Let $H$ have order $m$ and $K$ have order $n$, where $m$ and $n$ are relatively prime. Then $H \cap K=\{e\}$
Solution 1:
We have that $ H \cap K \leqslant H \Rightarrow |H \cap K| \mid |H|$ (Lagrange) and similarly $ H \cap K \leqslant K \Rightarrow |H \cap K| \mid |K|.$ Hence $|H \cap K|$ is a common divisor of both $|H|$ and $|K|.$ Since $\gcd (|H|,|K|) = 1\Rightarrow |H \cap K | = 1 \Rightarrow H \cap K = \{1\}. \text {} \Box$