Cyclic Group Supplement

Theorem 1. Let g be an element of a group G and write

hgi = ngk : k 2 Zo.

Then hgi is a subgroup of G.

Proof. Since 1 = g0, 1 2 hgi. Suppose a, b 2 hgi. Then a = gk, b = gm and ab = gkgm = gk+m.

Hence ab 2 hgi (note that k + m 2 Z). Moreover, a−1 = (gk)−1 = g−k and −k 2 Z, so that

a−1 2 hgi. Thus, we have checked the three conditions necessary for hgi to be a subgroup of G.

Definition 2. If g 2 G, then the subgroup hgi = {gk : k 2 Z} is called the cyclic subgroup of G

generated by g, If G = hgi, then we say that G is a cyclic group and that g is a generator of

G.

Examples 3. 1. If G is any group then {1} = h1i is a cyclic subgroup of G.

2. The group G = {1, −1, i, −i} µ C¤ (the group operation is multiplication of complex numbers)

is cyclic with generator i. In fact hii = {i0 = 1, i1 = i, i2 = −1, i3 = −i} = G. Note that

−i is also a generator for G since h−ii = {(−i)0 = 1, (−i)1 = −i, (−i)2 = −1, (−i)3 = i} = G.

Thus a cyclic group may have more than one generator. However, not all elements of G need

be generators. For example h−1i = {1,−1} 6= G so −1 is not a generator of G.

3. The group G = Z¤

7 = the group of units of the ring Z7 is a cyclic group with generator 3.

Indeed,

h3i = {1 = 30, 3 = 31, 2 = 32, 6 = 33, 4 = 34, 5 = 35} = G.

Note that 5 is also a generator of G, but that h2i = {1, 2, 4} 6= G so that 2 is not a generator

of G.

4. G = h¼i = {¼k : k 2 Z} is a cyclic subgroup of R¤.

5. The group G = Z¤

8 is not cyclic. Indeed, since Z¤

8 = {1, 3, 5, 7} and h1i = {1}, h3i = {1, 3},

h5i = {1, 5}, h7i = {1, 7}, it follows that Z¤

8 6= hai for any a 2 Z¤

8.

If a group G is written additively, then the identity element is denoted 0, the inverse of a 2 G

is denoted −a, and the powers of a become na in additive notation. Thus, with this notation,

the cyclic subgroup of G generated by a is hai = {na : n 2 Z}, consisting of all the multiples of

a. Among groups that are normally written additively, the following are two examples of cyclic

groups.

6. The integers Z are a cyclic group. Indeed, Z = h1i since each integer k = k · 1 is a multiple

of 1, so k 2 h1i and h1i = Z. Also, Z = h−1i because k = (−k) · (−1) for each k 2 Z.

7. Zn is a cyclic group under addition with generator 1.

Theorem 4. Let g be an element of a group G. Then there are two possibilities for the cyclic

subgroup hgi.

Case 1: The cyclic subgroup hgi is finite. In this case, there exists a smallest positive integer

n such that gn = 1 and we have

(a) gk = 1 if and only if n|k.

(b) gk = gm if and only if k ´ m (mod n).

1

Cyclic Group Supplement

(c) hgi = {1, g, g2, . . . , gn−1} and the elements 1, g, g2, . . . , gn−1 are distinct.

Case 2: The cyclic subgroup hgi is infinite. Then

(d) gk = 1 if and only if k = 0.

(e) gk = gm if and only if k = m.

(f ) hgi = {. . . , g−3, g−2, g−1, 1, g, g2, g3, . . .} and all of these powers of g are distinct.

Proof. Case 1. Since hgi is finite, the powers g, g2, g3, . . . are not all distinct, so let gk = gm with

k < m. Then gm−k = 1 where m − k > 0. Hence there is a positive integer l with gl = 1. Hence

there is a smallest such positive integer. We let n be this smallest positive integer, i.e., n is the

smallest positive integer such that gn = 1.

(a) If n|k then k = qn for some q 2 n. Then gk = gqn = (gn)q = 1q = 1. Conversely, if gk = 1,

use the division algorithm to write k = qn + r with 0 · r < n. Then gr = gk(gn)−q = 1(1)−q = 1.

Since r < n, this contradicts the minimality of n unless r = 0. Hence r = 0 and k = qn so that n|k.

(b) gk = gm if and only if gk−m = 1. Now apply Part (a).

(c) Clearly, {1, g, g2, . . . , gn−1} µ hgi. To prove the other inclusion, let a 2 hgi. Then a = gk

for some k 2 Z. As in Part (a), use the division algorithm to write k = qn+r, where 0 · r · n−1.

Then

a = gk = gqn+r = (gn)qgr = 1qgr = gr 2 {1, g, g2, . . . , gn−1}

which shows that hgi µ {1, g, g2, . . . , gn−1}, and hence that

hgi = {1, g, g2, . . . , gn−1}.

Finally, suppose that gk = gm where 0 · k · m · n−1. Then gm−k = 1 and 0 · m−k < n. This

implies that m − k = 0 because n is the smallest positive power of g which equals 1. Hence all of

the elements 1, g, g2, . . . , gn−1 are distinct.

Case 2. (d) Certainly, gk = 1 if k = 0. If gk = 1, k 6= 0, then g−k = (gk)−1 = 1−1 = 1, also.

Hence gn = 1 for some n > 0, which implies that hgi is finite by the proof of Part (c), contrary to

our hypothesis in Case 2. Thus gk = 1 implies that k = 0.

(e) gk = gm if and only if gk−m = 1. Now apply Part (d).

(f) hgi = {gk : k 2 Z} by definition of hgi, so all that remains is to check that these powers are

distinct. But this is the content of Part (e).

Recall that if g is an element of a group G, then the order of g is the smallest positive integer

n such that gn = 1, and it is denoted o(g) = n. If there is no such positive integer, then we say

that g has infinite order, denoted o(g) = 1. By Theorem 4, the concept of order of an element

g and order of the cyclic subgroup generated by g are the same.

Corollary 5. If g is an element of a group G, then o(t) = |hgi|.

Proof. This is immediate from Theorem 4, Part (c).

If G is a cyclic group of order n, then it is easy to compute the order of all elements of G. This

is the content of the following result.

Theorem 6. Let G = hgi be a cyclic group of order n, and let 0 · k · n − 1. If m = gcd(k, n),

then o(gk) =

n

m

.

2

Cyclic Group Supplement

Proof. Let k = ms and n = mt. Then (gk)n/m = gkn/m = gmsn/m = (gn)s = 1s = 1. Hence n/m

divides o(gk) by Theorem 4 Part (a). Now suppose that (gk)r = 1. Then gkr = 1, so by Theorem

3 Part (a), n | kr. Hence

n

m¯¯¯

k

m

r

and since n/m and k/m are relatively prime, it follows that n/m divides r. Hence n/m is the

smallest power of gk which equals 1, so o(gk) = n/m.

Theorem 7. Let G = hgi be a cyclic group where o(g) = n. Then G = hgki if and only if

gcd(k, n) = 1.

Proof. By Theorem 6, if m = gcd(k, n), then o(gk) = n/m. But G = hgki if and only if o(gk) =

|G| = n and this happens if and only if m = 1, i.e., if and only if gcd(k, n) = 1.

Example 8. If G = hgi is a cyclic group of order 12, then the generators of G are the powers gk

where gcd(k, 12) = 1, that is g, g5, g7, and g11. In the particular case of the additive cyclic group

Z12, the generators are the integers 1, 5, 7, 11 (mod 12).

Now we ask what the subgroups of a cyclic group look like. The question is completely answered

by Theorem 10. Theorem 9 is a preliminary, but important, result.

Theorem 9. Every subgroup of a cyclic group is cyclic.

Proof. Suppose that G = hgi = {gk : k 2 Z} is a cyclic group and let H be a subgroup of G. If

H = {1}, then H is cyclic, so we assume that H 6= {1}, and let gk 2 H with gk 6= 1. Then, since H

is a subgroup, g−k = (gk)−1 2 H. Therefore, since k or −k is positive, H contains a positive power

of g, not equal to 1. So let m be the smallest positive integer such that gm 2 H. Then, certainly

all powers of gm are also in H, so we have hgmi µ H. We claim that this inclusion is an equality.

To see this, let gk be any element of H (recall that all elements of G, and hence H, are powers of

g since G is cyclic). By the division algorithm, we may write k = qm + r where 0 · r < m. But

gk = gqm+r = gqmgr = (gm)qgr so that

gr = (gm)−qgk 2 H.

Since m is the smallest positive integer with gm 2 H and 0 · r < m, it follows that we must have

r = 0. Then gk = (gm)q 2 hgmi. Hence we have shown that H µ hgmi and hence H = hgmi. That

is H is cyclic with generator gm where m is the smallest positive integer for which gm 2 H.

Theorem 10 (Fundamental Theorem of Finite Cyclic Groups). Let G = hgi be a cyclic

group of order n.

1. If H is any subgroup of G, then H = hgdi for some d|n.

2. If H is any subgroup of G with |H| = k, then k|n.

3. If k | n, then hgn/ki is the unique subgroup of G of order k.

Proof. 1. By Theorem 9, H is a cyclic group and since |G| = n < 1, it follows that H = hgmi

where m > 0. Let d = gcd(m, n). Since d | n it is sufficient to show that H = hgdi. But d|m

also, so m = qd. Then gm = (gd)q so gm 2 hgdi. Hence H = hgmi µ hgdi. But d = rm + sn,

where r, s 2 Z, so

gd = grm+sn = grmgsn = (gm)r(gn)s = (gm)r(1)s = (gm)r 2 hgmi = H.

This shows that hgdi µ H and hence hgdi = H.

3

Cyclic Group Supplement

2. By Part (a), H = hgdi where d | n. Then k = |H| = n/d so k|n.

3. Suppose that K is any subgroup of G of order k. By Part (a), let K = hgmi where m | n.

Then Theorem 6 gives k = |K| = |gm| = n/m. Hence m = n/k, so K = hgn/ki. This proves

(c).

Remark 11. Part (b) of Theorem 10 is actually true for any finite group G, whether or not it is

cyclic. This result is Lagrange’s Theorem (Theorem 5.2.3, Page 219 of your text).

The subgroups of a group G can be diagrammatically illustrated by listing the subgroups, and

indicating inclusion relations by means of a line directed upward from H to K if H is a subgroup

of K. Such a scheme is called the lattice diagram for the subgroups of the group G. We will

illustrate by determining the lattice diagram for all the subgroups of a cyclic group G = hgi of

order 12. Since the order of g is 12, Theorem 10 (c) shows that there is exactly one subgroup hgdi

for each divisor d of 12. The divisors of 12 are 1, 2, 3, 4, 6, 12. Then the unique subgroup of G of

each of these orders is, respectively,

{1} = hg12i, hg6i, hg4i, hg3i, hg2i, hgi = G.

Note that hgmi µ hgki if and only if k | m. Hence the lattice diagram of G is:

G

hg2i hg3i

hg4i hg6i

h1i

4