A minimal normal subgroup of a solvable group is abelian.
I need to prove that this stands. However, I don't seem to understand the necessity of the group being solvable. Could you point out my mistake?
My proof goes as follows:
Let $H\triangleleft G$ be a minimal normal subgroup of $G$. Suppose by contradiction $H$ isn't abelian. Then there exist $h_{1},h_{2}\in H$ s.t. $[H,H]\ni\left[h_{1},h_{2}\right]\neq e$. Since the commutator subgroup is always normal in the group, we get that $\{e\}\neq [H,H]\triangleleft H$ in contradiction to $H$ being a minimal subgroup.
Note – our exercises tend to be imprecise, so the question itself might be faulty.
Best Answer
In any group $G$, if $H$ is a minimal normal subgroup of $G$ then your argument yields that $[H,H]=\{e\}$ or $[H,H]=H$.
If $[H,H]=\{e\}$ then $H$ is abelian; if $[H,H]=H$ then $H$ is perfect, so it is not solvable.
Since a subgroup of a solvable group is solvable, the latter possibility is excluded, so we conclude that $H$ is abelian.