2MA105 Algebraic Structures I

Per-Anders Svensson

http://homepage.lnu.se/staff/psvmsi/2MA105.html

Lecture 3

Cyclic groups Definition and Examples

Properties of cyclic groups

Subgroups of cyclic groups

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Cyclic groups Definition and Examples

ExampleWe investigate the group of units modulo 7, i.e. Z7 = {1, 2, 3, 4, 5, 6},where the binary operation is given by multiplication modulo 7. Wefind that

31 = 3 3 (mod 7)32 = 9 2 (mod 7)

33 = 3 32 3 2 6 (mod 7)34 = 3 33 3 6 4 (mod 7)35 = 3 34 3 4 5 (mod 7)36 = 3 35 3 5 1 (mod 7)

This means that each element of Z7 can be written in the form 3k forsome integer k .

ExerciseShow that it is also possible to write all elements i Z7 in the form 5k ,but not in the form 2k .

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ExampleWe repeat the previous example on the group of units modulo 8, i.e.Z8 = {1, 3, 5, 7}. In this group we have

11 = 1, 12 = 1, 13 = 1, 14 = 1, . . .31 = 3, 32 = 1, 33 = 3, 34 = 1, . . .51 = 5, 52 = 1, 53 = 5, 54 = 1, . . .71 = 7, 72 = 1, 73 = 7, 74 = 1, . . .

There is no element a Z8 such that the set of all powers ak becomesthe whole group Z8. But in each case above, we obtain a subgroupof Z8; for instance {3k | k Z} = {1, 3} is a subgroup having theCayley table

1 31 1 33 3 1

If G is a group and a G , is then H = {ak | k Z} always asubgroup of G?

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TheoremLet G be a group and suppose a G. Then

H = {ak | k Z}

is a subgroup of G. Moreover, H is the smallest subgroup of G thatcontains a, in the sense that every other subgroup that contains a,contains H as well.

Proof.Let us check the criteria closure, identity element, and inverse:

Closure If h1, h2 H , then there are k1, k2 Z such thath1 = ak1 and h2 = ak2 . Therebyh1h2 = ak1ak2 = ak1+k2 H .

Identity element The identity element 1G belongs to H , since 1G = a0.Inverse If h H then h = ak for some k Z. Moreover

h1 = (ak )1 = ak , whence h1 H .Thus H G . If K is a subgroup of G such that a K , then therequirements of closure, identity element, and inverse implies thatevery ak , k Z, belongs to K . Thereby H is contained in K , provingthe minimality of H .

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Definition (Cyclic subgroup)Let G be a group and a G . Then the subgroup {ak | a Z} iscalled the cyclic subgroup generated by a and is denoted a.ExampleReferring to the example, where Z8 = {1, 3, 5, 7} was studied, we cansee that 1 = {1}, 3 = {1, 3}, 5 = {1, 5}, and 7 = {1, 7}.In the example where we studied Z7 we saw that 3 = Z7.ExampleIn the group (Z,+) we use additive notation, so ak corresponds to

k a = a + a + + a k terms

,

so for instance 7 = {7k | k Z} contains all integer multiples of 7.We denote this subgroup by 7Z.In Z we have in general

n = {nk | k Z} = nZ.

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Definition (Cyclic group, Generator)A group G is called cyclic, if G = a for some a G . This a is thencalled a generator of G .

ExampleFrom what we previously have seen, Z7 is cyclic, having 3 as agenerator. On the other hand, Z8 is not cyclic, since none of its cyclicsubgroups equal the whole of Z8.

ExampleThe group Z of integers is cyclic, having 1 as a generator, since

1 = {k 1 | k Z} = {k | k Z} = Z.

There is one more generator of this group which one?

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ExampleThe groups Zn under addition modulo n are all cyclic. We canchoose 1 as a generator in each one of these groups, but there are ingeneral more generators, besides 1. For example, in Z5 = {0, 1, 2, 3, 4},all elements but 0 are generators, since

0 1 = 0 1 1 = 1 2 1 = 2 3 1 = 3 4 1 = 40 2 = 0 1 2 = 2 2 2 = 4 3 2 = 1 4 2 = 30 3 = 0 1 3 = 3 2 3 = 1 3 3 = 4 4 3 = 20 4 = 0 1 4 = 4 2 4 = 3 3 4 = 2 4 4 = 1

In each one of these cases, we run through all elements Z5 if wecompute integer multiples (modulo 5) of 1, 2, 3, and 4, respectively.

In the group Z4 = {0, 1, 2, 3} it is not the case that all elements but 0generates the whole group, since 2 = {0, 2} 6= Z4. On the otherhand, 3 is a generator.

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We have earlier defined the order of a group as the number ofelements in the group (if this is finite). We can also define the order ofan element in a group.

Definition (Order of an Element)Let G be a group and a an element of G . If a is a finite subgroupof G , then the order of a, denoted o(a), as the order of the group a,i.e. o(a) = |a|. If a is an infinite group, then a is said to be ofinfinite order.

ExampleIn the group of units Z8 = {1, 3, 5, 7} modulo 8 we haveo(3) = o(5) = o(7) = 2 (see an earlier example). Moreover o(1) = 1,since 1 = {1k | k Z} = {1} contains one single element only.ExampleIn Z5 we have o(a) = 5 for all a 6= 0, and o(0) = 1.In Z4 we have o(0) = 1, o(1) = o(3) = 4, and o(2) = 2.

ExampleIn the cyclic group Z, all elements but 0 is of infinite order.

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Properties of cyclic groups

TheoremIf a group is cyclic, then it is Abelian.

Proof.Suppose G = a is cyclic. Pick any x , y G . We wish to show thatxy = yx . Since G = a, there are integers m and n such that x = amand y = an . Therefore

xy = aman = am+n = an+m = anam = yx .

Does the converse of the theorem above hold, i.e. are every Abeliangroup cyclic?

No, Z8 is a counterexample.

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What can you say about the structure of a cyclic group G = a?We have two cases to consider, depending on whether G is finite ornot.

TheoremSuppose G = a is a finite cyclic group of order m. Then m is thesmallest positive integer such that am = 1G . Moreover,

G = {1G , a, a2, a3, . . . , am1}.

Thus, if a generates a finite cyclic group, it could happen thatas = at , even though s 6= t . For instance, 32 = 38 = 2 in Z7. But ifthe order of a is infinite, this cannot happen:

TheoremIf G = a is an infinite cyclic group, then

as = at = s = t .

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Subgroups of cyclic groups

TheoremIf G is a cyclic group, then any of its subgroups is cyclic as well.

ExampleThe group of integers Z is a cyclic group. Hence each subgroup of Z iscyclic, and thus has to be in the form

n = {nk | k Z} = nZ

for some integer n.

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We have seen in some examples, that different elements can do as agenerator for a cyclic group. For example, 1 = 3 = Z4 anda = Z5 for all a 6= 0 in Z5. Is there a way to find all possiblegenerators of a cyclic group?

TheoremLet G = a be a cyclic group of order n. Then

o(ak ) = ngcd(n, k)

for all integers k.

CorollaryLet G = a be a cyclic group of order n and k an integer. Then theelement ak generates G, if and only if n and k are relatively prime,i.e. gcd(n, k) = 1.

For the special case when G = Zn (which is a cyclic group of order n,in which additive notation is used) we obtain that k Zn generates acyclic subgroup of order n/ gcd(n, k), and that k = Zn if and only ifgcd(n, k) = 1.

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ExampleThe cyclic group Z12 has four generators, namely those elementsin Z12 that are relatively prime to 12, i.e. 1, 5, 7, and 11.The element 8 generates a cyclic subgroup of order

12

gcd(12, 8) =12

4= 3.

To be precise, we have 8 = {0, 4, 8}.ExampleThe group Z11 = {1, 2, 3, . . . , 10} of units modulo 11 turns out to becyclic, and 2 is one of its generators. We have |Z11| = 10, which meansthat any element in Z11 in the form 2k , where gcd(k , 10) = 1, i.e.23 = 8, 27 = 7, and 29 = 6 are the remaining generators of the group.The element 24 = 5 generates a cyclic subgroup of order

10

gcd(10, 4) =10

2= 5.

Actually, 5 = {1, 5, 52, 53, 54} = {1, 5, 3, 4, 9}.November ,

TheoremLet G be a cyclic group of order n. Then for each positive divisor dof n there is a uniquely determined subgroup in G of order d.

ExampleThe group Z100 is a cyclic group of order 100. According to thetheorem above, it contains exactly one subgroup each, of the orders1, 2, 4, 5, 10, 20, 25, 50, and 100, respectively, since these numbers arethe different positive divisors of 100.The uniquely determined subgroup of order 20 in Z100 is generatedby 5, since

o(5) = 100gcd(100, 5) =

100

5= 20.

ExerciseRegarding the previous example, which elements in 5 generates thissubgroup, besides 5? In other words, which elements in Z100 are oforder 20?

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Cyclic groups Definition and ExamplesProperties of cyclic groupsSubgroups of cyclic groups