The Structure of Finite Abelian Groups

A First Course in Number Theory

4.1 The Structure of Finite Abelian Groups

This chapter introduces analysis on finite abelian groups and their char- acters. We begin by using elementary number theory to determine the structure of finite abelian groups.

Let G be an abelian group, written additively, and let A1, . . . , Ak be subsets ofG. Thesum of these sets is the set

A1+· · ·+Ak={a1+· · ·+ak :ai∈Ai fori= 1, . . . , k}.

If G₁, . . . , G_k are subgroups of G, then the sumset G₁ +· · · +G_k is a subgroup ofG(Exercise 2). We say thatGis thedirect sumof the subgroups G₁, . . . , G_k, writtenG=G₁⊕· · ·⊕Gk, if every elementg∈Gcan be written uniquely in the form g =g1+· · ·+gk, where gi ∈Gi fori = 1, . . . , k. If G=G1⊕ · · · ⊕Gk, then |G|=|G1| · · · |Gk|(Exercise 3).

The order of an element g in an additive group is the smallest positive integer dsuch thatdg= 0. By Theorem 2.16, the order of an element of a finite group divides the order of the group.

Letpbe a prime number. Ap-groupis a group each of whose elements has an order that is a power of p. For every prime number p, let G(p) denote the set of all elements ofGwhose order is a power ofp. ThenG(p) is a subgroup of the abelian group G(Exercise 6).

Theorem 4.1 Let Gbe a finite abelian group, written additively, and let

|G| = m. For every prime number p, let G(p) be the set of all elements

g∈Gwhose order is a power of p. Then

G=

p|m

G(p).

Proof. Let m = k

i=1p^r_iⁱ be the standard factorization of m, and let mi = mp⁻_i^ri for i = 1, . . . , k. Then (m1, . . . , mk) = 1 by Exercise 15 in Section 1.4, and so there exist integersu1, . . . , uk such that

m1u1+· · ·+mkuk= 1.

Let g ∈ G, and define g_i = m_iu_ig ∈ G for i = 1, . . . , k. Since p^r_iⁱg_i = mu_ig= 0, it follows thatg_i∈G(p). Moreover,

g = (m₁u₁+· · ·+m_ku_k)g=m₁u₁g+· · ·+m_ku_kg

= g1+· · ·+gk ∈G(p1) +· · ·+G(pk), and so

G=G(p1) +· · ·+G(pk).

Suppose that

g1+· · ·+gk = 0,

wheregi∈G(pi) fori= 1, . . . , k. There exist nonnegative integersr1, . . . , rk

such thatg_i has orderp^r_iⁱ fori= 1, . . . , k. Let dj =

k

i=1 i=j

p^r_iⁱ.

If gj = 0, then djgj = 0. Since djgi = 0 for i= 1, . . . , k, i=j, it follows that

0 =d_j(g₁+· · ·+g_k) =d_jg_j,

and so g_j= 0 for allj= 1, . . . , k. Thus, 0 has no nontrivial representation inG=G(p₁) +· · ·+G(p_k). By Exercise 4, we conclude thatGis the direct sum of the subgroupsG(p_i).2

Lemma 4.1 Let Gbe a finite abelian p-group. Let g₁ ∈G be an element of maximum order p^r1, and let G₁=g1 be the cyclic subgroup generated by g1. Consider the quotient group G/G1. Let h ∈G. If h+G1 ∈ G/G1

has orderp^r, then there exists an elementg∈Gsuch thatg+G1=h+G1

andg has orderp^r inG.

Proof. If h+G1 has order p^r in G/G1, then the order of h in G is at most p^r1 (since p^r1 is the maximum order in G) and at least p^r (by

4.1 The Structure of Finite Abelian Groups 123 Exercise 7). SinceG1=p^r(h+G1) =p^rh+G1, it follows that p^rh∈G1, and sop^rh=ug1for some positive integeru≤p^r1(sinceg1has orderp^r1).

Writeu=p^sv, where (p, v) = 1 and 0 ≤s≤r1. Then vg1 also has order p^r1,and sop^svg1 has orderp^r1−s inG. Thenp^rh=p^svg1 has orderp^r1−s in G, and sohhas orderp^r1+r−s≤p^r1. It follows thatr≤s, and

p^rh=p^svg1=p^r(p^s−rvg1) =p^rg₁, where

g₁=p^s−rvg1∈G1. Let

g=h−g₁. Then

g+G1=h+G1.

Moreover, p^rg =p^rh−p^rg₁ = 0, and so the order of g is at most p^r. On the other hand, g+G1 has order p^r in the quotient groupG/G1, and so the order ofg is at leastp^r. Therefore,g has orderp^r.2

Theorem 4.2 Every finite abelianp-group is a direct sum of cyclic groups.

Proof. The proof is by induction on the cardinality of G. Let Gbe a finite abelian p-group. If G is cyclic, we are done. If G is not cyclic, let g₁ ∈ G be an element of maximum order p^r1, and let G₁ be the cyclic subgroup generated by g₁. The quotient group G/G₁ is a finite abelian p-group, and

1<|G/G1|= |G|

p^r1 <|G|.

Therefore, the induction hypothesis holds forG/G₁, and so G/G₁=H₂⊕ · · · ⊕H_k,

whereHi is a cyclic subgroup ofG/G1of order p^ri fori= 2, . . . , k. More- over,

|G|

p^r1 =|G/G1|= k i=2

p^ri.

By Lemma 4.1, for each i = 2, . . . , k there exists an elementgi ∈G such that gi+G1 generates Hi and gi has orderp^ri in G. LetGi be the cyclic subgroup of Ggenerated bygi. Then|Gi|=p^ri for i= 1, . . . , k. We shall prove thatG=G1⊕ · · · ⊕Gk.

We begin by showing that G=G1+· · ·+Gk. Ifg ∈G, theng+G1∈ G/G1, and there exist integersu2, . . . , uk such that

0≤ui≤p^ri−1 fori= 2, . . . , k

and

g+G1=u2(g2+G1)⊕ · · · ⊕uk(gk+G1) = (u2g2+· · ·+ukgk) +G1. It follows that

g−(u₂g₂+· · ·+u_kg_k) =u₁g₁∈G₁ for some integeru1 such that

0≤u1≤p^r1−1, and so

g=u1g1+u2g2+· · ·+ukgk ∈G1+· · ·+Gk. Therefore,G=G₁+· · ·+G_k. Since

|G|=|G1+· · ·+Gk| ≤ |G1| · · · |Gk|= k i=1

p^ri =|G|,

it follows that every element ofGhas a unique representation as an element in the sumsetG1+· · ·+Gk, and soG=G1⊕+· · ·+⊕Gk. This completes the proof.2

Theorem 4.3 Every finite abelian group is a direct sum of cyclic groups.

Proof. This follows immediately from Theorem 4.1 and Theorem 4.2.

2

LetG1, . . . , Gkbe abelian groups, written additively. Theirdirect product is the group

G1× · · · ×Gk={(g1, . . . , gk) :gi∈Gi fori= 1, . . . , k}, with addition defined by

(g₁, . . . , g_k) + (g₁, . . . , g_k) = (g₁+g₁, . . . , g_k+g_k).

IfG₁, . . . , G_k are subgroups of an abelian groupGand ifG=G₁⊕· · ·⊕Gk, thenG∼=G1× · · · ×Gk (Exercise 5).

LetG1, . . . , Gk be abelian groups, written multiplicatively. Theirdirect product is the group G1× · · · ×Gk consisting of all k-tuples (g1, . . . , gk) withgi∈Gi fori= 1, . . . , kand multiplication defined coordinate-wise by (g1, . . . , gk)(g₁, . . . , g_k) = (g1g₁, . . . , gkg_k).

4.1 The Structure of Finite Abelian Groups 125

Exercises

1. Let G=Z/12Zbe the additive group of congruence classes modulo 12. ComputeG(2) andG(3) and show explicitly that G(2)∼=Z/4Z, G(3)∼=Z/3Z, and

Z/12Z∼=Z/4Z⊕Z/3Z.

2. LetGbe an abelian group, written additively, and letG1, . . . , Gk be subgroups ofG. Prove thatG1+· · ·+Gk is a subgroup of G.

3. Let G be an abelian group, written additively, and let G1, . . . , Gk

be subgroups ofG such thatG= G1+· · ·+Gk. Prove that |G| ≤

|G1| · · · |Gk|. Prove that G = G1⊕ · · · ⊕Gk if and only if |G| =

|G1| · · · |Gk|.

4. Let G be an abelian group, written additively, and let G1, . . . , Gk

be subgroups of G such that G = G1+· · ·+Gk. Prove that G = G1⊕ · · · ⊕Gk if and only if the only representation of 0 in the form 0 =g1+· · ·+gk withgi∈Gi isg1=· · ·=gk= 0.

5. Let G1, . . . , Gk be subgroups of an abelian group G such thatG = G1⊕ · · · ⊕Gk. Prove thatG∼=G1× · · · ×Gk.

6. Let G be an additive abelian group. For every prime numberp, let G(p) denote the set of all elements ofGwhose order is a power ofp.

Prove thatG(p) is a subgroup ofG.

7. Letf :G→H be a group homomorphism, and letg∈G. Prove that the order off(g) inH divides the order ofg inG. Prove that ifGis ap-group and f is surjective, thenH is ap-group.

8. Let G be a finite abelianp-group. If r1, . . . , rk are positive integers with r1 ≥ · · · ≥ rk, then we say that G is of type (p^r1, . . . , p^rk) if G ∼= G1⊕ · · · ⊕Gk, where Gi is a cyclic group of order p^ri for i= 1, . . . , k. We shall prove that every finite abelian p-group has a unique type.

LetpG={pg:g∈G}.

(a) Prove thatpGis a subgroup of G.

(b) Prove that ifGis of type (p^r1, . . . , p^rk) withrj ≥2 andrj+1=

· · ·rk = 1, thenpGis of type (p^r1−1, . . . , p^rj−1).

(c) Prove that

|G|=p^k|pG|.

(d) Prove that ifGis of type (p^r1, . . . , p^rk) and also of type (p^s1, . . . , p^s), thenk=.

(e) Prove that if the finite abelianp-groupGis of type (p^r1, . . . , p^rk) and of type (p^s1, . . . , p^sk), thenri=si fori= 1, . . . , k.

Hint: Use induction on the cardinality of G. Let j and be the greatest integers such that r_j ≥2 and s≥2, respectively.

Apply the induction hypothesis to pG to show that j = and ri=si fori= 1, . . . , j.