I was hunting an example of a non-trivial finite group in which

1) All non-trivial normal subgroup are non-abelian.

2) There exists a nontrivial subnormal abelian subgroup.

Is there any hope to find this out?


A subgroup $H$ of a given group $G$ is a subnormal subgroup of $G$ if there is a finite chain of subgroups of the group, each one normal in the next, beginning at H and ending at $G$.

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    $\begingroup$ It would be good to include the definition of "subnormal", as it's a simple concept, but probably not a familiar term to everybody. $\endgroup$ – Slade Oct 24 '13 at 23:56
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    $\begingroup$ This would be an interesting question for infinite groups! $\endgroup$ – Derek Holt Oct 25 '13 at 10:33
  • $\begingroup$ @DerekHolt In fact I ask below, as a comment to the answer, what will happen in infinite case. It will be better to open a new question or edit this one? $\endgroup$ – W4cc0 Oct 25 '13 at 10:38

In a finite group $G$, a subgroup $H\le G$ is contained in the Fitting subgroup (denoted $\mathbf{F}(G)$) if and only if $H$ is both subnormal and nilpotent.

If a finite group $G$ has an abelian subnormal subgroup $K\le G$, then $K\le \mathbf{F}(G)$. Because $\mathbf{F}(G)$ is non-trivial, we also have its center is non-trivial: $Z(\mathbf{F}(G))\neq\lbrace1\rbrace$. This is because $\mathbf{F}(G)$ is nilpotent.

But then $Z(\mathbf{F}(G))$ is an abelian normal subgroup of $G$. Hence no example - as outlined in the question - exists.

  • $\begingroup$ Oh thanks Steve D! Can I ask you also an example in the infinite case? $\endgroup$ – W4cc0 Oct 25 '13 at 8:10
  • $\begingroup$ @W4cc0 Maybe you should ask that as a new question. $\endgroup$ – Alexander Gruber Oct 26 '13 at 1:31

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