= Solution
Both <subgroups> contain the <identity element>, so $H\cap K$ is nonempty. If $x,y\in H\cap K$, then $xy^{-1}$ belongs to both $H$ and $K$, hence to their intersection. This gives closure under inverses by setting $x=e$, and closure under products by using the inverse of $y$. Therefore \b[$H\cap K$ is a subgroup of $G$], and also of both $H$ and $K$.
<Lagrange's theorem for finite groups> states that the order of a <subgroup> divides the order of its finite ambient <group>; more precisely $|L|=[L:M]|M|$ for $M\le L$. Applying it twice gives
$$
|H\cap K|\mid p,\qquad |H\cap K|\mid q.
$$
Since $p$ and $q$ are <coprime>, their only positive common divisor is one. Thus
$$
\boxed{H\cap K=\{e\}}.
$$
Back to article page
