Finite Abelian Groups

0. Introduction

Though it might be bad form, we’ll start by stating the big theorem that we want to prove. We’ll then work on the proof throughout the next few sections. Ultimately, this is a generalization of the ∼ fact that Zmn = Zm × Zn if and only if m, n ∈ Z. The Fundamental Theorem of Finite Abelian Groups Every ﬁnite abelian group is isomorphic to a direct product of cyclic groups of prime power order.

The Fundamental Theorem implies that every ﬁnite abelian group can be written (up to isomor- phism) in the form

α1 × α2 × · · · × αn , Zp1 Zp2 Zpn with pi prime (not necessarily distinct) and αi ∈ N.

Example. Every ﬁnite abelian group of order 540 = 22 · 33 · 5 is isomorphic to exactly one of the following:

(1) Z2 × Z2 × Z3 × Z3 × Z3 × Z5 (4) Z4 × Z3 × Z3 × Z3 × Z5 (2) Z2 × Z2 × Z3 × Z9 × Z5 (5) Z4 × Z3 × Z9 × Z5 (3) Z2 × Z2 × Z27 × Z5 (6) Z4 × Z27 × Z5

These notes are derived primarily from Abstract Algebra, Theory and Applications by Thomas Judson (16ed). Most of this material is drawn from Chapter 13. Last Updated: January 24, 2020 1 1. p-groups

Our goal will be to take an arbitrary ﬁnite abelian group and decompose it in a manner according to the fundamental theorem. This requires ﬁrst building up the theory of p-groups.

Deﬁnition 1. Let p be a prime. A group G is a p-group if every element in G has order a power of p.

Deﬁnition 2. The groups Z2 Z4 and Z2 ×Z2 are all 2-groups, as is D4. The group Z27 is a 3-group.

By Lagrange’s Theorem, every group of order pn, p a prime, is automatically a p-group since the order of every element must divide pn. We will prove a converse to this for ﬁnite abelian groups. The proof of the next lemma is an easy exercise.

α1 α2 αn Lemma 1. Let G be a ﬁnite abelian group and write |G| = p1 p2 ··· pn with the pi distinct k primes. The set Gi = {g ∈ G : |g| = pi , k ∈ Z} is a subgroup of G. Lemma 2. Let G be a ﬁnite abelian group of order n. If p is a prime dividing n, then G contains an element of order p

Proof. The proof is by strong induction. The case n = 1 is trivial so assume the lemma holds for all groups of order k, 1 ≤ k ≤ n, and suppose p is a prime dividing n. If G has no proper nontrivial subgroups, then G = hai for any e 6= a ∈ G and n must be prime (exercise). Thus, |a| = p.

Suppose G contains a nontrivial proper subgroup H. Because G is abelian, then H is a normal subgroup and |G| = |H| · |G/H|. Thus, p | |G| or p | |G/H|. Suppose |H| = r, so 1 < r < n.

If p | r, then p | |H| and so H contains an element h of order p by the inductive hypothesis and h ∈ G. For the other case, p - r and so p divides |G/H|. By the inductive hypothesis, G/H contains an element aH of order p. Then H = (aH)p = apH and so ap ∈ H but a∈ / H. We claim |ar| = p.

Because ap ∈ H, then e = (ap)r = (ar)p. Thus, |ar| | p so |ar| = 1 or p. Therefore it suﬃces to show that ar 6= e. Since p and r are relatively prime, there exist integers r, s such that sp + tr = 1. If ar = e, then

a = asp+tr = aspatr = (ap)s(ar)t = (ap)s ∈ H.

r r This contradicts a∈ / H. Thus, a 6= e so |a | = p. Lemma 3. A ﬁnite abelian group is a p-group if and only if its order is a power of p.

Proof. If |G| is a power of p, then by Lagrange’s theorem, so is every element of G. Conversely, if |G| is not a power of p, then there exists a prime q | |G| and by the previous lemma, G contains an element of order q, a contradiction.

One immediate consequence of the above lemma is that each Gi is a p-group. 2 2. Proof of the Fundamental Theorem (Part I)

In this section, we prove the Fundamental Theorem for ﬁnite p-groups. The proof will conclude in the next section wherein we decompose a ﬁnite abelian group into a direct product of p-groups.

We begin with a technical result that will help in the proof of the ﬁrst proposition.

Lemma 4. Let G be a ﬁnite abelian p-group that is not cyclic. Suppose that g ∈ G has maximal order. If h ∈ G\hgi has smallest possible order, then |h| = p.

Proof. Let g ∈ G be of maximal order in G, say |g| = pm for some m ≤ n. Since G is not cyclic, G 6= hgi. Choose h ∈ G\hgi where h has smallest possible order, say |h| = p`. Since e ∈ hgi, then h 6= e and so ` > 0. But |hp| = p`−l (exercise) and so |hp| has smaller order than |h|, whence hp ∈ hgi. That is, hp = gr for some r. By the above,

`−1 `−1 ` (gr)p = (hp)p = hp = e.

Then |gr| ≤ p`−1 < pm and so gr is not a generator for hgi. Write r = ps and deﬁne a = g−sh, so a∈ / hgi. Then ap = g−sphp = g−rhp = h−php = e.

Since a has minimal non-trivial order and a∈ / hgi, then |h| = p.

Lemma 5. Let G be a ﬁnite group, N a normal subgroup of G, and g ∈ G an element of maximal order in G. If hgi ∩ N = {e}, then |gN| = |g| and so gN is an element of maximal order in G/N.

Proof. First note that if a ∈ G has order m, then (aN)m = amN = N and so |aN| divides m. That is, the order of a coset aN is at most the order of a.

Now if g ∈ G has maximal order n in G, then |gN| ≤ |g| by the above. On the other hand, if |gN| = k, then N = (gN)k = gkN and so k ∈ N. Thus, gk ∈ N and so gk = e by hypothesis. That is, |g| divides k and so |gN| = |g|. By the above argument, no element of G/N has order greater than the maximal order in G and so the conclusion follows.

Proposition 6. Let G be a ﬁnite abelian p-group and suppose that g ∈ G has maximal order. Then G is the internal direct product of hgi and some subgroup K.

n Proof. By Lemma 3, |G| = p for some n ∈ N. The case n = 1 (or n = 0) implies G is cyclic in which case this result is trivial. Assume now that n > 1 and that G is not cyclic. We inductively assume that the lemma holds for all k such that 1 ≤ k < n.

Let g ∈ G be of maximal order in G, say |g| = pm for some m ≤ n. Choose h∈ / hgi where h has smallest possible order and set H = hhi. By Lemma 4, |H| = |h| = p. Thus, |G/H| = |G|/|H| = pn/p = pn−1 and so we can apply the inductive hypothesis to G/H. 3 Since H has no nontrivial subgroups, then hgi ∩ H = {e}. By Lemma 5, gH is an element of G/H of maximal order. By the inductive hypothesis, G/H is the internal direct product of hgHi and some subgroup K0 of G/H. The Correspondence Theorem now implies that K0 = K/H for some subgroup K of G containing H. We claim that G is the internal direct product of hgi and K.

t t 0 Let b ∈ G. Then bH = (gH) y = g Hy for some t ∈ N and y ∈ K. Then there exists h, h ∈ H such that bh = gth0y and since H ⊂ K, b = gt(h0yh−1) ∈ hgiK. Thus, G = hgiK.

The group G is abelian so it is left only to prove that hgi∩K = ∅. Suppose there exists b ∈ hgi∩K. Then bH ∈ hgHi ∩ K/H = H, so b ∈ H. But this implies that b ∈ hgi ∩ H = {e}. It follows that G is the internal direct product of hgi and K.

Corollary 7. Let G be a ﬁnite abelian p-group and suppose that g ∈ G has maximal order. Then G is isomorphic to hgi × K for some subgroup K.

4 3. Proof of the Fundamental Theorem (Part II)

Thus, the following deﬁnition is just a generalization of our previous one, as is the subsequent proposition whose proof is left as an exercise.

Deﬁnition 3. A group G is the internal direct product of subgroups H1,H2,...,Hn provided

(1) G = H1H2 ··· Hn = {h1h2 ··· hn : hi ∈ Hi} (as sets), S (2) Hi ∩ h j6=i Hji = {e}, and (3) hihj = hjhi for all hi ∈ Hi, hj ∈ Hj, i 6= j.

Proposition 8. If a group G is the internal direct product of subgroups H1,H2,...,Hn, then ∼ G = H1 × H2 × · · · × Hn and each g ∈ G can be written uniquely as h1h2 ··· hn, hi ∈ Hi.

Lemma 9. Let G be a ﬁnite abelian group. Then G is the internal direct product of p-groups.

α1 α2 αn Proof. Write |G| = p1 p2 ··· pn with the pi distinct primes. By Lemma 1, the sets Gi are subgroup of G. We will show that G is an internal direct product of the Gi.

β1 β2 βn If g ∈ G, then by Lagrange’s Theorem, |g| = p1 p2 ··· pn where 0 ≤ βi ≤ αi for all i. Let β ai = |g|/p i . The ai are relatively prime so there exist bi such that a1b1 + a2b2 + ··· + anbn = 1. Then g = ga1b1+a2b2+···+anbn = ga1b1 ga2b2 ··· ganbn . For each i, we have pβi gaibi = gbi|g| = e.

a b Thus, g i i ∈ Gi for all i. It follows that G = G1G2 ··· Gn. S k S Let h ∈ Gi ∩ h j6=i Gji. Because h ∈ Gi, h = pi for some k ∈ Z. But if h ∈ h j6=i Gji, then h is the product of elements whose orders are not divisible by pi. Hence, |h| is not divisible by pi (because the Gi are all abelian). Thus, k = 0, so h = e.

Theorem 10 (The Fundamental Theorem of Finite Abelian Groups). Every ﬁnite abelian group is isomorphic to a direct product of cyclic groups of prime power order.

Proof. Let G be a ﬁnite abelian group. By Lemma 9, G is the internal (and hence the external) direct product of the Gi, which are all p-groups by Lemma 3. Applying Corollary 7, inductively, we get that each of the Gi decompose (isomorphically) into the direct product of cyclic groups. The result follows.

5 4. Finitely generated abelian groups

Deﬁnition 4. Let G be a group and X ⊂ G. The smallest subgroup containing X is the subgroup generated by X, denoted hXi. If hXi = G, then G is said to be generated by X. If in addition |X| < ∞, then G is said to be ﬁnitely generated.

Example. (1) If G is generated by X and |X| = 1, then G is cyclic.

(2) Dn is generated by X = {r, s}. (3) Finite groups are ﬁnitely generated, just take X = G. n o (4) is not ﬁnitely generated. To see this, let X = p1 , p2 ,..., pn be a generating set for Q q1 q2 qn G with all fractions written in lowers terms. Choose p ∈ Z such that p - qi for all i. Then p∈ / hXi.

Proposition 11. If X is a set of generators for G, then every element in G can be written as a product of (powers of) the elements of X.

Theorem 12 (The Fundamental Theorem of Finitely Generated Abelian Groups). Every ﬁnite abelian group is isomorphic to a direct product of cyclic groups of the form

α1 × α2 × · · · × αn × × · · · × , Zp1 Zp2 Zpn Z Z with pi prime (not necessarily distinct) and αi ∈ N.

We won’t prove this theorem here.