Exercise 14.1.3

Proof. By (14.2),

According the Third Sylow Theorem the number $N$ of p-Sylow subgroups of G satisfies

so that $N=1+\mathit{kp},k\ge 0$, thus $N\wedge p=1$, and $N\mid \mathit{pm}$, therefore $N\mid m$. If $k\ne 0$, then $N>p$, but $N\mid m>0$, which implies $N\le m<p$. This contradiction shows that $k=0$, and $N=1$, i.e. there is exactly one $p$-Sylow subgroup $H$ of $G$.

For all $g\in G$, $\mathit{gH}{g}^{-1}$ is also a $p$-Sylow subgroup of $G$, hence $\mathit{gH}{g}^{-1}=H$ for all $g\in G$: $H$ is normal in $G$. □