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

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Math 371 Lecture #37 §8.2 (Ed.2), 9.2 (Ed.3): Finite Abelian Groups Math 371 Lecture #37 x8.2 (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 = fm + n : m 2 M; n 2 Ng, 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) = fa 2 G : jaj = pn for some n ≥ 0g: Lemma. For an abelian group G, the set G(p) is a subgroup of G. Proof. For a; b 2 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 2 G, and hence G is a subgroup. Lemma 9.4. Let G be an abelian group and a 2 G an element of finite order. Then a = a1 + a2 + ··· + at for ai 2 G(pi), i = 1; 2; : : : ; t where p1; p2; : : : ; pt are the distinct positive primes that divide jaj. The proof is in the Appendix of this Lecture Note. The proof uses induction on the number of primes that divide jaj. 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 jGj. The proof is in the Appendix. It uses Lemma 9.4 to establish the existence of a = a1 + a2 + ··· + at with ai 2 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 jbj ≤ jaj for all b 2 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 2 B. Proof. If jaj = pn is an element of maximal order in a p-group B, then each b 2 B has order jbj = 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 2 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 = f0g, 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 jGj. We show that every finite abelian p-group H is a direct sum of cyclic groups by induction on the order of H. ∼ When jHj = 2, we have H = Z2 which is cyclic. Now assume the assertion is true for all groups whose order is less than jHj, 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) 2 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 j t and k j t. Since (m; k) = 1, we have that mk j t, so that mk ≤ t. We have that mk(1; 1) = (mk; mk) = (0; 0) because m j mk and k j mk. Since t is the smallest positive integer for which t(1; 1) = (0; 0), we must have mk = t = j(1; 1)j. ∼ 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 j 450, i.e., 450 = 3 · 15, and 450 j 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 j m2, m2 j m3,..., mt−1 j 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 jGj. 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 jaj is divisible by at most one prime p1, then jaj is a power of p1 (no other prime divides jaj), whence a 2 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 2 G(pi), i = 1; : : : ; k−1. If jaj 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 jaj = p1 p2 ··· pk . r2 rk r1 Let m = p2 ··· pk and n = p1 , for which jaj = mn.
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