Proof. Let |G| = p. Let a ∈ G with a = 0G , and let k be the order of a. As

the order of an element divides the order of the group, we have k | p, and

so k = 1 or k = p. Since a = 0G , we must have k = 1, and so k = p, which

implies that a generates G. 2

Theorem 8.33. If G1 and G2 are ¬nite cyclic groups of relatively prime

order, then G1 — G2 is also cyclic.

Proof. This follows from Example 8.47, together with our classi¬cation of

cyclic groups. 2

Theorem 8.34. Any subgroup of a cyclic group is cyclic.

Proof. This is just a restatement of part (ii) of Theorem 8.30 and part (ii)

of Theorem 8.31 2

Theorem 8.35. If ρ : G ’ G is a group homomorphism, and G is cyclic,

then img(G) is cyclic.

206 Abelian groups

Proof. If G is generated by a, then it is easy to see that the image of ρ is

generated by ρ(a). 2

The next three theorems are often useful in calculating the order of a

group element.

Theorem 8.36. Let G be an abelian group, let a ∈ G be of ¬nite order n,

and let m be an arbitrary integer. Then the order of ma is n/ gcd(m, n).

Proof. By our classi¬cation of cyclic groups, we know that the subgroup a

is isomorphic to Zn , where under this isomorphism, a corresponds to [1]n and

ma corresponds to [m]n . The theorem then follows from the observations in

Example 8.46. 2

Theorem 8.37. Suppose that a is an element of an abelian group, and for

some prime p and integer e ≥ 1, we have pe a = 0G and pe’1 a = 0G . Then

a has order pe .

Proof. If m is the order of a, then since pe a = 0G , we have m | pe . So

m = pf for some f = 0, . . . , e. If f < e, then pe’1 a = 0G , contradicting the

assumption that pe’1 a = 0G . 2

Theorem 8.38. Suppose G is an abelian group with a1 , a2 ∈ G such that

a1 is of ¬nite order n1 , a2 is of ¬nite order n2 , and gcd(n1 , n2 ) = 1. Then

the order of a1 + a2 is n1 n2 .

Proof. Let m be the order of a1 + a2 . It is clear that n1 n2 (a1 + a2 ) = 0G ,

and hence m divides n1 n2 .

We claim that a1 © a2 = {0G }. To see this, suppose a ∈ a1 © a2 .

Then since a ∈ a1 , the order of a must divide n1 . Likewise, since a ∈ a2 ,

the order of a must divide n2 . From the assumption that gcd(n1 , n2 ) = 1,

it follows that the order of a must be 1, meaning that a = 0G .

Since m(a1 + a2 ) = 0G , it follows that ma1 = ’ma2 . This implies that

ma1 belongs to a2 , and since ma1 trivially belongs to a1 , we see that

ma1 belongs to a1 © a2 . From the above claim, it follows that ma1 = 0G ,

and hence n1 divides m. By a symmetric argument, we see that n2 divides

m. Again, since gcd(n1 , n2 ) = 1, we see that n1 n2 divides m. 2

For an abelian group G, we say that an integer k kills G if kG = {0G }.

Consider the set KG of integers that kill G. Evidently, KG is a subgroup of

Z, and hence of the form mZ for a uniquely determined non-negative integer

m. This integer m is called the exponent of G. If m = 0, then we see that

m is the least positive integer that kills G.

We ¬rst state some basic properties.

8.5 Cyclic groups 207

Theorem 8.39. Let G be an abelian group of exponent m.

(i) For any integer k such that kG = {0G }, we have m | k.

(ii) If G has ¬nite order, then m divides |G|.

(iii) If m = 0, then for any a ∈ G, the order of a is ¬nite, and the order

of a divides m.

(iv) If G is cyclic, then the exponent of G is 0 if G is in¬nite, and is |G|

is G is ¬nite.

Proof. Exercise. 2

The next two theorems develop some crucial properties about the struc-

ture of ¬nite abelian groups.

Theorem 8.40. If a ¬nite abelian group G has exponent m, then G contains

an element of order m. In particular, a ¬nite abelian group is cyclic if and

only if its order equals its exponent.

Proof. The second statement follows immediately from the ¬rst. For the

¬rst statement, assume that m > 1, and let m = r pei be the prime

i=1 i

factorization of m.

First, we claim that for each i = 1, . . . , r, there exists ai ∈ G such that

(m/pi )ai = 0G . Suppose the claim were false: then for some i, (m/pi )a = 0G

for all a ∈ G; however, this contradicts the minimality property in the

de¬nition of the exponent m. That proves the claim.

Let a1 , . . . , ar be as in the above claim. Then by Theorem 8.37, (m/pei )ai

i

ei

has order pi for each i = 1, . . . , r. Finally, by Theorem 8.38, the group

element

(m/pe1 )a1 + · · · + (m/per )ar

r

1

has order m. 2

Theorem 8.41. Let G be a ¬nite abelian group of order n. If p is a prime

dividing n, then G contains an element of order p.

Proof. We can prove this by induction on n.

If n = 1, then the theorem is vacuously true.

Now assume n > 1 and that the theorem holds for all groups of order

strictly less than n. Let a be any non-zero element of G, and let m be the

order of a. Since a is non-zero, we must have m > 1. If p | m, then (m/p)a is

an element of order p, and we are done. So assume that p m and consider

the quotient group G/H, where H is the subgroup of G generated by a.

Since H has order m, G/H has order n/m, which is strictly less than n,

208 Abelian groups

and since p m, we must have p | (n/m). So we can apply the induction

hypothesis to the group G/H and the prime p, which says that there is an

element b ∈ G such that b + H ∈ G/H has order p. If is the order of b,

then b = 0G , and so b ≡ 0G (mod H), which implies that the order of

b + H divides . Thus, p | , and so ( /p)b is an element of G of order p. 2

As a corollary, we have:

Theorem 8.42. Let G be a ¬nite abelian group. Then the primes dividing

the exponent of G are the same as the primes dividing its order.

Proof. Since the exponent divides the order, any prime dividing the exponent

must divide the order. Conversely, if a prime p divides the order, then since

there is an element of order p in the group, the exponent must be divisible

by p. 2

Exercise 8.23. Let G be an abelian group of order n, and let m be an

integer. Show that mG = G if and only if gcd(m, n) = 1.

Exercise 8.24. Let G be an abelian group of order mm , where

gcd(m, m ) = 1. Consider the map ρ : mG — m G to G that sends (a, b)

to a + b. Show that ρ is a group isomorphism.

Exercise 8.25. Let G be an abelian group, a ∈ G, and m ∈ Z, such that

m > 0 and ma = 0G . Let m = pe1 · · · per be the prime factorization of m.

r

1

For i = 1, . . . , r, let fi be the largest non-negative integer such that fi ¤ ei

and m/pfi · a = 0G . Show that the order of a is equal to pe1 ’f1 · · · per ’fr .

r

1

i

Exercise 8.26. Show that for ¬nite abelian groups G1 , G2 whose exponents

are m1 and m2 , the exponent of G1 — G2 is lcm(m1 , m2 ).

Exercise 8.27. Give an example of an abelian group G whose exponent is

zero, but where every element of G has ¬nite order.

Exercise 8.28. Show how Theorem 2.11 easily follows from Theorem 8.31.

8.6 The structure of ¬nite abelian groups (—)

We next state a theorem that classi¬es all ¬nite abelian groups up to iso-

morphism.

Theorem 8.43 (Fundamental theorem of ¬nite abelian groups). A

¬nite abelian group (with more than one element) is isomorphic to a direct