Group Chains, Commutator Subgroups and Solvability

Commutator Subgroups and Solvability

Just as f′′ is the derivative of f′, so G′′ is the commutator subgroup of G′. And we can continue taking commutator subgroups, just as we took third and fourth derivatives etc.

Of course G′′ is normal in G′, with an abelian quotient group. As it turns out, G′′ is also normal in G. As described in the previous section, any automorphism of G permutes the commutators, which are the generators of G′. So the automorphism on G induces an automorphism on G′. The automorphism of G′, in turn, maps commutators to commutators, which are the generators of G′′. Thus every automorphism of G induces an automorphism on G′′, and G′′ is normal in G. By induction, this continues forever. The k^th commutator subgroup is normal in G.

If the k^th commutator subgroup equals e, we have a normal series for G, and G is solvable. After all, the quotient at each level is abelian.

Conversely, assume G is solvable. We want to show the k^th order commutator subgroup is e. This is true if the series consists of G and e, whence G is already abelian. Proceed by induction on the length of the solvable series.

Let N₀ = G, and let N₁ be the first proper normal subgroup in the series. The series ends at N_k = e. Since N₀/N₁ is abelian, N₁ contains G′. Restrict N₁, and the rest of the series, to G′, to find a smaller solvable group. By induction on the length of the series, the commutator subgroup of G′ drops to e after k-1 iterations. Step back to G, and its commutator subgroup drops to e after k iterations.

A group is solvable iff it has a finite chain of commutator subgroups. Since each subgroup is normal in G, this is a normal series, not just a subnormal series.