Let $H, K\le G$ be finite subgroups. If $|H|$ and $|K|$ are relatively prime, show that $H\cap K=\{e\}.$ I'm a little stuck on a question from a textbook that I'm self-studying group theory from, and could use a little bit of help.
Let $G$ be a group containing finite subgroups $H$ and $K$. If $|H|$ and $|K|$ are relatively prime, show that $H \cap K =\{e\}. $
At this point, the concepts of cosets and Lagrange's Theorem have been introduced, but nothing else like the Sylow Theorems. I am a little confused as to how to do this, so any help would be great. Cheers.
 A: Since $H$ and $K$ are both subgroups, so is $H\cap K$ (which is a finite subgroup of the finite subgroups $H$ and $K$). By the Lagrange theorem, the order of $H\cap K$ must divide both $|H|$ and $|K|$. Since the orders of $H$ and $K$ are relatively prime, necessarily $|H\cap K|=1$. Therefore, $H\cap K=\{e\}.$
A: Since $H,K\le G$, both $H$ and $K$ are groups. (Why?)

Note that $H\cap K\le H$ and $H\cap K\le K$.
Proof: Without loss of generality, I will show $H\cap K\le H$. I will use the one-step subgroup test.
Since $e\in H$ and $e\in K$, we have $e\in H\cap K$, so $H\cap K\neq \varnothing$.
Suppose $r\in H\cap K$. Then, by definition, we have $r\in H$ and $r\in K$, so, in particular, $r\in H$. Hence $H\cap K\subseteq H$.
Suppose $x,y\in H\cap K$. Then $x,y\in H$ and $x,y\in K$, so, since $H, K\le G$, we have $xy^{-1}\in H$ and $xy^{-1}\in K$. Hence $xy^{-1}\in H\cap K$.
Hence $H\cap K\le H$.$\square$

Now apply Lagrange's Theorem twice.
A: Since $H$ and $K$ are finite subgroups of $G$ then $H\cap K$ is also a subgroup of $G$ (because subgroups are closed under intersection) but furthermore, $H\cap K$ is also a subgroup of $H$ and $K$ respectively. So by Lagrange's theorem $|H\cap K|$ must divide $|H|$ and $|K|$, but since $|H|$ and $|K|$ are relatively prime, the only possible value for $|H\cap K|$ is $1$ and every subgroup must contain the identity, hence $H\cap K=\langle e \rangle$.
A: Note
Just a note to add that the answers here assume that the intersection of two subgroups of a group is itself a subgroup which, to me, is not obvious.
Either the question should be amended to say that "You may assume that the intersection of two subgroups of a group is itself a group", or proving that first should be part of the proof.
To see proofs of this assumption see, for example, "Groups and Symmetry" by MA Armstrong (pages 23 and 24), or this excellent hint towards a proof posted on MSE by @amWhy
