## |HK|=|H||K|/|H intersect K|

Finally, the LaTeX path not specified problem has been solved by WordPress!

This post is about how to prove that $|HK|=\frac{|H||K|}{|H\cap K|}$, where $H$ and $K$ are finite subgroups of a group $G$.

A tempting thing to do is to use the “Second Isomorphism Theorem”, $HK/H\cong K/(H\cap K)$. However that would be a serious mistake since the conditions for the Second Isomorphism Theorem are not met. In fact $HK$ may not even be a group.

The correct way is to note that $HK=\bigcup_{h\in H}hK$.

Therefore $|HK|=|K|\times |\{hK:h\in H\}|$. For $h_1,h_2\in H$, we have:

\begin{aligned}h_1K=h_2K&\iff h_1h_2^{-1}\in K\\ &\iff h_1h_2^{-1}\in H\cap K\\ &\iff h_1(H\cap K)=h_2(H\cap K) \end{aligned}

Therefore $|\{hK:h\in H\}|=|\{h(H\cap K):h\in H\}|$, i.e. the number of distinct cosets $h(H\cap K)$. Since $H\cap K$ is a subgroup of $H$, applying Lagrange’s Theorem gives the number of distinct cosets $h(H\cap K)$ to be $\frac{|H|}{|H\cap K|}$.

Thus, we have $|HK|=\frac{|H|}{|H\cap K|}\cdot |K|$.