Show that $AB$ is a subgroup of $G$

Question

I need to prove the following:

Let $G$ be a group with subgroups $A$ and $B$ with the property that $A$ is normalized by $B.$ Then the subset $AB:=\{ab:a\in A, b\in B\}$ is a subgroup of G.

Closure is easy: for $a_1b_1,a_2b_2 \in AB, a_1b_1a_2b_2=a_1(b_1a_2b_1^{-1})b_1b_2.$ By Normality of $A$ the term inside the parenthesis is in $A.$ Hence, $a_1(b_1a_2b_1^{-1})\in A.$ It is clear that $b_1b_2\in B.$ Hence the product is in $AB.$

I cannot prove the existence of identity and inverse although it seemed easy at first.

Answer 1

Hints:

Prove the lemma:

If $\;A,B\;$ subgroups of $\;G\;$ , then $\;AB\;$ is a subgroup iff $\;AB=BA\;$.

In our case:

$$\forall\,a\in A\;\forall\,b\in B\;\exists\, a'\in A\;\;s.t.\;\;b^{-1}ab=a'\implies ab=ba' $$

Develop a little the above to show that in this case $\;AB=BA\;$ and then apply the lemma.