[See here]
Thompson's thesis result. Let G be a finite group admitting a fixed-point-free automorphism of prime order. Then G is nilpotent.
[See here]
I found here a nice proof of the following result using the previous two. I will write it down with some details.
Herstein Theorem. Let the finite group G admit an abelian maximal subgroup. then G is solvable.
Proof. Let M be an abelian maximal subgroup of G. Since the normal core
of M in G is abelian, we are reduced to show that 
is solvable, hence we may assume that is trivial. Let p be a prime divisor of 
, and let P be a Sylow p-subgroup of G containing a Sylow p-subgroup of M. Suppose 
. Since P is a p-group, there exists 
which normalizes 
. Since M is abelian and maximal in G, 
, contradicting 
. Therefore 
. This proves that M is a Hall subgroup of G, i.e. 
. By Burnside normal p-complement theorem, every Sylow p-subgroup of M admits a normal complement in G, and the intersection of these normal complements gives a normal complement N of M in G. Let 
be an element of prime order. Since M centralizes m, it normalizes 
, thus 
. If 
then the maximality of M implies that 
, i.e. 
, contradicting . Therefore 
, so m induces a fixed-point-free automorphism of N of prime order, and by Thompson's result N is nilpotent. Since 
is abelian, this implies that G is solvable.
