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 2 Z. The Fundamental Theorem of Finite Abelian Groups Every finite abelian group is isomor- phic 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 2 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 jGj = p1 p2 ··· pn with the pi distinct k primes. The set Gi = fg 2 G : jgj = pi ; k 2 Zg 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 2 G and n must be prime (exercise). Thus, jaj = p. Suppose G contains a nontrivial proper subgroup H. Because G is abelian, then H is a normal subgroup and jGj = jHj · jG=Hj. Thus, p j jGj or p j jG=Hj. Suppose jHj = r, so 1 < r < n. If p j r, then p j jHj and so H contains an element h of order p by the inductive hypothesis and h 2 G. For the other case, p - r and so p divides jG=Hj. By the inductive hypothesis, G=H contains an element aH of order p. Then H = (aH)p = apH and so ap 2 H but a2 = H. We claim jarj = p. Because ap 2 H, then e = (ap)r = (ar)p. Thus, jarj j p so jarj = 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 2 H: r r This contradicts a2 = H. Thus, a 6= e so ja j = p. Lemma 3. A finite abelian group is a p-group if and only if its order is a power of p. Proof. If jGj is a power of p, then by Lagrange's theorem, so is every element of G. Conversely, if jGj is not a power of p, then there exists a prime q j jGj 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 2 G has maximal order. If h 2 Gnhgi has smallest possible order, then jhj = p. Proof. Let g 2 G be of maximal order in G, say jgj = pm for some m ≤ n. Since G is not cyclic, G 6= hgi. Choose h 2 Gnhgi where h has smallest possible order, say jhj = p`. Since e 2 hgi, then h 6= e and so ` > 0. But jhpj = p`−l (exercise) and so jhpj has smaller order than jhj, whence hp 2 hgi. That is, hp = gr for some r. By the above, `−1 `−1 ` (gr)p = (hp)p = hp = e: Then jgrj ≤ p`−1 < pm and so gr is not a generator for hgi. Write r = ps and define a = g−sh, so a2 = hgi. Then ap = g−sphp = g−rhp = h−php = e: Since a has minimal non-trivial order and a2 = hgi, then jhj = p. Lemma 5. Let G be a finite group, N a normal subgroup of G, and g 2 G an element of maximal order in G. If hgi \ N = feg, then jgNj = jgj and so gN is an element of maximal order in G=N. Proof. First note that if a 2 G has order m, then (aN)m = amN = N and so jaNj divides m. That is, the order of a coset aN is at most the order of a. Now if g 2 G has maximal order n in G, then jgNj ≤ jgj by the above. On the other hand, if jgNj = k, then N = (gN)k = gkN and so k 2 N. Thus, gk 2 N and so gk = e by hypothesis. That is, jgj divides k and so jgNj = jgj. 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 2 G has maximal order. Then G is the internal direct product of hgi and some subgroup K. n Proof. By Lemma 3, jGj = p for some n 2 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 2 G be of maximal order in G, say jgj = pm for some m ≤ n. Choose h2 = hgi where h has smallest possible order and set H = hhi. By Lemma 4, jHj = jhj = p. Thus, jG=Hj = jGj=jHj = 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 = feg. 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 2 G. Then bH = (gH) y = g Hy for some t 2 N and y 2 K. Then there exists h; h 2 H such that bh = gth0y and since H ⊂ K, b = gt(h0yh−1) 2 hgiK. Thus, G = hgiK. The group G is abelian so it is left only to prove that hgi\K = ;. Suppose there exists b 2 hgi\K. Then bH 2 hgHi \ K=H = H, so b 2 H. But this implies that b 2 hgi \ H = feg. 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 2 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 = fh1h2 ··· hn : hi 2 Hig (as sets), S (2) Hi \ h j6=i Hji = feg, and (3) hihj = hjhi for all hi 2 Hi, hj 2 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 2 G can be written uniquely as h1h2 ··· hn, hi 2 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 jGj = p1 p2 ··· pn with the pi distinct primes.