# Subgroups 0f abelian-by-finite groups are abelian-by-finite!

I already have this answer: If $\mathcal X$ and $\mathcal Y$ are classes of groups that are closed under taking subgroups (such as abelian groups and finite groups: any subgroup of an abelian group is abelian, and any subgroup of a finite group is finite) then the class of $\mathcal X$-by-$\mathcal Y$ groups (i.e., groups $G$ with a normal subgroup $N$ such that $N$ is in $\mathcal X$ and $G/N$ is in $\mathcal Y$) is closed under taking subgroups.

For let $H$ be a subgroup of such a group $G$. Then $H∩N$ is a normal subgroup in $\mathcal X$, since it's a subgroup of $N$, and $H/(H∩N)$ is in $\mathcal Y$, since it's isomorphic to a subgroup of $G/N$.

Can you please explain why: $H\cap N$ is a normal subgroup in $\mathcal X$ and why $H/(H \cap N)$ is finite and why it is isomorphic to a subgroup of $G/N$??

Hint: apply the second isomorphism theorem: $HN/N \cong H/(H \cap N)$. So $HN/N$ being a subgroup of $G/N$ is finite. In addition, $H \cap N$ is normal in $H$, since $N$ is normal in $G$. And of course $H \cap N$ is abelian, being a subgroup of $N$.