Math 371 Lecture #37 §8.2 (Ed.2), 9.2 (Ed.3): Finite Abelian Groups

The customary notation for abelian groups is the additive notation: a + b for the binary operation, 0 for the identity of the addition, and ka for a added to itself k times. We will use the results of direct products here, written in the additive notation: M +N = {m + n : m ∈ M, n ∈ N}, direct sum M ⊕ N, and M is a direct summand M ⊕ N. The classification of finite abelian groups begins with identifying certain subgroups. Definition. For an abelian group G and a positive prime p, the set G(p) denotes the set of elements of G whose order is some power of p:

G(p) = {a ∈ G : |a| = pn for some n ≥ 0}.

Lemma. For an abelian group G, the set G(p) is a subgroup of G. Proof. For a, b ∈ G(p), we have n, m ≥ 0 such that pna = 0, pmb = 0, so that pm(−b) = −pmb = −0 = 0, whence a − b satisfies

pnpm(a − b) = pm(pna) − pn(pmb) = 0 − 0 = 0.

Thus a − b ∈ G, and hence G is a subgroup.  Lemma 9.4. Let G be an abelian group and a ∈ G an element of finite order. Then a = a1 + a2 + ··· + at for ai ∈ G(pi), i = 1, 2, . . . , t where p1, p2, . . . , pt are the distinct positive primes that divide |a|. The proof is in the Appendix of this Lecture Note. The proof uses induction on the number of primes that divide |a|.

Theorem 9.5. If G is a finite abelian group, then G = G(p1) ⊕ G(p2) ⊕ · · · ⊕ G(pt) where p1, p2, . . . , pt are the distinct positive primes divisors of |G|. The proof is in the Appendix. It uses Lemma 9.4 to establish the existence of a = a1 + a2 + ··· + at with ai ∈ G(pi), i = 1, 2, . . . , t, and then shows that this is unique. Definitions. For a positive prime p, a group in which every element has an order that is a power of p is called a p-group. The group G(p) is a p-group. An element a of a p-group B is called an element of maximal order if |b| ≤ |a| for all b ∈ B. Elements of maximal order always exist in a finite p-group. Lemma. If a is an element of maximal order pn in a p-group B, then pnb = 0 for all b ∈ B. Proof. If |a| = pn is an element of maximal order in a p-group B, then each b ∈ B has order |b| = pj for some 0 ≤ j ≤ n. n j n−j n n−j j n−j Since p = p p , then p b = p p b = p 0 = 0.  The next step in classifying finite abelian groups is the show that a finite abelian p-group has a cyclic direct summand. Lemma 9.6. Let G be a finite abelian p-group and a ∈ G an element of maximal order. Then there is a subgroup K of G such that G = hai ⊕ K. The long, convoluted proof of this is in the Appendix. In it we find a subgroup K such that G = hai + K and hai ∩ K = {0}, and then use Theorem 9.3 (Ed.3). The Fundamental Theorem of Finite Abelian Groups. Every finite abelian group G is the direct sum of cyclic groups, each of prime power order. Proof. By Theorem 9.5 (Ed.3), a finite abelian group G is the direct sum of its p- subgroups G(p) for the distinct positive primes p that divide |G|. We show that every finite abelian p-group H is a direct sum of cyclic groups by induction on the order of H. ∼ When |H| = 2, we have H = Z2 which is cyclic. Now assume the assertion is true for all groups whose order is less than |H|, and let a be an element of maximal order pn in H. By Lemma 9.6 (Ed.3), there exists a subgroup K of H such that H = hai ⊕ K. By the induction hypothesis, the subgroup K is a direct sum of cyclic subgroups, each with order a power of p. This implies that H = hai ⊕ K is also a direct sum of cyclic groups, each with order a power of p.  Example. The integer 100 can be written as a product of prime powers in four ways:

2 · 2 · 5 · 5, 2 · 2 · 52, 24 · 5 · 5, 22 · 52.

The Fundamental Theorem of Finite Abelian Groups tells us that there every abelian group of order 100 must be isomorphic to one of

Z2 ⊕ Z2 ⊕ Z5 ⊕ Z5, Z2 ⊕ Z2 ⊕ Z25, Z4 ⊕ Z5 ⊕ Z5, Z4 ⊕ Z25. No two of these finite abelian groups are isomorphic (as they have different numbers of elements of order 2 and 5).

Why is Z100 not on the list? Or it is Z4 ⊕ Z25 in disguise? ∼ Lemma 9.8. For positive integers m and k, if (m, k) = 1, then Zmk = Zm ⊕ Zk.

Proof. The order of (1, 1) ∈ Zm ⊕ Zk is the smallest positive integer t such that (t, t) = t(1, 1) = (0, 0). Thus we have t ≡ 0 (mod m) and t ≡ 0 (mod k). Thus implies that m | t and k | t. Since (m, k) = 1, we have that mk | t, so that mk ≤ t. We have that mk(1, 1) = (mk, mk) = (0, 0) because m | mk and k | mk. Since t is the smallest positive integer for which t(1, 1) = (0, 0), we must have mk = t = |(1, 1)|. ∼ This says that Zm ⊕ Zk has an element of order mk, so that Zm ⊕ Zk = Zmk.  We can extend Lemma 9.8 to more than one direct sum.

n1 n2 nt Theorem 9.9. If n = p1 p2 ··· pt for distinct primes p1, p2, . . . , pt and positive integers n1, n2, . . . , nt, then ∼ n = n1 ⊕ n2 ⊕ · · · ⊕ nt . Z Zp1 Zp2 Zpt Proof. The conclusion of the theorem is true for finite abelian groups of order 2. Now assume inductively that it is true for finite abelian groups of order less than n.

n1 n2 nt Applying Lemma 9.8 (Ed.3) with m = p1 and k = p2 ··· pt (which are relatively prime ∼ by the distinctness of the primes p1, p2, . . . , pt), we obtain n = n1 ⊕ k. Z Zp1 Z ∼ The induction hypothesis shows that k = n2 ⊕ · · · ⊕ nt . Z Zp2 Zpt  We can combine Theorem 9.9 and the Fundamental Theorem of Finite Abelian Groups to give a second way of expressing a finite abelian group as a direct sum of cyclic subgroups. Example. For the finite abelian group, given as the direct sum of cyclic groups of prime power orders, G = Z2 ⊕ Z4 ⊕ Z3 ⊕ Z9 ⊕ Z81 ⊕ Z25 ⊕ Z125, we arrange the prime power orders of the cyclic factors by size, with one row for each prime as such: 2 22 3 32 34 52 53. Then we use the columns (and Theorem 9.9) to express

G = Z3 ⊕ (Z2 ⊕ Z32 ⊕ Z52 ) ⊕ (Z22 ⊕ Z34 ⊕ Z53 ) ∼ = Z3 ⊕ Z450 ⊕ Z40500. Notice that 3 | 450, i.e., 450 = 3 · 15, and 450 | 40500, i.e., 40500 = 450 · 90. This can always be done. Theorem 9.10. Every finite abelian group is the direct sum of cyclic groups of orders m1, m2, . . . , mt where m1 | m2, m2 | m3,..., mt−1 | mt. ∼ Definitions. For a finite abelian group G = Zm1 ⊕ Zm2 ⊕ · · · ⊕ Zmt , the integers m1, m2, . . . , mt of Theorem 9.10 are called the invariant factors of G. For a finite abelian group G expressed as a direct sum of cyclic groups of prime power orders, the prime powers are called the elementary divisors of G. The product of the invariant factors and the product of the elementary divisors both equal |G|. Example (Continued). There are four isomorphism classes of finite abelian groups of order 100. For G = Z2 ⊕ Z2 ⊕ Z5 ⊕ Z5, the elementary divisors are 2, 2, 5, 5, the invariant factors are 10, 10, and G is isomorphic to Z10 ⊕ Z10.

For G = Z2 ⊕Z2 ⊕Z25, the elementary divisors are 2, 2, 25, the invariant factors are 2, 50, and G is isomorphic to Z2 ⊕ Z50.

For G = Z4 ⊕ Z5 ⊕ Z5, the elementary divisors are 4, 5, 5, the invariant factors are 5, 20, and G is isomorphic to Z5 × Z20.

For G = Z4 ⊕ Z25, the elementary divisors are 4, 25, the invariant factor is 100, and G is isomorphic to Z100. The elementary divisors are the determining characteristics of finite abelian groups. Theorem 9.12. Two finite abelian groups of the same order are isomorphic if and only if they have the same elementary divisors. The proof of this is very long and technical. Appendix. Proof of Lemma 9.4. We use induction on the number of distinct positive primes that divide elements of G of finite order.

If |a| is divisible by at most one prime p1, then |a| is a power of p1 (no other prime divides |a|), whence a ∈ G(p1). Now suppose for every element b whose finite order is divisible by at most k − 1 distinct positive primes p1, p2, . . . , pk−1, that b = b1 +b2 +···+bk−1 for bi ∈ G(pi), i = 1, . . . , k−1.

If |a| is divisible by at most k distinct positive primes p1, p2, . . . , pk, then there are positive r1 r2 rk integers r1, r2, . . . , rk such that |a| = p1 p2 ··· pk . r2 rk r1 Let m = p2 ··· pk and n = p1 , for which |a| = mn.

By distinction of the primes p1, p2, . . . , pk we have that (m, n) = 1, so there are integers u, v such that 1 = mu + nv. Hence a = 1a = (mu + nv)a = mua + nva.

r1 r1 Because |a| = mn with n = p1 , we have p1 (mua) = (nm)ua = u(mna) = u0 = 0, so that a ∈ G(p1).

k1 r2 k−1 Similarly, m(nva) = v(mna) = v0 = 0, so the order of nva divides m = p1 p2 ··· pk−1, an integer with at most k − 1 distinct prime divisors.

By the induction hypothesis, we have that nva = a2 + a3 + ··· + ak for ai ∈ Gi, i = 2, 3, . . . , k.

With a1 = mua ∈ G(p1), we have by a = mua + nva that a = a1 + a2 + ··· + ak.  Proof of Theorem 9.5. For a ∈ G we have |a| divides |G|.

By Lemma 9.4, we then have that a = a1 + a2 + ··· + at with ai ∈ G(pi), i = 1, 2, . . . , t (where ai = 0 if pi - |a|).

To prove the expression of a is unique, we suppose that a = b1 +b2 +···+bt for bi ∈ G(pi). Since G is a abelian we obtain

a1 − b1 = (b2 − a2) + (b3 − a3) + ··· + (bt − at).

Since ai, bi ∈ G(pi) and G(pi) is a group, we have bi − ai ∈ G(pi) for all i = 1, . . . , t.

ri Hence there exist ri ≥ 0 such that |bi − ai| = pi . r2 r3 rt With m = p2 p3 ··· pt we then have that m(bi − a)i) for all i = 2, 3, . . . , t, and hence that m(a1 − b1) = 0.

So |a1 − b1| must divide |a1 − b1.

Since a1 − b1 ∈ G(p1), we have that |a1 − b1| is a power of p1.

r1 r3 rt 0 The only power of p1 that divides m = p2 p3 ··· pt is p1 = 1.

Thus a1 − b1 = 0, so that a1 = b1.

Repeating the argument on a2 + a3 + ··· + at = b2 + b3 + ··· bt gives a2 = b2, a3 = b3, and on to at = bt. Thus every element of a ∈ G can be written uniquely at a1 + a2 + ··· + at for ai ∈ G(pi), and hence G = G(p1) ⊕ G(p2) ⊕ · · · ⊕ G(pt).  Proof of Lemma 9.6. Consider the set of subgroups H of G such that hai ∩ H = {0}. This set is nonempty since H = {0} satisfies hai ∩ H = {0}. This set is finite because G is finite. In this set there is then a largest subgroup K of G (i.e., with the largest order) for which hai ∩ K = {0}. We are to show that G = hai + K. Suppose to the contrary that G 6= hai+K, so there is nonzero b ∈ G such that b ∈ hai+K Since G is a p-group, there is an integer j ≥ 0 such that pjb = 0. Thus there is a smallest positive integer l ≤ j such that plb ∈ hai+K because 0 ∈ h0i+K. Then the element c = pl−1b is not in hai + K by the minimality of l. However, pc = plb is in hai + K, so there are t ∈ Z and k ∈ K such that pc = ta + k. Since a has maximal order pn = |a| then pnx = 0 for all x ∈ G (by the previous Lemma). Thus pn−1(ta + k) = pn−1(ta + k) = pn−1pc = 0. This means that pn−1ta = −pn−1k where pn−1ta ∈ hai and pn−1k ∈ K. Since hai ∩ K = {0}, then pn−1ta = 0. With |a| = pn we have that pn | pn−1t, so that pn−1t = pnm for some m ∈ Z. Cancellation of pn−1 on both sides gives t = pm. This gives pc = ta + k = pma + k, so that k = pc − pma = p(c − ma). Set d = c − ma. Then pd = p(c − ma) = pk ∈ K, but d 6∈ K since if k0 = c − ma ∈ K would imply that c = ma + k0 ∈ hai + K, contradicting c 6∈ hai + K. The set H = {x + zd : x ∈ K, z ∈ Z} is a subgroup of G that satisfies K ⊂ H. Since d = 0 + 1d ∈ H but d 6∈ K, we have H 6= K, so that K is a proper subset of H. Since K is the largest subgroup of G for which hai ∩ K = {0}, then hai ∩ H 6= {0}. Let w be a nonzero element of hai ∩ H.

Then sa = w = k1 + rd for k1 ∈ K and r, s ∈ Z. We show that (p, r) = 1 but supposing (p, r) > 1 and reaching a contradiction. If (p, r) > 1, then as p is prime, we have r = py for some y ∈ Z, then as pd ∈ K we have 0 6= w = sa = k1 + pyd ∈ hai ∩ K, a contradiction. Consequently, we have that (p, r) = 1. Then there are integers u, v such that pu + rv = 1. Therefore

c = 1c = (pu + rv)c = u(pc) + v(rc) = u(ta + k) + v(r(d + ma)) = u(ta + k) + v(rd + rma)

= u(ta + k) + v(sa − k1 + rma)

= (ut + vs + rm)a + (uk − vk1) ∈ hai + K.

This contradicts c 6∈ hai + K. Thus G = hai ⊕ K by Theorem 9.3 (Ed.3).