## Finite subgroups with relatively prime orders intersect trivially

Let $G$ be a group and let $H, K \leq G$ be finite subgroups of relatively prime order. Prove that $H \cap K = 1$.

Let $|H|=p$ and $|K|=q$. We saw in a previous exercise that $H \cap K$ is a subgroup of both $H$ and $K$; by Lagrange’s Theorem, then, $|H \cap K|$ divides $p$ and $q$. Since $\mathsf{gcd}(p,q) = 1$, then, $|H \cap K| = 1$. Thus $H \cap K = 1$.