Let G be a finite group, and suppose that A and B are subgroups of G, prove that $\displaystyle |AB| = \frac { |A||B|} { |A \bigcap B |} $

Proof so far:

Now, by the second iso theorem, $\displaystyle \frac {|AN|}{|N|} = \frac {|A|}{|A \bigcap N|} $ if N is a normal subgroup of AN.

So since B is a normal subgroup of AB, this would be true, right?

This thing look too simple, so I probably did something wrong, please check..

Thank you.