TITLE: Example of a Non-Abelian Group
QUESTION [5 upvotes]: 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 $G$.

REPLY [7 votes]: 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.