Exercise 4.5.3 (Proof of Cauchy's Theorem with Sylow's Theorem)

Proof. Suppose that $p$ divides the order of $G$. Then $|G|=p^{\alpha }m$, where $\alpha >1$ and $p\nmid m$. By Sylow’s Theorem, there is a subgroup $H$ of $G$ of order $p^{\alpha }$.

Moreover, by Exercise 4.3.29, the $p$-group $H$ has a subgroup of order $p^{\beta }$ for every $\beta$ with $0\le \beta \le \alpha$.

In particular, for $\beta =1$, $H$ has a subgroup $K$ of order $p$, which is a subgroup of $G$. This proves Cauchy’s Theorem. □