Show that the p-Sylow subgroup is normal in $G$

Question

Let $G$ be a finite group and suppose that $\phi$ is an automorphism of $G$ such that $\phi^3$ is the identity automorphism. Suppose further that $\phi(x) = x$ implies that $x = e$. Prove that for every prime $p$ which divides $o(G)$, the $p$-Sylow subgroup is normal in $G$.

Answer 1

Definition: Let $A$ act on $G$ via Automorphisms.The action of $A$ on $G$ is said to be Frobenius if $ \phi(g)\neq g$ for every nonidendity $\phi\in A$ and $g\in G$.

Now, let $A=<\phi>$ then clearly $A$ acts on $G$ via Automorphisms.It is given that $\phi(g)\neq g$.Thus,we need to show that $\phi^2(g)\neq g$ for nonidendity $g\in G$.

Assume $\phi^2(g)= g\implies \phi^4(g)=\phi^2 (g) $ since $\phi^3=I$ $$\phi(g)=\phi^2(g)=g $$ contradiction.

so it is a Frobenius action which means that Frobenius kernel $G$ is nilpotent(Thompson theorem).Hence $G$ has normal Sylow-$p $subgroup.

http://for.mat.bham.ac.uk/P.J.Flavell/research/publications/frobenius.pdf