8.5 Cyclic groups

Let G be an abelian group. For a ∈ G, de¬ne a := {za : z ∈ Z}. It is

easy to see that a is a subgroup of G ”indeed, it is the image of the group

homomorphism discussed in Example 8.35. Moreover, a is the smallest

subgroup of G containing a; that is, a contains a, and any subgroup H

of G that contains a must also contain a . The subgroup a is called the

subgroup (of G) generated by a. Also, one de¬nes the order of a to be

the order of the subgroup a .

More generally, for a1 , . . . , ak ∈ G, we de¬ne a1 , . . . , ak := {z1 a1 + · · · +

zk ak : z1 , . . . , zk ∈ Z}. One also veri¬es that a1 , . . . , ak is a subgroup

of G, and indeed, is the smallest subgroup of G that contains a1 , . . . , ak .

The subgroup a1 , . . . , ak is called the subgroup (of G) generated by

a1 , . . . , ak .

An abelian group G is said to be cyclic if G = a for some a ∈ G, in

which case, a is called a generator for G. An abelian group G is said to

be ¬nitely generated if G = a1 , . . . , ak for some a1 , . . . , ak ∈ G.

Multiplicative notation: if G is written multiplicatively, then a := {az :

z ∈ Z}, and a1 , . . . , ak := {az1 · · · azk : z1 , . . . , zk ∈ Z}; also, for emphasis

1 k

and clarity, we use the term multiplicative order of a.

Classi¬cation of cyclic groups. We can very easily classify all cyclic

groups. Suppose that G is a cyclic group with generator a. Consider the

map ρ : Z ’ G that sends z ∈ Z to za ∈ G. As discussed in Example 8.35,

this map is a group homomorphism, and since a is a generator for G, it must

be surjective.

Case 1: ker(ρ) = {0}. In this case, ρ is an isomorphism of Z with G.

Case 2: ker(ρ) = {0}. In this case, since ker(ρ) is a subgroup of Z di¬erent

from {0}, by Theorem 8.8, it must be of the form nZ for some n > 0.

Hence, by Theorem 8.26, the map ρ : Zn ’ G that sends [z]n to za

¯

is an isomorphism of Zn with G.

So we see that a cyclic group is isomorphic either to the additive group Z

8.5 Cyclic groups 203

or the additive group Zn , for some positive integer n. We have thus classi¬ed

all cyclic groups “up to isomorphism.” From this classi¬cation, we obtain:

Theorem 8.29. Let G be an abelian group and let a ∈ G.

(i) If there exists a positive integer m such that ma = 0G , then the least

such positive integer n is the order of a; in this case, we have:

“ for any integer z, za = 0G if and only if n | z, and more

generally, for integers z1 , z2 , z1 a = z2 a if and only if z1 ≡

z2 (mod n);

“ the subgroup a consists of the n distinct elements

0 · a, 1 · a, . . . , (n ’ 1) · a.

(ii) If G has ¬nite order, then |G| · a = 0G and the order of a divides |G|.

Proof. Part (i) follows immediately from the above classi¬cation, along with

part (vi) of Theorem 8.20. Part (ii) follows from part (i), along with La-

grange™s theorem (Theorem 8.16), since a is a subgroup of G. 2

Example 8.45. The additive group Z is a cyclic group generated by 1. The

only other generator is ’1. More generally, the subgroup of Z generated by

m ∈ Z is mZ. 2

Example 8.46. The additive group Zn is a cyclic group generated by [1]n .

More generally, for m ∈ Z, the subgroup of Zn generated by [m]n is equal

to mZn , which by Example 8.21 has order n/ gcd(m, n). In particular, [m]n

generates Zn if and only if m is relatively prime to n, and hence, the number

of generators of Zn is φ(n). 2

Example 8.47. Consider the additive group G := Zn1 — Zn2 , and let ± :=

([1]n1 , [1]n2 ) ∈ Zn1 — Zn2 . For m ∈ Z, we have m± = 0G if and only if

n1 | m and n2 | m. This implies that ± generates a subgroup of G of order

lcm(n1 , n2 ).

Suppose that gcd(n1 , n2 ) = 1. From the above discussion, it follows that

G is cyclic of order n1 n2 . One could also see this directly using the Chinese

remainder theorem: as we saw in Example 8.42, the Chinese remainder

theorem gives us an isomorphism of G with the cyclic group Zn1 n2 .

Conversely, if d := gcd(n1 , n2 ) > 1, then all elements of Zn1 — Zn2 have

order dividing n1 n2 /d, and so Zn1 — Zn2 cannot be cyclic. 2

Example 8.48. For a, n ∈ Z with n > 0 and gcd(a, n) = 1, the de¬nition

in this section of the multiplicative order of ± := [a]n ∈ Z— is consistent

n

204 Abelian groups

with that given in §2.5, and is also the same as the multiplicative order of a

modulo n. Indeed, Euler™s theorem (Theorem 2.15) is just a special case of

part (ii) of Theorem 8.29. Also, ± is a generator for Z— if and only if a is a

n

primitive root modulo n. 2

Example 8.49. As we saw in Example 8.24, all elements of Z— have mul-

15

tiplicative order dividing 4, and since Z— has order 8, we conclude that Z—

15 15

is not cyclic. 2

Example 8.50. The group Z— is cyclic, with [2] being a generator:

5

[2]2 = [4] = [’1], [2]3 = [’2], [2]4 = [1]. 2

Example 8.51. Based on the calculations in Example 2.6, we may conclude

that Z— is cyclic, with both [3] and [5] being generators. 2

7

The following two theorems completely characterize the subgroup struc-

ture of cyclic groups. Actually, we have already proven the results in these

two theorems, but nevertheless, these results deserve special emphasis.

Theorem 8.30. Let G be a cyclic group of in¬nite order.

(i) G is isomorphic to Z.

(ii) The subgroups of G are in one-to-one correspondence with the non-

negative integers, where each such integer m corresponds to the cyclic

group mG.

(iii) For any two non-negative integers m, m , mG ⊆ m G if and only if

m | m.

Proof. That G ∼ Z was established in our classi¬cation of cyclic groups, it

=

su¬ces to prove the other statements of the theorem for G = Z. It is clear

that for any integer m, the subgroup mZ is cyclic, as m is a generator. This

fact, together with Theorem 8.8, establish all the other statements. 2

Theorem 8.31. Let G be a cyclic group of ¬nite order n.

(i) G is isomorphic to Zn .

(ii) The subgroups of G are in one-to-one correspondence with the positive

divisors of n, where each such divisor d corresponds to the subgroup

dG; moreover, dG is a cyclic group of order n/d.

(iii) For each positive divisor d of n, we have dG = G{n/d}; that is, the

kernel of the (n/d)-multiplication map is equal to the image of the

d-multiplication map; in particular, G{n/d} has order n/d.

8.5 Cyclic groups 205

(iv) For any two positive divisors d, d of n, we have dG ⊆ d G if and only

if d | d.

(v) For any positive divisor d of n, the number of elements of order d in

G is φ(d).

(vi) For any integer m, we have mG = dG and G{m} = G{d}, where

d := gcd(m, n).

Proof. That G ∼ Zn was established in our classi¬cation of cyclic groups,

=

and so it su¬ces to prove the other statements of the theorem for G = Zn .

The one-to-one correspondence in part (ii) was established in Theorem 8.9.

The fact that dZn is cyclic of order n/d can be seen in a number of ways;

indeed, in Example 8.44 we constructed an isomorphism of Zn/d with dZn .

Part (iii) was established in Example 8.21.

Part (iv) was established in Theorem 8.9.

For part (v), the elements of order d in Zn are all contained in Zn {d},

and so the number of such elements is equal to the number of generators of

Zn {d}. The group Zn {d} is cyclic of order d, and so is isomorphic to Zd ,

and as we saw in Example 8.46, this group has φ(d) generators.

Part (vi) was established in Example 8.21. 2

Since cyclic groups are in some sense the simplest kind of abelian group,

it is nice to have some su¬cient conditions under which a group must be

cyclic. The following theorems provide such conditions.