Let $G$ and $M$ be as in exam. Let $p$ be a prime number such that $|G|=p^m$. Compute the stabilizer $M_0$ and conclude that $G$ contains an element of order $p$. This is known as Cauchy's Theorem.

$|M|=|G|^{p-1}=0\,\modul(p)$ and $(g_0,\ldots,g_{p-1})\in M_0$ iff for all $k\in\Z_p$: $g_k=g_0$, i.e. $g_0^p=e$. Since $(e,\ldots,e)\in M_0$ and $|Z_p|=p$ it follows by exam that $M_0$ contains at least $p-1$ elements $g\neq e$ such that $g^p=e$. If the order $q$ of such an element $g$ is smaller than $p$, then $q|p$, which is impossible.