Let $G$ be a group and $H_1$, $H_2$ finite subgroups of $G$. Prove that $|H_1H_2| = |H_1||H_2| /|H_1 \cap H_2| .$

Question

Let $G$ be a group and $H_1$, $H_2$ finite subgroups of $G$. Prove that $|H_1H_2| = \dfrac{|H_1||H_2|}{ |H_1 \cap H_2| }.$

I tried googling the problem to give me a hint on how to proceed and I saw a proof http://zimmer.csufresno.edu/~sdelcroix/sol251home6.pdf where they use equivalence relations. But I don't see how to do the proof with that. Can someone give me a hint or explain the reasoning behind the proof in the link?

Answer 1

Hint:

For $\;h,h_1\in H\;,\;\;k,k_1\in K\;$ , we have that

$$hk=h_1k_1\iff h_1^{-1}h=k_1k^{-1}\in H\cap K$$