I'm interested in knowing whether certain groups $G$ are metabelian.
In general, my groups $G$ have the following form: there is an exact sequence $1\to N\to G\to Q\to 1$ where $N$ is abelian, and $Q=K\rtimes H$ with $K$ and $H$ abelian.
Clearly $G$ is soluble of length 3. Moreover I know that the derived subgroup $G'$ centralizes $N$ and that the derived subgroup $Q'$ centralizes $K$.
My obvious idea is to change the above exact sequence obtaining an abelian group $A$ (containing $G'$) such that $G/A$ is abelian. Can someone please give me a hand on how to proceed?
Thanks in advance.