So I'm guessing AB is just defined as $\displaystyle AB : = \{ ab \, \, | \, \, a \in A, b \in B \}$If A is an abelian group with $\displaystyle A \unlhd G$ and B is any subgroup of G, prove that $\displaystyle A \cap B \unlhd AB$

I started off with:

Let $\displaystyle x \in A \cap B, a \in A, b \in B$

We must show that $\displaystyle (ab) x (ab)^{-1} \in A \cap B$

$\displaystyle (ab) x (ab)^{-1} = ab x b^{-1} a^{-1}$

And now here I'm stuck...