Suppose that $G$ is a group and $A \lhd G$. Prove that if $A$ is abelian and $G=AH$ for some subgroup $H$ of $G$, then $A \cap H \lhd G$

Question

Suppose that $G$ is a group and $A \lhd G$. Prove that if $A$ is abelian and $G=AH$ for some subgroup $H$ of $G$, then $A \cap H \lhd G$

So, I need to prove that $H \cap A \lhd G$

How do I go about showing that $N_G(H \cap A)= G$ and then I'm assuming that I would have to show $A \subseteq N_G(H \cap A)$,and $H \subseteq N_G(H \cap A)$

Answer 1

First of all note that $A\cap H$ is a subgroup of $G$ (intersections of subgroups are always subgroups). Now let $g\in G$. Then $g=ah$ for some $a\in A$ and some $h\in H$. Now let $x\in A\cap H$. Then $$g^{-1}xg=h^{-1}a^{-1}xah$$ Now $a^{-1}xa=x$, since $x\in A$ and $A$ is abelian. Also since $x\in H$, $h^{-1}xh\in H$, and since $A$ is normal, $h^{-1}xh\in A$. So $$g^{-1}xg=h^{-1}xh\in H\cap A$$ Hence $H\cap A$ is normal in $G$.

Answer 2

$\bullet$ If $G = AH$, then we can write every element as a product $ah$, where $a \in A$, and $h \in H$.

$\bullet$ If something is in $A \cap H$, it's in $H$.

So, see what happens when you pick elements $a \in A$ and $h_1, h_2 \in H$ and compute $h_1^{ah_2}$. Can you guarantee it's in $H$? Can you guarantee it's in $A$ as well? (For this one, you'll need to remember that $h \in A\cap H$, hence $A$, to begin with).