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Subject : MATHEMATICS Subject : MATHEMATICS Paper 1 : ABSTRACT ALGEBRA Chapter 4 : Nilpotent and solvable groups Module 2 : Solvable groups - I Anjan Kumar Bhuniya Department of Mathematics Visva-Bharati; Santiniketan West Bengal 1 Solvable groups - I Learning outcomes: 1. Solvable groups. 2. Subgroup and homomorphic image of a solvable group is solvable. Historically, solvable groups arose in the context of Galois theory. Galois introduced this notion to investigate solvability by radicals of the quintics over a field. Now solvable groups are an interesting class of groups in their own right, specially in connection to the structure theory of finite groups. Here we give a purely group theoretic treatment of the solvable groups. Definition 0.1. Let G be a group. Then a chain G = H0 ⊇ H1 ⊇ · · · Hn−1 ⊇ Hn = feg of subgroups is called a solvable series for G if Hi+1 is normal in Hi and Hi=Hi+1 is commutative for every i = 0; 1; ··· ; n − 1. A group G is called a solvable group if G has a solvable series. Every abelian group is solvable. For, if G is abelian, then G = H0 ⊇ H1 = feg is a solvable series for G. Example 0.2. (i) S3 = H0 ⊇ fe; (123); (132)g ⊇ feg is a solvable series for S3. Thus S3 is a solvable group. Recall that S3 is not nilpotent. (ii) S4 is solvable. Theorem 0.3. Every nilpotent group is solvable. Proof. Let G be a nilpotent group. Then Zn(G) = G for some positive integer n. Then the series of normal subgroups G = Zn(G) ⊇ Zn−1(G) ⊇ · · · ⊇ Z0(G) = feg is a solvable series, since Zi(G)=Zi−1(G) = Z(G=Zi−1(G)) which is abelian for every i = 1; 2; ··· ; n. Hence G is a solvable group. Now we show that the class of all solvable groups is closed under finite direct product, taking subgroups and homomorphic images. Theorem 0.4. Every finite direct product of solvable groups is solvable. 2 Proof. This is sufficient to prove the result for the direct product of two solvable groups. Let G and H be two solvable groups. Then both G and H have solvable series of subgroups, say G = G0 ⊇ G1 ⊇ · · · ⊇ Gn = feg and H = H0 ⊇ H1 ⊇ · · · ⊇ Hm = feg: Consider following series of subgroups of G × H: G × H =G0 × H0 ⊇ G1 × H0 ⊇ · · · ⊇ feg × H0 ⊇ feg × H1 ⊇ · · · ⊇ feg × Hm ⊇ feg: This is a solvable series for G × H, since Gi+1 × H0 is normal in Gi × H0 and feg × Hi+1 is normal Gi×H0 fe}×Hi in feg × Hi, and both ' Gi=Gi+1 and ' Hi=Hi+1 are abelian. Gi+1×H0 fe}×Hi+1 Hence G × H is a solvable group. Theorem 0.5. Let G be a solvable group. Then every subgroup of G is solvable. Proof. Let H be a subgroup of G. Since G is solvable, so it has a solvable series, say G = G0 ⊇ G1 ⊇ · · · ⊇ Gn = feg. Then Gi+1 is a normal subgroup of Gi which implies that H \ Gi+1 is a normal subgroup of H \ Gi. Denote Hi = H \ Gi and consider the series H = H0 ⊇ H1 ⊇ · · · ⊇ Hn = feg of subgroups of H. Now, Hi+1 = H \ Gi+1 = H \ Gi \ Gi+1 = Hi \ Gi+1 which implies, by the second isomorphism theorem, that Hi = Hi ' HiGi+1 . Since the Hi+1 Hi\Gi+1 Gi+1 HiGi+1 quotient group is a subgroup of the abelian group Gi=Gi+1, so it is abelian. Thus Gi+1 H = H0 ⊇ H1 ⊇ · · · ⊇ Hn = feg is a solvable series for H and the subgroup H is a solvable group. Theorem 0.6. Let G be a solvable group. Then the quotient group G=H is solvable for every normal subgroup H of G. Proof. Let G = G0 ⊇ G1 ⊇ · · · ⊇ Gn = feg be a solvable series of G. Since H is a normal subgroup, so GiH is a subgroup of G and H ⊆ GiH for every i. Consider the series G=H = G0=H ⊇ G1H=H ⊇ G2H=H ⊇ · · · ⊇ GnH=H = fHg of subgroups of G=H. Also Gi+1 is normal in Gi and hence Gi+1H is a normal subgroup of GiH. Thus Gi+1H=H is a normal subgroup of GiH=H. Now, by the third isomorphism theorem 3 GiH=H ' GiH=Gi+1H. Gi+1H=H Define : Gi=Gi+1 −! GiH=Gi+1H by (aGi+1) = aGi+1H. Then Gi+1 ⊆ Gi+1H implies that is well-defined and for every ah 2 GiH, ahGi+1H = (aGi+1H)(hGi+1H) = aGi+1H = (aGi+1) implies that is onto. Also is a homomorphism. Since Gi=Gi+1 is abelian it follows that GiH=Gi+1H = (Gi=Gi+1) is abelian. Thus the series G=H = G0=H ⊇ G1H=H ⊇ G2H=H ⊇ · · · ⊇ GnH=H = fHg is a solvable series and G=H is a solvable group. Corollary 0.7. Every homomorphic image of a solvable group is solvable. Proof. Let G be a solvable group and f : G −! G be an epimorphism. Then by the first isomor- phism theorem G=kerf ' G. Since G is solvable, the quotient G=kerf is a solvable group. Hence G is solvable. Theorem 0.8. Let H be a normal subgroup of a group G. If both H and G=H are solvable, then G is solvable. Proof. The correspondence theorem implies that every subgroup of G=H is of the form K=H, where K is a subgroup of G such that H ⊆ K, and K=H is normal in G=H if and only if K is normal in G. Since G=H is a solvable group, so it has a solvable series, say G=H = K0=H ⊇ K1=H ⊇ · · · ⊇ Km=H = fHg: Ki=H Since Ki+1=H is normal in Ki=H, so Ki+1 is normal in Ki and ' Ki=Ki+1, by the third Ki+1=H isomorphism theorem. Thus Ki=Ki+1 is abelian. Since H is a solvable group, it has a solvable series, say H = H0 ⊇ H1 ⊇ · · · ⊇ Hn = feg: Hence G = K0 ⊇ K1 ⊇ · · · ⊇ Km = H ⊇ H1 ⊇ · · · ⊇ Hn = feg is a solvable series for G and G is solvable. Now we have the following two interesting consequences. Corollary 0.9. Let G be a group, H and K be two subgroups of G and H be normal in G. If both H and K are solvable then HK is solvable. 4 Proof. First note that HK is a subgroup of G, since H is normal in G. Now, by the second isomorphism theorem, we have HK=H ' K=H \ K. Since H \ K is a subgroup of the solvable group K, it is solvable; and so K=H \K is a solvable group. Then both H and HK=H are solvable, whence HK is solvable. Corollary 0.10. Let G be a group and H, K be two normal subgroups of G such that both the quotients G=H and G=K are solvable. Then G is solvable if and only if H \ K is solvable. Proof. Let G be a solvable group. Since H \ K is a subgroup of G, so it is solvable. Conversely, assume that H \ K is a solvable group. Both H and K are normal. Now by the second isomorphism theorem, HK=H ' K=H \ K. Since HK=H is a subgroup of the solvable group G=H, so it is and hence K=H \ K is solvable. Then it follows that K is solvable, since H \ K is solvable. Thus both G=K and K are solvable, and hence G is solvable. Example 0.11. Let G be a group of order 14 = 7 × 2. Then it follows from the Sylow Theorems that G has a normal Sylow 7-subgroup H. Since jHj = 7 which is a prime, so H is a solvable group. Similarly, jG=Hj = 2 implies that the quotient G=H is solvable. Hence G is solvable. Example 0.12. Let G be a group of order 2002 = 143 × 7 × 2. Then it follows from the Sylow Theorems that G has a normal Sylow 143-subgroup H. Since jHj = 143 which is a prime, so H is a solvable group. Now jG=Hj = 14 and hence it is solvable. Thus G is a solvable group. 1 Summary • Let G be a group. Then a chain G = H0 ⊇ H1 ⊇ · · · Hn−1 ⊇ Hn = feg of subgroups is called a solvable series for G if Hi+1 is normal in Hi and Hi=Hi+1 is commutative for every i = 0; 1; ··· ; n − 1. A group G is called a solvable group if G has a solvable series. • Every abelian group is solvable. For, if G is abelian, then G = H0 ⊇ H1 = feg is a solvable series for G. • Every nilpotent group is solvable. • Every finite direct product of solvable groups is solvable. • Let G be a solvable group. Then every subgroup of G is solvable. 5 • Let G be a solvable group. Then the quotient group G=H is solvable for every normal subgroup H of G. • Every homomorphic image of a solvable group is solvable. • Let H be a normal subgroup of a group G. If both H and G=H are solvable, then G is solvable. • Let G be a group, H and K be two subgroups of G and H be normal in G. If both H and K are solvable then HK is solvable. • Let G be a group and H, K be two normal subgroups of G such that both the quotients G=H and G=K are solvable. Then G is solvable if and only if H \ K is solvable.
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