Let $A$ and $B$ be two finite subgroups of a group $G$. Show that $|AB|=\frac{|A| \times |B|}{|A \cap B|}$ Let $A$ and $B$ be two finite subgroups of a group $G$. Show that $$|AB|=\frac{|A| \times |B|}{|A \cap B|}$$ I have no idea how to start. Anyone can help ? I think of divisibility to prove this, but I got nowhere.
 A: One way to prove it is by showing that
$$
\lvert AB : B \rvert = \lvert A : A \cap B \rvert.
$$
So you should set up a (well defined,) 1-1 correspondence between the cosets of $B$ in (the subset) $AB$ and those of $A\cap B$ in $A$.
Spoiler

 Try $aB \mapsto a(A \cap B)$. Remember to show it's well defined first of all.

A: Let $a_1,a_2\in A$ and $b_1,b_2\in B$ and $$a_1.b_1=a_2.b_2...........(1)$$
$$\Rightarrow a_2^{-1}a_1=b_2.b_1^{-1}$$
$$\Rightarrow a_2^{-1}a_1\in (A\cap B)\text{ and }b_2.b_1^{-1}\in (A\cap B)$$
$$\Rightarrow a_1(A\cap B)=a_2(A\cap B) \text{ and } b_1(A\cap B)=b_2(A\cap B)$$
And similarly the other way round (i.e. if they belong to the same coset then (1) follows).
From this reasoning the formula follows.This is because if for some $a_1,a_2\in A$ and $b_1,b_2\in B$, $a_1.b_1=a_2.b_2$ then $a_1,a_2$ must be in the same coset of $(A\cap B)$ so here for finding the total no. of elements in $AB$ we have to count the total no. of elements of the form $ab, a\in A,b\in B$ we have to devide it by the total no. of its repeted representations.
$$|AB|=(|A| \times |B|)\times \frac{1}{\text{no of repeated reresentation of each}}={(|A| \times |B|)}\times \frac{1}{|A\cap B|}$$
