Prove that every finite group G contains a (unique) soluble normal subgroup N such that G/N has no nontrivial abelian normal subgroups. > Prove that every finite group $G$ contains a (unique) soluble normal subgroup $N$ such that $G/N$ has no nontrivial abelian normal subgroups. Thanks a lot. To show that $G/N$ has no nontrivial abelian normal subgroup, I think we can use the fact that if $H$ is a group such that $N$ is normal in $H$ with the property that $H/N$ and $N$ are soluble then $H$ is soluble. Choose $H$ such that $H/N$ is abelian.

If $M$ and $N$ are normal soluble subgroups of a group $G$, then $MN$ is again a soluble normal subgroup. For $MN/N \cong M/(M \cap N)$, and $M/(M \cap N)$ is soluble as quotient of the soluble $M$, so $MN/N$ is soluble, $N$ is soluble, so (also according to your valid remark) $MN$ must be soluble. This shows that in every group there is a unique maximal soluble normal subgroup (it could be trivial). As you did call this group $N$.

Now assume $K/N$ is a non-trivial abelian normal subgroup of $G/N$. Then $N \lt K \lhd G$, with the commutator subgroup $K' \subseteq N$. Since $N$ is soluble, $K'$ is soluble and hence $K$ must be soluble. By the soluble normal maximality of $N$, $K \subseteq N$, which is a contradiction.