Let $G_1, G_2$ be subgroups of a group $G$ and $G_1G_2 = \{g_1g_2: g_1 \in G_1, g_2 \in G_2\}$. Prove that $|G_1| \cdot |G_2| = |G_1G_2| \cdot |G_1 \cap G_2|$.
I've thought about cosets and considered sets $g_1G_2$ for all $g_1 \in G_1$. Sum of all these sets is equal to $G_1G_2$ and each element is counted the same number of times, $|G_1 \cap G_2|$ presumably, but I can't seem to prove this fact nor see a connection.
Define $\varphi: G_1\times G_2\to G_1G_2$ by $\varphi(g_1,g_2)=g_1g_2$. The map $\varphi$ is obviously surjective. Now let $g_1\in G_1,g_2\in G_2$. For all $x\in G_1\cap G_2$ have $\varphi(g_1x,x^{-1}g_2)=g_1g_2$. The other direction is also correct: if $\varphi(h_1,h_2)=g_1g_2$ then there must be some $x\in G_1\cap G_2$ such that $(h_1,h_2)=(g_1x,x^{-1}g_2)$. Why is it true? Well, assume that $\varphi(h_1,h_2)=g_1g_2$. It means that $h_1h_2=g_1g_2$, and hence $g_1^{-1}h_1=g_2h_2^{-1}$. So now let $x=g_1^{-1}h_1=g_2h_2^{-1}$. This element is in $G_1\cap G_2$ (because $g_1^{-1}h_1\in G_1$ and $g_2h_2^{-1}\in G_2$), and indeed $h_1=g_1x, h_2=x^{-1}g_2$. Alright, so this proves that $|\varphi^{-1}(g_1g_2)|=|G_1\cap G_2|$. In other words, for all $y\in G_1G_2$ we have $|\varphi^{-1}(y)|=|G_1\cap G_2|$. The result follows.
Hint: Given $g_1g_2\in G_1G_2$, the same element is counted as all different possible $(g_1g)(g^{-1}g_2)$ for each distinct $g\in G_1\cap G_2$. That's the connection.
