Is the non-split central extension of $\mathbb{Z}^2$ by $\mathbb{Z}_2$ virtually $\mathbb{Z}^2$? Recall that $H^2(\mathbb{Z}^2;\mathbb{Z_2})=\mathbb{Z}_2$. Using the theory of group extensions  we can construct a non-split central extension $\mathbb{Z}_2\hookrightarrow\Gamma\twoheadrightarrow\mathbb{Z}^2$.  Here the $\mathbb{Z}_2$ is central in $\Gamma$.
The following is a presentation for $\Gamma$,
$\langle a,b,x\ |\ [a,x]=[b,x]=x^2=1, [a,b]=x\rangle$.
Is $\Gamma$ virtually $\mathbb{Z}^2$? If so what is a set of generators for such a subgroup and how can we determine them?
 A: Firstly, note that the subgroup $N=\langle a^2, b^2\rangle$ is free abelian of rank two. to prove this, it is enough to show that $a^2$ and $b$ commute. So note that $axax=a^2$, and we have:
$$
\begin{align*}
a^2b&=a\cdot ab\\
&=a\cdot bax\\
&=ab\cdot ax\\
&=bax\cdot ax\\
&=ba^2
\end{align*}
$$
as required.
Lets call $G:=\langle a, b, x\rangle$. We now need to prove that $N$ is of finite index in $G$. Firstly, recall that $a^2$ and $b$ commute, and also (by symmetry) $a$ and $b^2$ commute. This implies that $N$ in closed under conjugation by $a$ and by $b$, while it is closed under conjugation by $x$ as $x$ commutes with both $a^2$ and $b^2$. Therefore, $N$ is a normal subgroup of $G$. Then $G/N$ is a finite group of order at most $8$: it has presentation
$$\langle a, b, x\mid a^2, b^2, x^2, [a, b]=x, [a, x], [b, x]\rangle$$
and elements of this group may be written in the form $a^pb^qx^r$, as $a$ and $b$ commute at the cost of an $x$. As there are $2$ choices for each of $p$, $q$ and $r$, we have that $|G/N|\leq8$. [I guess it is actually $8$ on the nose, but I'll leave you to think about that bit :-)]
