

2 Structure of cyclic groups

3 Subgroups of finite cyclic groups

Cyclic groups

Definition1 A group G is cyclic if G = 〈a〉 = {an : n ∈ Z} for some

a ∈ G. We say the element a is a generator of G.2 Let G be a group and a ∈ G. The subgroup〈a〉 = {an : n ∈ Z} is the cyclic subgroup of G generated bya.

3 If 〈a〉 has a finite number of element, then the order of a isthe order |〈a〉| of this subgroup.

4 If 〈a〉 has infinitely many element, we say a is of infiniteorder.

Remark

If a is of finite order, then the order of a is exactly the smallestpositive integer m such that am = e.

Cyclic groups

Remark

Cyclic groups

Remark

Cyclic groups

Remark

Cyclic groups

Remark

Cyclic groups

Remark

Cyclic groups

Theorem (6.1)

Every cyclic group G is abelian.

Proof.

We have G = 〈a〉 for some a ∈ G. If x , y ∈ G, then x = ar ,y = as for some integers r and s. Now

xy = ar as = ar+s = as+r = asar = yx .

Thus G is abelian.

Cyclic groups

Theorem (6.1)

Every cyclic group G is abelian.

Proof.

We have G = 〈a〉 for some a ∈ G. If x , y ∈ G, then x = ar ,y = as for some integers r and s. Now

xy = ar as = ar+s = as+r = asar = yx .

Thus G is abelian.

Cyclic groups

Theorem (6.6)

A subgroup of a cyclic group is cyclic.

Proof.

Let G = 〈a〉 be a cyclic group, and H be a subgroup.Case H = {e}. Then H = 〈e〉 and H is cyclic.Case H 6= {e}. Then an ∈ H for some positive integer n. Letm be the smallest positive integer such that am ∈ H, and setc = am. We claim that H = 〈c〉. Let b ∈ H. SinceH ⊂ G = 〈a〉, b = an for some n ∈ Z. By the divisionalgorithm, there exist integers q and r with 0 ≤ r < m such thatn = qm + r . Now ar = an−qm = bc−q ∈ H because b, c ∈ H.By the assumption that m is the smallest positive integer suchthat am ∈ H, we must have r = 0. That is, b = cq ∈ 〈c〉.

Cyclic groups

Theorem (6.6)

Proof.

Cyclic groups

Theorem (6.6)

Proof.

Cyclic groups

Theorem (6.6)

Proof.

Cyclic groups

Theorem (6.6)

Proof.

Cyclic groups

Theorem (6.6)

Proof.

Cyclic groups

Theorem (6.6)

Proof.

Cyclic groups

Theorem (6.6)

Proof.

Cyclic groups

Theorem (6.6)

Proof.

Cyclic groups

Theorem (6.6)

Proof.

Cyclic groups

Theorem (6.6)

Proof.

Cyclic groups

Theorem (6.6)

Proof.

Subgroups of Z

Corollary (6.7)

The subgroups of Z under addition are precisely the groups nZunder addition for n ∈ Z.

In-class exercises1 Let m and n be two integers, and let

H = {am + bn : a, b ∈ Z} .

Prove that H is a subgroup of Z.2 Consider m = 6, n = 4. Find an integer k such that

H = kZ.3 Make a general conjecture about the relation between m,

n, and k .

Subgroups of Z

Corollary (6.7)

H = {am + bn : a, b ∈ Z} .

n, and k .

Subgroups of Z

Corollary (6.7)

H = {am + bn : a, b ∈ Z} .

n, and k .

Subgroups of Z

Corollary (6.7)

H = {am + bn : a, b ∈ Z} .

n, and k .

Subgroups of Z

Corollary (6.7)

H = {am + bn : a, b ∈ Z} .

n, and k .

Subgroups of Z

Theorem

Let m and n be two integers, and set d = gcd(m, n) andH = {am + bn : a, b ∈ Z}. Then H = dZ.

Proof.We will showH ⊂ dZ: Since d |m and d |n, we have d |(am + bn) for alla, b ∈ Z. Therefore H ⊂ dZ.dZ ⊂ H: It suffices to show that d = am + bn for someintegers a, b ∈ Z. This follows from the Euclidean algorithm inhigh school algebra.

Subgroups of Z

Theorem

Subgroups of Z

Theorem

Subgroups of Z

Theorem

Subgroups of Z

Theorem

Subgroups of Z

Theorem

Structure of cyclic groups

Theorem (6.10)

Let G be a cyclic group with generator a. If the order of G isinfinite, then G is isomorphic to 〈Z,+〉. If G is has finite order n,then G is isomorphic to 〈Zn,+n〉.

Proof for the case am 6= e for all m 6= 0

Define φ : Z 7→ G by φ(m) = am. We claim that φ is anisomorphism. We check1-to-1: If φ(m) = φ(n), then am = an and am−n = e. Byassumption, this happends only when m − n = 0. Therefore,φ is one-to-one.Onto: Given b ∈ G, we have b = am for some m ∈ Z. Thenφ(m) = b.Homomorphy: φ(m + n) = am+n = aman = φ(m)φ(n).

Theorem (6.10)

Proof for the case am = e for some m ∈ ZLet n be the smallest positive integer such that an = e. Defineφ : Zn 7→ G by φ(m̄) = am. We need to check four things.Well-defined: Suppose that m̄ = k̄ . We need to show thatam = ak . Now m̄ = k̄ implies m = k + pn for some integer p.Thus, am = ak+pn = ak (an)p = ak .1-to-1: Suppose that φ(m̄) = φ(k̄). Then am = ak . It followsthat am−k = e. Since n is the smallest positive integer suchthat an = e, we have n|(m − k). (cf. proof of Theorem 6.6.)Thus, m̄ = k̄ .Onto: Clear.Homomorphy: Easy.

A remark on isomorphic groups

Remark

Suppose that φ : G1 7→ G2 is an isomorphism between twogroups. Then there is a one-to-one correspondence betweenthe subgroups of G1 and those of G2. The correspondence isgiven by H ↔ φ(H) for H < G.This means that isomorphic groups have the same number ofsubgroups and the same subgroup structures. Thus, theabove theorem tells that to understand the structures of cyclicgroups, it suffices to study those of Z and Zn.

Remark

In-class exercise

Let φ : G1 7→ G2 be given as above.1 Let H be a subgroup of G1. Prove that φ(H) is a subgroup

of G2.2 Prove that if φ is an isomorphism, then H is isomorphic to

φ(H).3 Prove that if H and H ′ are subgroups of G1 such that

H 6= H ′, then φ(H) 6= φ(H ′).

In-class exercise

H 6= H ′, then φ(H) 6= φ(H ′).

In-class exercise

H 6= H ′, then φ(H) 6= φ(H ′).

Theorem (6.14)

Let G be a cyclic group of n elements generated by a. Letb ∈ G and let b = as, and d = gcd(n, s). Then

〈b〉 ={

a0 = e, ad , a2d , . . . , an−d}

Corollary

Let G be a cyclic group of n elements generated by a. Then〈ar 〉 = 〈as〉 if and only if gcd(n, r) = gcd(n, s).

Theorem (6.14)

Let G be a cyclic group of n elements generated by a. Letb ∈ G and let b = as, and d = gcd(n, s). Then

〈b〉 ={

a0 = e, ad , a2d , . . . , an−d}

Corollary

Let G be a cyclic group of n elements generated by a. Then〈ar 〉 = 〈as〉 if and only if gcd(n, r) = gcd(n, s).

Corollary (6.16)

If a is a generator of a finite cyclic group of order n, then theother generators of G are the elements of the form ar , where ris relatively prime to n.

Corollary

Let G be a cyclic group of n elements generated by a. Then thenumber of subgroups of G is equal to the number of divisors ofn. They are

{e, ad , a2d , . . . , an−d},

where d runs over all divisors of n.

Corollary (6.16)

If a is a generator of a finite cyclic group of order n, then theother generators of G are the elements of the form ar , where ris relatively prime to n.

Corollary

Let G be a cyclic group of n elements generated by a. Then thenumber of subgroups of G is equal to the number of divisors ofn. They are

{e, ad , a2d , . . . , an−d},

where d runs over all divisors of n.

Additive version of Theorem 6.14

Remark

The above results are stated in terms of multiplicative groups.In terms of additive groups, such as Zn, Theorem 6.14 saysthat the subgroup of Zn generated by s̄ is

{0̄, d̄ , 2d , . . . , n − d},

where d = gcd(n, s).

Example

Find all subgroups of Z7.

Solution

By Theorem 6.6, every subgroup of a cyclic group is cyclic.Thus, the subgroups of Z7 are 〈0〉, 〈1〉, . . ., 〈6〉. Now,gcd(7, 1) = gcd(7, 2) = · · · = gcd(7, 6) = 1. Thus,〈1〉 = 〈2〉 = · · · = 〈6〉 = {0, 1, . . . , 6} = Z7. Also,gcd(7, 0) = 7. Thus, 〈0〉 = {0}.Conclusion: There are two subgroups {0} and Z7.

Example

Solution

Example

Solution

Example

Solution

Example

Solution

Example

Solution

Proof of Theorem 6.14

Proof of Theorem 6.14.

We have ar ∈ 〈b〉 ⇔ ar = bk = aks for some integer k . Sincen is the smallest positive integer such that an = e (see Proof ofTheorem 6.10), ar = aks if and only if r = ks + `n for someinteger `. That is, ar ∈ 〈b〉 ⇔ r = ks + `n for some integers kand `. As k and ` run over all integers, r runs over allmultiples of d = gcd(n, s). Therefore,

〈b〉 = {e, ad , a2d , . . . , an−d}.

More examples

Example

Draw the subgroup diagram of Z4.Solution. The divisors of 4 are 1, 2, and 4. The subgroupsgenerated by them are {0, 1, 2, 3}, {0, 2}, and {0}, respectively.So the subgroup diagram is

Z4 = {0, 1, 2, 3}

〈2〉 = {0, 2}

〈0〉 = {0}

More examples

Example

Z4 = {0, 1, 2, 3}

〈2〉 = {0, 2}

〈0〉 = {0}

More examples

Example

Z4 = {0, 1, 2, 3}

〈2〉 = {0, 2}

〈0〉 = {0}

More examples

Example

Z4 = {0, 1, 2, 3}

〈2〉 = {0, 2}

〈0〉 = {0}

More examples

Draw the subgroup diagram of Z12.Solution. The divisors of 12 are 1, 2, 3, 4, 6, 12. Thesubgroup diagram is

〈1〉 = Z12

〈3〉〈2〉

〈6〉〈4〉

〈12〉 = {0}

��

More examples

〈1〉 = Z12

〈3〉〈2〉

〈6〉〈4〉

〈12〉 = {0}

��

More examples

〈1〉 = Z12

〈3〉〈2〉

〈6〉〈4〉

〈12〉 = {0}

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Exercises

In-class exercise1 What is wrong with the statement “an element a of a group

G has order n ∈ Z+ if and only if an = e”?2 Find all the generators of Z6.3 Determine the number of subgroups of Z36.4 Draw the subgroup diagram of Z18.

Homework

Problems 18, 20, 24, 28, 32, 36, 37, 40, 44, 49, 50, 55 ofSection 6.

Exercises

In-class exercise1 What is wrong with the statement “an element a of a group

G has order n ∈ Z+ if and only if an = e”?2 Find all the generators of Z6.3 Determine the number of subgroups of Z36.4 Draw the subgroup diagram of Z18.

Homework

Problems 18, 20, 24, 28, 32, 36, 37, 40, 44, 49, 50, 55 ofSection 6.

section 6 -- cyclic groupsjupiter.math.nctu.edu.tw/~weng/courses/alg_2007/algebra...

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