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


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*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 $G/N$.
