Group Theory

4 Solvable groups

Definition 4.1. A group \(G\) is solvable iff there is a sequence \({1}=G_0\triangleleft G_1,\dots ,\triangleleft G_n = G\) with each quotient \(G_i/G_{i-1}\) abelian.

Definition 4.2. The commutator subgroup \([G,G]\) is the subgroup of \(G\) generated by all elements of the form \(aba^{-1}b{-1}\).

More generally, \([S,P]\) is the subgroup generated by the commutators of elements of \(S\) and \(P\) when \(S,P\) are subsets of \(G\). The commutator subgroup is a characteristic subgroup, ie. it is invariant under all automorphisms (so is the center). In particular it is normal, and the quotient \(G/[G,G]\) is abelian since \([c][d] = [cd][d^{-1}c^{-1}dc] = [d][c]\). Moreover \([G,G]\) is in the kernel of any homomorphism to an abelian group, so we have the universal property:

Proposition 4.3 (Universal property of the commutator subgroup). Given a homomorphism from \(G\) to an abelian group, \(H\), it uniquely factors through \(G/[G,G]\).

Proposition 4.4. Let \(N\triangleleft G\) be a normal subgroup. Then \(G\) is solvable iff \(G/N\) and \(N\) are solvable.

Proof. If \(G\) is solvable, then looking at \(N\cap G_i\) shows \(N\) is solvable, and looking at \(G_iN/N\) shows that \(G/N\) is solvable. Conversely if \(G/N\) and \(N\) are solvable, we can use the Lattice Isomorphism theorem to tack together the \((G/N)_i\) and the \(N_j\) to see that \(G\) is solvable. □

Proposition 4.5. Let \(G^{(i)} = [G^{(i-1)},G^{(i-1)}]\), \(G^0 = G\). Then \(G\) is solvable iff some \(G^{(j)}\) is trivial.

Proof. Certainly if some \(G^{(j)}\) is trivial, the sequence \(G^{(i)}\) shows that \(G\) is solvable. conversely, if \(G\) is solvable via the sequence \(G_i\), then \(G^{(i)} \subset G_i\), which is seen by induction and Proposition 4.3. □

The series \(G^{(i)}\) is known as the derived series.