By minimal non-solvable group, I mean a non-solvable group whose all proper subgroups are solvable.
Let $G$ be a minimal non-solvable group. If $G$ is not a simple group, then it can be easily verified that $G$ contains only one maximal normal subgroup say $N$. Moreover, in this case we can prove that for each nontrivial normal subgroup $H$ of $G$, we have $N\cap H=N$ or $H$, since otherwise $NH=G$ which contradicts nonsolvablity of $G$. More precisely, any other possible proper normal subgroup of $G$ is incuded in $N$. Now my question:
Does there exist any example of a minimal nonsolvable group $G$ such that $G$ has two nontrivial proper normal subgroups?
Best Answer
A natural place to look for examples would be to look for a minimal simple group with non-cyclic Schur multiplier.
The Schur multiplier of the Suzuki group $S = {}^2B_2(8)$ is elementary abelian of order $4$.
It is possible to verify (for example by constructing this group in GAP or Magma) that the covering group $G = 2^2.S$ is minimal non-solvable. (Probably there are non-computational ways to do this which are not too difficult)
In any case here the center is elementary abelian of order $4$, which gives you four nontrivial proper normal subgroups.
EDIT: Here is a general argument. Let $S$ be a finite simple group, and let $G$ be a perfect central extension $$1 \rightarrow Z \rightarrow G \rightarrow S \rightarrow 1.$$ Consider a maximal subgroup $M < G$. If $Z \not\leq M$, then $G = MZ$. Since $Z$ is central, then $M$ is normal. But then $M$ being a maximal subgroup implies $G/M$ is cyclic of prime order, contradicting the assumption that $G$ is perfect.
So every maximal subgroup $M < G$ satisfies $Z < M < G$ and $M/Z$ is maximal in $G/Z \cong S$.
Therefore if $G$ is a perfect central extension of a minimal simple group $S$, then $G$ is a minimal non-solvable group.
Then for the examples that the question asks for, just find a minimal simple group $S$ such that the Schur multiplier is not cyclic of prime order, and not trivial.