Subject : MATHEMATICS Paper 1 : ABSTRACT ALGEBRA Chapter 4 : Nilpotent and solvable groups Module 1 : Nilpotent groups Anjan Kumar Bhuniya Department of Mathematics Visva-Bharati; Santiniketan West Bengal 1 Nilpotent groups Learning Outcomes: 1. Nilpotent groups. 2. Product of two nilpotent groups is nilpotent. 3. Converse of the Lagrange's Theorem holds on nilpotent groups. Converse of the Lagrange's Theorem holds on commutative groups. In this module we charac- terize a more general class of groups, namely nilpotent groups where the converse of the Lagrange's Theorem holds. Also every nilpotent group is a direct product of its Sylow subgroups and conversely. Definition 0.1. (i) Let G be a group. Denote Z0(G) = feg; Z1(G) = Z(G); and Zi+1(G) is the unique normal subgroup of G such that Zi(G) ⊆ Zi+1(G) and Zi+1(G)=Zi(G) = Z(G=Zi(G)): Then the chain of normal subgroups feg ⊆ Z1(G) ⊆ Z2(G) ⊆ · · · is called the upper central series or ascending central series of G. (ii) A group G is defined to be nilpotent if Zn(G) = G for some n 2 N. Example 0.2. Every abelian group G is a nilpotent group, since Z1(G) = Z(G) = G. But the converse is not true, which we show in the following result. Theorem 0.3. Every finite p-group is nilpotent. Proof. Let G be a finite p-group. If jGj = 1, then G is nilpotent. Suppose that jGj > 1. Then G being a nontrivial p-group, it follows that Z1(G) = Z(G) 6= feg. If Z1(G) 6= G, then G=Z1(G) is a nontrivial p-group and jZ(G=Z1(G))j > 1. This implies Z2(G) contains Z1(G) properly, that is Z1(G) ( Z2(G). If Z2(G) 6= G, then similarly Z2(G) ( Z3(G). If for all n 2 N, Zn(G) 6= G, then we get an infinite strictly ascending chain of subgroups feg ( Z1(G) ⊆ Z2(G) ( ··· 2 But this can not be , since G is a finite group. Thus there is n 2 Z such that Zn(G) = G and G is nilpotent. Now we an example of a subgroup which is not nilpotent. Example 0.4. We know that Z(S3) = feg. Thus Z1(S3) = feg and S3=Z1(S3) ' S3. This implies Z(S3=Z1(S3)) = ffegg. Since Z2(S3) is the unique normal subgroup of S3 such that Z2(S3)=Z1(S3) = Z(S3=Z1(S3)) = ffegg, so it follows that Z2(S3) = feg. Similarly, Zn(S3) = feg for all n 2 N. Hence S3 is not nilpotent. Theorem 0.5. The direct product of a finite number of nilpotent groups is nilpotent. Proof. This is sufficient to prove the result for the direct product of two groups. Let H and K be two nilpotent groups and denote G = H × K. First we prove that Zn(G) = Zn(H) × Zn(K) for all n 2 N by the induction on n. We have already proved in the Module 1 on the external direct product of groups that the result holds for n = 1. Assume that Zn(G) = Zn(H) × Zn(K) for some n ≥ 1. Since Zn(H × K) = Zn(H) × Zn(K), so there is an isomorphism : H=Z (H) × K=Z (K) −! H×K . n n Zn(H×K) Now we have, Z(G=Zn(G)) =Z(H × K=Zn(H × K)) =Z( (H=Zn(H) × K=Zn(K))) = (Z(H=Zn(H) × K=Zn(K))) = (Z(H=Zn(H)) × Z(K=Zn(K))) = (Zn+1(H)=Zn(H) × Zn+1(K)=Zn(K)) =Zn+1(H) × Zn+1(K)=Zn(H) × Zn(K) =Zn+1(H) × Zn+1(K)=Zn(H × K) Since Zn+1(G) is the unique normal subgroup of G such that Zn+1(G)=Zn(G) = Z(G=Zn(G)), so it follows that Zn+1(G) = Zn+1(H) × Zn+1(K). Thus, by the principle of mathematical induction, Zn(G) = Zn(H) × Zn(K) for all n = 0; 1; 2; ··· . Since both H and K are nilpotent, there exists a positive integer n such that Zn(H) = H and Zn(K) = K. Hence Zn(H × K) = Zn(H) × Zn(K) = H × K. and H × K is nilpotent. 3 Now we give an alternative equivalent characterization of nilpotent groups, namely by de- scending central series. Let G be a group and H; K be two subgroups of G. Denote by [H; K] the subgroup generated by all elements of the form hkh−1k−1 for all h 2 H and k 2 K. Then we have Lemma 0.6. Let G be a group, H, K be two subgroups of G and K normal in G. Then [H; G] ⊆ K if and only if HK=K ⊆ Z(G=K). Definition 0.7. Let G be a group. Define inductively G[1] = G; G[i+1] = [G[i];G] for all i ≥ 1: Then the chain of normal subgroups G = G[1] ⊇ G[2] ⊇ · · · is called the descending central series of G. Theorem 0.8. Let G be a group. Then G is nilpotent if and only if G[n+1] = feg for some integer n ≥ 0. Proof. First assume that there is an integer n ≥ 0 such that G[n+1] = feg. Consider the series feg = G[n+1] ⊆ G[n] ⊆ · · · ⊆ G[1] = G: [i] [i+1] [i+1] [n+1−i] It is easy to check that G =G ⊆ Z(G=G ) for all i = 1; 2; ··· ; n and G ⊆ Zi(G) for all [1] i = 0; 1; ··· ; n. Thus G = G ⊆ Zn(G), and so G is nilpotent. Conversely, suppose that G is nilpotent. Then there is an integer n ≥ 1 such that Zn(G) = G. Thus we have a series of normal subgroups feg = Z0(G) ⊆ Z1(G) ⊆ · · · ⊆ Zn(G) = G: [i] [n+1] Then it follows that G ⊆ Zn+1−i(G) for all i = 1; 2; ··· ; n + 1. Thus G ⊆ Z0(G) = feg, and so G[n+1] = feg. Theorem 0.9. Every subgroup of a nilpotent group is nilpotent. Proof. Let G be a group and H be a subgroup of G. Since G is nilpotent, so there is an integer n ≥ 0 such that G[n+1] = feg. Now we show that H[i] ⊆ G[i] by the induction on i. Now H[1] = H ⊆ G = G[1]. Assume that H[i] ⊆ G[i]. Then H[i+1] = [H[i];H] ⊆ [G[i];G] ⊆ G[i+1] and hence, by the principle of mathematical induction, H[i] ⊆ G[i] for all i = 1; 2; ··· ; n + 1. Hence H[n+1] ⊆ G[n+1] = feg, and so H is nilpotent. 4 Example 0.10. Let G = S3 × Z. Since S3 is not nilpotent, so G is not nilpotent. Thus G is an infinite group which is not nilpotent. Theorem 0.11. A finite group G is nilpotent if and only if it is the direct product of its Sylow subgroups. Proof. Let G be a nilpotent group. First we show that every Sylow subgroup of G is normal. If possible, let P be a Sylow subgroup of G such that N(P ) 6= G. Denote H = N(P ). Since G is nilpotent, there is n 2 N such that feg = Z0(G) ⊆ Z1(G) ⊆ · · · ⊆ Zn(G) = G Since H is proper and Z0(G) ⊆ H, so there is the largest index m < n such that Zm(G) ⊆ H. Then Zm+1(G) ( H. Consider a 2 Zm+1(G) n H, then Zm+1(G)=Zm(G) implies that aZm(G) 2 −1 −1 Z(G=Zm(G)) and so haZm(G) = ahZm(G) for every h 2 H. Hence h a ha 2 Zm(G) ⊆ H, which implies that a−1ha = h(h−1a−1ha) 2 H. Thus a−1Ha ⊆ H and a 2 N(H). Therefore H ( N(H), that is N(P ) ( N(N(P )), which contradicts that P is a Sylow subgroup. Thus N(P ) = G and every Sylow subgroup of G is normal. Hence, G is a direct product of its Sylow subgroups. Conversely, let G be a direct product of its Sylow subgroups P1;P2; ··· ;Pk. Since Pi is a pi-group for some prime pi, so Pi is nilpotent for every i = 1; 2; ··· ; k. Hence their direct product G is also nilpotent. Now we have the following consequence that shows that the converse of the Lagrange's Theorem holds on every nilpotent group. Corollary 0.12. Let G be a finite nilpotent group. Then for every positive divisor m of jGj, G has a subgroup of order m. n1 n1 nk Proof. Let jGj = p1 p2 ··· pk . Since G is nilpotent, so G = P1 ×P2 ×· · ·×Pk, where Pi is the Sylow ki pi-subgroup of G for i = 1; 2; ··· ; k. Also jPij = pi . Let m be a positive integer such that m j jGj. m1 m2 mk ni Then m = p1 p2 ··· pk , where 0 ≤ mi ≤ ni. Since Pi is pi-group of order pi and mi ≤ ni, so Pi mi has a subgroup Hi of order pi , by the Sylow's first theorem. Then H = H1 × H2 × · · · × Hk is a m1 m2 mk subgroup of G of order jH1jjH2j · · · jHkj = p1 p2 ··· pk = m. 1 Summary • (i) Let G be a group. Denote Z0(G) = feg; Z1(G) = Z(G); and Zi+1(G) is the unique normal subgroup of G such that Zi(G) ⊆ Zi+1(G) and Zi+1(G)=Zi(G) = Z(G=Zi(G)): 5 Then the chain of normal subgroups feg ⊆ Z1(G) ⊆ Z2(G) ⊆ · · · is called the upper central series or ascending central series of G.