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 = {e} 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 = {e} is a solvable series for G.

Example 0.2. (i) S3 = H0 ⊇ {e, (123), (132)} ⊇ {e} 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) = {e} 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 = {e}

and H = H0 ⊇ H1 ⊇ · · · ⊇ Hm = {e}.

Consider following series of subgroups of G × H:

G × H =G0 × H0 ⊇ G1 × H0 ⊇ · · · ⊇ {e} × H0 ⊇ {e} × H1 ⊇ · · · ⊇ {e} × Hm ⊇ {e}.

This is a solvable series for G × H, since Gi+1 × H0 is normal in Gi × H0 and {e} × Hi+1 is normal Gi×H0 {e}×Hi in {e} × Hi, and both ' Gi/Gi+1 and ' Hi/Hi+1 are abelian. Gi+1×H0 {e}×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 = {e}.

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 = {e} 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 = {e} 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 = {e} 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 = {H} 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 ψ : Gi/Gi+1 −→ GiH/Gi+1H by

ψ(aGi+1) = aGi+1H.

Then Gi+1 ⊆ Gi+1H implies that ψ is well-deﬁned and for every ah ∈ GiH,

ahGi+1H = (aGi+1H)(hGi+1H) = aGi+1H = ψ(aGi+1) implies that ψ is onto. Also ψ is a homomorphism. Since Gi/Gi+1 is abelian it follows that

GiH/Gi+1H = ψ(Gi/Gi+1) is abelian. Thus the series

G/H = G0/H ⊇ G1H/H ⊇ G2H/H ⊇ · · · ⊇ GnH/H = {H} 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 isomorphism 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 = {H}.

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 = {e}.

Hence

G = K0 ⊇ K1 ⊇ · · · ⊇ Km = H ⊇ H1 ⊇ · · · ⊇ Hn = {e} 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 |H| = 7 which is a prime, so H is a solvable group. Similarly, |G/H| = 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 |H| = 143 which is a prime, so H is a solvable group. Now |G/H| = 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 = {e}

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 = {e} 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.