Define the commutator and the derived series. State and prove two nontrivial theorems about derived series.
Solution
For \(x,y\in G\), the commutator is \[ [x,y]=xyx^{-1}y^{-1}. \] The commutator subgroup is \[ G'=[G,G]=\langle[x,y]:x,y\in G\rangle. \] The derived series is defined recursively by \[ G^{(0)}=G, \qquad G^{(n+1)}=[G^{(n)},G^{(n)}]. \] A group is solvable exactly when \(G^{(r)}=1\) for some \(r\).
1. Homomorphisms preserve the derived series.
If \(\phi:G\to H\) is a group homomorphism, then \[ \phi(G^{(n)})=\phi(G)^{(n)} \] for every \(n\ge0\).
Indeed, \[ \phi([x,y])=[\phi(x),\phi(y)], \] so \[ \phi([A,A])=[\phi(A),\phi(A)] \] for every subgroup \(A\le G\). Induction on \(n\) proves the formula.
In particular, every \(G^{(n)}\) is characteristic in \(G\): for an automorphism \(\phi\), \[ \phi(G^{(n)})=G^{(n)}. \]
2. Solvability is closed under extensions.
Let \(N\trianglelefteq G\). If both \(N\) and \(G/N\) are solvable, then \(G\) is solvable.
Suppose \[ (G/N)^{(r)}=1. \] By the quotient map and 1, \[ G^{(r)}N/N=1, \] so \[ G^{(r)}\le N. \] If \(N^{(s)}=1\), then induction gives \[ G^{(r+j)}\le N^{(j)} \] for every \(j\ge0\). Hence \[ G^{(r+s)}\le N^{(s)}=1. \] Therefore \(G\) is solvable.