The intersection of an abelian normal subgroup $A$ and subgroup $B$ is normal in $AB$ 
Let $G$ be a finite group with two subgroups $A$ and $B$. Show that if $A$ is Abelian and is normal in $G$, then $A\cap B$ is normal in $AB.$ 

It's been awhile since I did Abstract Algebra, (this is a prelim question) the above seems easy but still I want to check my proof:
We will first show that $A\cap B\leq AB.$ Observe that $A\cap B$ is a subset of $AB$ since if $x\in A\cap B$ then $x\in A$ and $x\in B$ and thus( for $x\in A $ and $e\in B$) $x\in AB=\{ab: a\in A \text{ and } b\in B\}.$ Clearly, $e \in A\cap B.$ Suppose that $x,y\in A\cap B.$ Then $x,y\in A$ and $x,y\in B.$ Since $A,B\leq G$, we have that $xy\in A$ and $xy\in B$, and thus $xy\in A\cap B.$ Finally, if $x\in A\cap B$ then $x^{-1}\in A\cap B$ since $x,x^{-1}\in A$ and $x,x^{-1}\in B.$ Now we will show that $A\cap B \triangleleft AB.$ Observe that if $k\in A\cap B$ and $g\in AB$ then it suffices to show that $gkg^{-1}\in A\cap B.$ Now consider, $\forall k\in A\cap B$ and $g\in AB$:
$$\begin{align*}
gkg^{-1}&=(ab)k(ab)^{-1} &&\text{ (where }a\in A \text{ and } b\in B)\\
&=a(bkb^{-1})a^{-1}\\
&=ak'a^{-1}&&\text{ (Since } A\triangleleft G \, , k'=bkb^{-1}\in A)\\
&=k'aa^{-1}&&\text{ (Since } A \text{ is Abelian )}\\
&=k'e\\
&=k'\in A\\
&\implies gkg^{-1}=k'\in A\cap B\\
\end{align*}$$
 A: The first part is too complicated: $A\subseteq AB$ (whether or not $AB$ is a group), and $A\cap B\subseteq A$; so you are done as far as inclusion; and intersection of subgroups is always a subgroup, so $A\cap B$ is a subgroup and is contained in $AB$. Of course, you should note that $AB$ actually is a subgroup, because $AB=BA$ by the normality of $A$.
Your final assertion does not follow. You have justified correctly that $k'\in A$; but you have not said a single word as to why it is in $B$. So how do you justify your assertion that $k'\in A\cap B$? 
The reason it is in $B$ is that $k\in A\cap B$, so $k\in B$. Therefore, $bkb^{-1}$, being a product of elements in $B$, must be in $B$. Since it is also in $A$ (by the reasons given), you h ave that $k'=bkb^{-1}\in A\cap B$.
A: Here is another solution: I don't mean that it's better, but I think you might pick up some tricks from it.
When you want to prove normality of a subgroup, it's sometimes useful (and certainly logical) to look at $N_G(H):=\{g\in G\mid gHg^{-1}\subseteq H \}$. (This is called the normalizer of $H$ in $G$.


*

*Lemma: $N_G(H)$ is a subgroup of $G$, and in fact it is the largest subgroup of $G$ in which $H$ is normal.

*Lemma: The product of a subgroup of $G$ and a normal subgroup of $G$ is a subgroup of $G$. Thus, $AB$ is a subgroup of $G$.

*Since $A\lhd G$, then certainly $A\lhd AB$. By an isomorphism theorem$^\dagger$, $A\cap B\lhd B$.

*$A\cap B$ is certainly normal in $A$ (and so is every other subgroup of $A$: Why?)

*Observe $A$ and $B$ are in $N_G(A\cap B)$, hence the subgroup $AB\subseteq N_G(A\cap B)$. Done! (by the first lemma!)
$\dagger$ The one I am thinking of says if $A\lhd G$ and $B<G$, then $A\cap B\lhd B$ and $AB/A\cong B/(A\cap B)$.
