9.4. MORE ON ALGEBRAIC FIELD EXTENSIONS 233

Now let H be a normal subgroup of G such that G/H is Abelian. Then if g,h ∈ G,

[gh] = [hg] , gh(hg)−1 = ghg−1h−1 ∈ H

Thus every commutator is in H and so H ⊇ G.The last assertion is obvious because G/{ι} is isomorphic to G. Also, to say that

G′ = {ι} is to say thataba−1b−1 = ι

which implies that ab = ba. ■Let G be a group and let G′ be its commutator subgroup. Then the commutator sub-

group of G′ is G′′ and so forth. To save on notation, denote by G(k) the kth commutatorsubgroup. Thus you have the sequence

G≡ G(0) ⊇ G(1) ⊇ G(2) ⊇ G(3) · · ·

each G(i) being a normal subgroup of G(i−1) although this does not say that G(i) is a normalsubgroup of G. Then there is a useful criterion for a group to be solvable.

Theorem 9.4.43 If G is a solvable group and Ĝ is a subgroup of G then Ĝ is also solvable.

Proof: Suppose G = H0 ⊇H1 ⊇ ·· · ⊇Hm = {ι} where the quotient groups are Abelianand the Hi are normal. Consider Hk∩ Ĝ. Would this be a normal subgroup of Ĝ? Let a ∈ Ĝand x ∈Hk∩ Ĝ. Is axa−1 ∈Hk∩ Ĝ? We know this product is in Ĝ because Ĝ is a subgroup.We know it is in Hk because Hk is normal. Thus the Hk∩Ĝ are normal. What of the quotientgroups

(Hk ∩ Ĝ

)/(Hk+1∩ Ĝ

)? Are these Abelian? If [x] , [y] are in

(Hk ∩ Ĝ

)/(Hk+1∩ Ĝ

)is [xy] = [yx]? Is (xy)−1 yx∈Hk+1∩Ĝ? This equals y−1x−1yx. However, xy,yx are both in Hk

and the quotient groups Hk/Hk+1 are Abelian, so (xy)−1 yx ∈Hk+1. But also (xy)−1 yx ∈ Ĝbecause Ĝ is a subgroup. Hence [x] [y] = [xy] = [yx] = [y] [x] and so the quotient groups areAbelian. Hence Ĝ = H0∩ Ĝ⊇ Ĝ∩H1 ⊇ ·· · ⊇ Ĝ∩Hm = {ι} and so Ĝ is solvable. ■

Theorem 9.4.44 Let G be a group. It is solvable if and only if G′ is solvable so G(k) = {ι}for some k.

Proof: If G(k) = {ι} then G is clearly solvable because G(k−1)/G(k) is Abelian byTheorem 9.4.42. The sequence of commutator subgroups provides the necessary sequenceof subgroups.

Next suppose that you have

G = H0 ⊇ H1 ⊇ ·· · ⊇ Hm = {ι}

where each is normal in the preceding and the quotient groups are Abelian. Then fromTheorem 9.4.42, G(1) ⊆H1. Thus H ′1 ⊇ G(2). But also, from Theorem 9.4.42, since H1/H2is Abelian,

H2 ⊇ H ′1 ⊇ G(2).

Continuing this way G(k) = {ι} for some k ≤ m.Alternatively, you could let Ĝ = G′. This is solvable by Theorem 9.4.43 since it is a

subgroup of G. ■

Theorem 9.4.45 If G is a solvable group and if H is a homomorphic image of G, then His also solvable.