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 $g\in AB$: $\begin{align*} gkg^{-1}&=(ab)k(ab)^{-1} &&\text{ (where }a\in A \text{ and } b\in B)\\ &=a(bkb^{-1})a^{-1}\\ &=ak'a^{-1}&&\text{ (Since } A\triangleleft G \, , k'=bkb^{-1}\in A)\\ &=k'aa^{-1}&&\text{ (Since } A \text{ is Abelian )}\\ &=k'e\\ &=k'\in A\\ &\implies gkg^{-1}=k'\in A\cap B\\ \end{align*}$