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Primary Decomposition Theorem for Finite Abelian Groups

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Primary Decomposition Theorem for Finite Abelian Groups Subject : MATHEMATICS Paper 1 : ABSTRACT ALGEBRA Chapter 1 : Direct Product of Groups Module 3 : Primary decomposition theorem for finite abelian groups Anjan Kumar Bhuniya Department of Mathematics Visva-Bharati; Santiniketan West Bengal 1 Primary decomposition theorem for finite abelian groups Learning outcomes: 1. p-primary abelian groups. 2. Primary Decomposition Theorem. 3. p-primary components. In this module we prove that every finite abelian group can be expressed as an internal direct product of a family of p-primary subgroups. In most of the results on the structure of finite abelian groups the feature that plays the most powerful role is that every finite abelian group is a Z-module. So, it is convenient to consider addition as the binary operation in the study of finite abelian groups. Internal direct product of a family of subgroups G1;G2; ··· ;Gk of a group G, if it is abelian, is called direct sum of G1;G2; ··· ;Gk and is denoted by G1 ⊕ G2 ⊕ · · · ⊕ Gk. Thus a finite abelian group G is a direct sum of subgroups G1;G2; ··· ;Gk if and only if G = G1 + G2 + ··· + Gk and Gi \(G1 +···+Gi−1 +Gi+1 +···+Gk) = f0g for every i = 1; 2; ··· ; k. In view of the fact that every internal direct sum of the subgroups G1;G2; ··· ;Gk is isomorphic to the external direct product of G1;G2; ··· ;Gk, we call both the internal direct sum and the external direct product simply as direct sum and use the notation G1 ⊕G2 ⊕· · ·⊕Gk to represent direct sum of groups G1;G2; ··· ;Gk. Definition 0.1. Let p be a prime. An abelian group G is called p-primary if every element of G is of order pm for some m ≥ 0, equivalently, if for each a 2 G there is m ≥ 0 such that pma = 0. Lagrange's Theorem implies that every finite abelian group G of order pn is a p-primary group. Following result shows that order of every p-primary abelian group is of this form. Lemma 0.2. Every finite p-primary abelian group is of order pn for some n ≥ 0. Proof. Let G = fg1; g2; ··· ; gkg be a p-primary abelian group. Then, for any gi 2 G, gi 2< gi >⊆< g1 > + < g2 > + ··· + < gk > implies that G ⊆< g1 > + < g2 > + ··· + < gk >. Also, < g1 > + < g2 > + ··· + < gk >⊆ G. Therefore G =< g1 > + < g2 > + ··· + < gk >. Now we show that jGj = pn by the induction on k, the number of elements of G. If k = 1, then n G =< g1 > implies that jGj = j < g1 > j = o(g1) = p . Suppose that G has k + 1 elements. Then 2 G =< g1 > + < g2 > + ··· + < gk+1 >, and we have j < g > + < g > + ··· + < g > jj < g > j jGj = 1 2 k k+1 j(< g1 > + < g2 > + ··· + < gk >)\ < gk+1 > j which implies that jGj divides j < g1 > + < g2 > + ··· + < gk > jj < gk+1 > j. Since j < g1 > r s + < g2 > + ··· + < gk > j = p , by induction hypothesis, and j < gk+1 > j = o(gk+1) = p , so jGj divides pr+s. Thus it follows that jGj = pn for some n ≥ 0. Thus the Klein's 4-group K4 is a 2-primary group. For a prime number p, the group Zp ⊕Zp2 ⊕ Zp3 is a p-primary group. But Z2 ⊕ Z3 is not a p-primary group. Let G be a finite abelian group and p a prime number. Denote m Gp = fa 2 G j p a = 0 for some integer m ≥ 0g: r Then Gp is the set of all elements of order p for some r ≥ 0 in G. For every prime number p, Gp is a subgroup of G. Example 0.3. Consider G = Z36. Then G2 = f[a] 2 Z36 j o([a]) = 1; 2; or 4g 36 = f[a] 2 j = 1; 2; or 4g Z36 gcd(a; 36) = f[a] 2 Z36 j gcd(a; 36) = 36; 18; or 9g = f[0]; [9]; [18]; [27]g ' Z4: Similarly, G3 = f[0]; [4]; [8]; [12]; [16]; [20]; [24]; [28]; [32]g which is isomorphic to Z9. Also we have G = G2 ⊕ G3. Theorem 0.4 (Primary Decomposition Theorem). 1. Let G be a finite abelian group and jGj = n1 n2 nk p1 p2 ··· pk . Then (a) G = Gp1 ⊕ Gp2 ⊕ · · · ⊕ Gpk , ni (b) jGpi j = pi for every i. 2. Let G and H be two finite abelian groups of the same order. Then G ' H if and only if Gp ' Hp for every prime p j jGj. 3 n1 n2 nk m1 m2 mk Proof. 1. (a) Let x 2 G. Then o(x) j p1 p2 ··· pk implies that o(x) = p1 p2 ··· pk where m1 m2 mk d mi 0 ≤ mi ≤ ni. Denote d = p1 p2 ··· pk and ri = mi for every 1 ≤ i ≤ k. Then pi (rix) = pi dx = 0 implies that rix 2 Gpi for every i. Since gcd(r1; r2; ··· ; rk) = 1, so there are si 2 Z such that 1 = s1r1 + s2r2 + ··· + skrk which implies that x = s1r1x + s2r2x + ··· + skrkx 2 Gp1 + Gp2 + ··· + Gpk . Therefore G = Gp1 + Gp2 + ··· + Gpk . To show that this sum is a direct li sum, consider x 2 Gpi \ (Gp1 + Gp2 + ··· Gpi−1 + Gpi+1 + ··· + Gpk ). Then pi x = 0 for some li ≥ 0 and x = x1 + x2 + ··· + xi−1 + xi+1 + ··· + xk where xi 2 Gpi . Then there is some lj ≥ 0 such that lj l1 l2 li−1 li+1 lk pj xj = 0 for all j = 1; 2; ··· ; i−1; i+1; ··· ; k, and hence ux = 0 where u = p1 p2 ··· pi−1 pi+1 ··· pk . li li li Since gcd(u; pi ) = 1, so there exist r; s 2 Z such that 1 = ru + spi , and so x = rux + spi x = 0. Therefore G = Gp1 ⊕ Gp2 ⊕ · · · ⊕ Gpk . mi (b) Every Gpi is a pi-primary abelian group and hence jGpi j = pi for some mi ≥ 0. Then m1 m2 mk n1 n2 nk m1 m2 mk G = Gp1 ⊕Gp2 ⊕· · ·⊕Gpk implies that jGj = p1 p2 ··· pk , that is, p1 p2 ··· pk = p1 p2 ··· pk . It follows from the Fundamental Theorem of Arithmetic that ni = mi for every i = 1; 2; ··· ; k. 2. Suppose that : G −! H is an isomorphism. Let p be a prime divisor of jGj. Then, for m m m every a 2 Gp, p a = 0 implies that p (a) = (p a) = (0) = 0, and so (Gp) ⊆ Hp. Thus jGp : Gp −! Hp is an one-to-one homomorphism. Since jGj = jHj, so jGpj = jHP j, by 1:(b). Hence jGp : Gp −! Hp is onto. Therefore Gp ' Hp. Converse follows trivially. Thus for every prime number p j jGj, the subgroups Gp are primary building blocks of the group G. Definition 0.5. Let G be a finite abelian group. Then for every prime divisor p of jGj, Gp is called the p-primary component of G. Example 0.6. Consider G = Z2 ⊕ Z23 ⊕ Z52 . Then the 2-primary component of G is Z2 ⊕ Z23 and the 5-primary component of G is Z52 . Example 0.7. Let G be a p-primary finite abelian group. We show that every nonidentity element of G is of order p if and only if G ' Zp ⊕ Zp ⊕ · · · ⊕ Zp. First assume that every nonidentity element of G is of order p. Since G is p-primary, so it follows that jGj = pn for some n ≥ 1. We prove the result by the principle of mathematical 4 n+1 induction on n. If n = 1, then jGj = p which implies that G ' Zp. Let jGj = p , n ≥ 1. Consider a nonidentity element a of G and denote H =< a >. Then jHj = o(a) = p which implies that H ' Zp. Now consider S = fK ⊆ G j K is a subgroup of G such that H \ K = f0gg: Then f0g is in S. Therefore S is nonempty. Also S is a partially ordered set with respect to set inclusion. Since S has only finite number of elements, so S has a maximal element, say K. Now we show that G = H ⊕ K. Otherwise, H \ K = f0g implies that G 6= H + K. Consider c 2 G n (H + K). Then c 62 K implies that K is properly contained in K+ < c >. Also if x 2 H \ (K+ < c >) then x = h = k + nc for some h 2 H, k 2 K and n 2 N, which implies that h − k = nc 2< c > \(H + K). Now o(c) = p implies that j < c > \(H + K)j = 1 or p. If j < c > \(H +K)j = p, then < c > \(H +K) =< c > and so < c >⊆ H +K which contradicts that c 62 H + K. Hence j < c > \(H + K)j = 1 and so < c > \(H + K) = f0g. Thus h − k = 0 which implies that x = h = k 2 H \ K = f0g. Thus H \ (K+ < c >) = f0g, and hence K+ < c >2 S which contradicts the maximality of K in S. Thus G = H ⊕ K. Then jKj = pn which implies that K ' Zp ⊕ Zp ⊕ · · · ⊕ Zp (n-times), by induction hypothesis. Thus G ' Zp ⊕ Zp ⊕ · · · ⊕ Zp (n + 1-times). Converse follows directly. 1 Summary • Let p be a prime. An abelian group G is called p-primary if every element of G is of order pm for some m ≥ 0, equivalently, if for each a 2 G there is m ≥ 0 such that pma = 0. • Every finite p-primary abelian group is of order pn for some n ≥ 0. n1 n2 nk • 1. Let G be a finite abelian group and jGj = p1 p2 ··· pk . Then (a) G = Gp1 ⊕ Gp2 ⊕ · · · ⊕ Gpk , ni (b) jGpi j = pi for every i.
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