Let $G$ be a group and let $H,K$ be subgroups of $G$ where $|H|=12$ and $|K|=5$. Show that $H\cap K = \{e\}$. 
Let $G$ be a group and let $H,K$ be subgroups of $G$ where $|H|=12$ and $|K|=5$. Show that  $H\cap  K = \{e\}$.

I used LaGrange's theorem to show that $|H|||G|$ and $|K|||G|$ so $12||G|$ and $5||G|$, and that $12$ and $5$ are relatively prime.  I'm not sure if this has gotten me closer to solving the problem, as I am stuck.  Any help is greatly appreciated.
NOTE: This problem provides does not specifically define $e$, so I'm attempting to prove this assuming $e$ is the identity of $G$.

We know that $H\cap K$ has at least the element $e$. suppose that $H\cap K$ has another element $g$ so that $g\in H$ and $g\in K$. the $|H\cap K|$ divides the order of $H$ and $K$ but $5$ and $12$ don't have common divisors. So the only element in $H\cap K$ is $e$.
