[Math pills] Abelian maximal subgroups

Burnside normal p-complement theorem. Let p be a prime number, G a finite group, and P a p-Sylow subgroup of G. If P is central in its normalizer then it admits a normal complement in G.
[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.