If a nilpotent group $A$ act on a group $G$ then $G$ is solvable.

Question

Theorem: If a nilpotent group $A$ act on $G$ by automorphism and $C_G(A)=e$ then $G$ is solvable.

I hope that the statement of theorem is true. It should belong to Hartley, but when I googled it I could not find the the theorem.

If someone provides the whole statement of the theorem with source, I would be thankful.

Answer 1

If $A$ is a subgroup of ${\rm Aut(G)}$ with $C_G(A)=1$, then one says that $G$ acts fixed-point-freely upon $G$ by automorphisms.

The fixed-point-free automorphism conjecture asserts that if a finite group $G$ admits a fixed-point-free automorphism group $A$ (and, if $A$ is noncyclic, further suppose that $gcd(|G|, |A|) = 1$), then G is soluble.

Several special cases are known, e.g., see the article Solubility of finite groups admitting a fixed-point-free automorphism of order $rst$ by Peter Rowley.

The nilpotent case has been proved quite recently in the article Nilpotent fixed-point-free automorphism groups and regular abelian Carter subgroups by Jabara and Spiga:

Theorem (Jabara, Spiga 2013): Let $G$ be a finite group and let $A$ be a nilpotent group acting on $G$ as a group of automorphisms with $C_G(A) = 1$. Then $G$ is soluble.