To show every sylow p-subgroup is normal in G 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.
Its an problem from Herstein, prob 19 in 2.12,2nd edition.I need some hint to solve this.Thank you.

This is not exactly an easy exercise! The proof outline below is from Burnside's book, "The Theory of Groups of Finite Order".


*

*Show that $G = \{x^{-1}\phi(x) : x \in G \}$.

*If $g$ and $\phi(g)$ are conjugate in $G$, then $g=1$. (Proof. Suppose $\phi(g) = h^{-1}gh$. By 1, $h = x^{-1}\phi(x)$ for some $x$, so $\phi(g) = \phi(x^{-1})xgx^{-1}\phi(x) \Rightarrow \phi(xgx^{-1})=xgx^{-1} \Rightarrow xgx^{-1}=1 \Rightarrow g=1$.)

*$g\phi(g)\phi^2(g)=1$ for all $g \in G$. (Let $x = g\phi(g)\phi^2(g)$. Tthen $x$ and $\phi(x)$ are conjugate, so it follows from 2.)

*Similarly $\phi^2(g)\phi(g)g = 1$, so $g$ and $\phi(g)$ commute for all $g \in G$.

*Any two conjugate elements in $G$ commute. (Proof. Let $g,h \in G$. By 1, $h=x^{-1}\phi(x)$ for some $x \in G$. Now by 4, $xgx^{-1}$ commutes with $\phi(xgx^{-1})$, so $g$ commutes with $x^{-1}\phi(x)\phi(g)\phi(x^{-1})x = h\phi(g)h^{-1}$ So $\phi(g)$ commutes with $h^{-1}gh$ and similarly $\phi^2(g)$ commutes with $h^{-1}gh$, and hence, by 3, so does $g$.)

*Now, if $p$ is a prime dividing $|G|$, and $g \in G$ has order $p$, then the conjugates of $g$ in $G$ generate a normal (abelian) subgroup $N$ of $G$ of order a power of $p$. Now the largest normal subgroup $O_p(G)$ of $G$ of order a power of $p$ is characteristic in $G$ and hence left invariant by $\phi$ so we can complete the proof by applying induction to $G/N$.
