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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.

  • 2
    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".2011-10-09
  • 0
    Good to know. Thank you Jack.2011-10-10

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