Lagrange's Theorem Proof Help

Question

Let H and K be subgroups of a group G with |H| = n and |K| = m, where gcd(n,m) = 1. Show that the intersection of H and K equals <1>.

Proof:
Let H and K be subgroups of a group G with |H| = n and |K| = m, where gcd(n,m) = 1. Suppose the intersection of H and K is a subgroup of H. Then, by Lagrange's Theorem, the intersection of H and K divides n. Suppose the intersection of H and K is a subgroup of K. Then, by Lagrange's Theorem, the intersection of H and K divides m.

I do not know where to go from this point.

Answer 1

You're just about finished here: $|H \cap K|$ is a divisor of both $n$ and $m$, but there aren't too many numbers dividing both $n$ and $m$.

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A different approach: Suppose that $g \in H \cap K$. Now note that $g$ is an element of two different (sub)groups: Namely, $H$ and $K$. How must the order of $g$ relate to $|H|$, and how must it relate to $|K|$?

Answer 2

You are done. Since $|H\cap K|$ divides both $n$ and $m$ it must divide $\gcd(n,m)=1$, so $|H\cap K|=1$.