Martin Isaacs's exercise 3.5 (character theory of finite groups)

Question

I need some help with this exercise:

Suppose $A\subseteq{G}$ is abelian, and $|G:A|$ is a prime power. Show that $G'\lt{G}$

Thank you very much in advance.

Answer 1

Hint: Reduce to a simple group and apply theorem 3.9.

Reduction to simple group:

Suppose by way of contradiction that $G'=G$. If $G$ is not simple, then $G$ has a proper non-identity normal subgroup $N$. If $AN=G$, then $G/N = AN/N \cong A/A\cap N$ is abelian, so $G=G' \leq N$, contradicting $N$ being proper. Hence $\bar G = G/N$ is a finite group with abelian subgroup $\bar A = AN/N \leq \bar G$, and $[\bar G:\bar A]$ (which divides $[G:A]$) is a prime power. However, $\bar G' = \bar G$, so we have a smaller counterexample. Continuing in this way, we may assume $G$ is simple (lest we find a new $N$).

Final contradiction:

However, for any non-identity element $a \in A$, $A \leq C_G(a)$ since $A$ is abelian, so $[G:C_G(a)]$ divides $[G:A]$, a prime power. This contradicts theorem 3.9 which says that the only $a \in G$ (for $G$ simple) with $[G:C_G(a)]$ a prime power is $a=1$ with $[G:C_G(a)]=1$. $\square$