Let $G= N \rtimes_{\varphi} Q$ be a semidirect product coming from a short exact sequence $$ 1 \rightarrow N \rightarrow G \rightarrow Q \rightarrow 1 $$and assume that $N$ and $Q$ are finitely generated abelian groups.
Suppose that there is $N_{0}$ a finite index subgroup in $N$ such that $\varphi_{|N_{0}}$ is the trivial homomorphism. That is, $Q$ acts trivially on $N_{0}$ by conjugation.
My question is whether $N \rtimes_{\varphi} Q $ is VIRTUALLY abelian.
My attempt is the following:
Suppose that $N= N_{0}g_{1} \dot{\cup} \cdots \dot{\cup} N_{0} g_{n}$ for some $g_{1},\dots,g_{n}\in N$ (this comes from the fact that $N_{0}$ has finite index in $N$).
Since $N_{0}$ is not necessarily normal in $G$, we cannot take the product $N_{0}Q$. Nevertheless, we can take the subgroup generated by $N_{0}$ and $Q$, $\langle N_{0}, Q \rangle$.
I would like to show that $\langle N_{0}, Q \rangle$ has finite index in $G$. Note that $G= NQ= QN$. Thus, for any $g\in G$, $$g= qn,$$ for some $q\in Q, n\in N$.
But since $n\in N$, $n= n_{0}g_{i}$ for some $n_{0}\in N_{0}$, $i\in \{1,\dots,n\}$. Thus, $$G= \langle N_{0},Q \rangle g_{1} \cup \dots \cup \langle N_{0}, Q \rangle g_{n}.$$
In conclusion, $\langle N_{0}, Q \rangle$ has finite index in $G$. In addition, $\langle N_{0}, Q \rangle$ is abelian because $N_{0}$ and $Q$ are abelian and $Q$ acts trivially on $N_{0}$.