# Example of a Non-Abelian Group

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?

Notation

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 $G$.

• It would be good to include the definition of "subnormal", as it's a simple concept, but probably not a familiar term to everybody. – Slade Oct 24 '13 at 23:56
• This would be an interesting question for infinite groups! – Derek Holt Oct 25 '13 at 10:33
• @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? – 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.