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 $G$.

Let $x\in H\cap K$ then by Lagrange theorem the order of $x$ divides the two coprime orders: $|H|$ and $|K|$ so $o(x)=1$ and then $x=e$. Conclude.
