Order of a product of subgroups. Prove that $o(HK) = \frac{o(H)o(K)}{o(H \cap K)}$. Let $H$, $K$ be subgroups of $G$. Prove that $o(HK) = \frac{o(H)o(K)}{o(H \cap K)}$.
I need this theorem to prove something. 

Firstly, it can be shown that $HK \le G \iff H \unlhd HK$ or $K \unlhd HK$. Without loss of generality, we can assume that $K \unlhd HK$.
Let $T = H\cap K$. Then $T \unlhd H$.
Consider the function $f: H/T \to HK/K$ where $f(hT)=hK$ for each left coset $hT \in H/T$.
Suppose $f(hT)=f(gT)$ for some $hT, gT \in H/T$. Then $hK=gK$. So $h^{-1}g \in K$. But since $h, g \in H, h^{-1}g \in H$. So $h^{-1}g \in T$. Then $hT=gT$. So $f$ is an injective function.
Now take $(hk)K \in HK/K$ where $h \in H$ and $k \in K$. Then $(hk)K=hK$. So there exists $hT \in H/T$ such that $f(hT)= (hk)K$. So $f$ is a surjective function.
Since $f$ is a bijective function, $|H/T|=|HK/K|$. Then $\frac {|H|}{|T|}= \frac {|HK|}{|K|}$. Thus $|HK|= \frac {|H||K|}{|T|} = \frac {|H||K|}{|H \cap K|}$.
