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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.

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G=SL(2,3) has N of order 2, K elementary abelian of order 4, and H of order 3. G' centralizes N=Z(G), and of course Q' centralizes K = Q', since K is abelian. However, G is not metabelian. There are many other counterexamples. Also, S4 has the form as in your exact sequence, but G' does not centralize G". –  Jack Schmidt Oct 9 '11 at 6:20
    
Good to know. Thank you Jack. –  Charles Oct 10 '11 at 3:43
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1 Answer

$G=\operatorname{SL}(2,3)$ has $N=Z(G)$ of order 2, $K=Q_8/N$ elementary abelian of order 4, and $H$ of order 3. $G'$ centralizes $N=Z(G)$, and of course $Q'=K$ centralizes $K = Q'$, since $K$ is abelian. However, $G$ is not metabelian.

There are many other counterexamples, so I don't see how to fix it.

Also, your “I know” refers to extra hypotheses: $S_4$ has the form specified in your exact sequence, but $G'$ does not centralize $G''$

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