TITLE: What happens when the order of two finite subgroups are relatively prime?
QUESTION [6 upvotes]: I'm returning from an exam on group-theory and there were 2 questions I couldn't solve (and still can't), so I'm asking here for any hint you could possibly give.

Let G be a group and H and K subgroups such that $|H| = n$, $|K| = m$ and $gcd(n, m) = 1$. Show that $H \cap K = \{e\}$.

I wish I could show you some of my attempts before hand, but they're all rubbish that didn't get me anywhere. Essentially, the only (and last) thing I remembered and thought it could be useful was to see if if H and K are partitions of G. I think I've read something similar somewhere but can't recall where, so, am uncertain about it.
The other question I couldn't solve is, I think, related to this, so I shall try it once I understand this one.
Thanks for taking the time to read! Any tip is appreciated.

REPLY [10 votes]: Let $g \in H\cap K$. By Lagrange's Theorem, the order of g divides m and n, but the greatest common divisor of m and n is 1, so g=e.