Prove that if $G$ is an $N$-group, then either (i) $G$ is solvable, or (ii) $G$ has a unique minimal normal subgroup $K$, the factor group $G/K$ is solvable, and $K$ is simple.
Suppose that $G$ is an $N$-group and $G$ is not solvable. Then we know that $G$ can not have a normal $P$ group. Also since for any non trivial normal subgroup $N$, let $P$ be a sylow subgroup of $N$, then by the Fratini Argument $G = N_G(P)N$. So by the second isomorphism theorem $G/N = N_G(P)N/N \cong N_G(P)/N \cap N_G(P)$. So since $N_G(P)$ is solvable and quotients of solvable groups are solvable $N_G(P)/N \cap N_G(P)$ is solvable, and hence $G/N$ is solvable.
So it suffices to show that if $G$ is an $N$-group and $G$ is not solvable that $G$ has a minimal normal subgroup. Since $G$ is not solvable we know that if we keep taking the commutator subgroup of the previous commutator subgroup we eventually get for some integer $n$ that $G^{n} = G^{n+ 1} \neq 1$ where $G^n$ is the $n^{th}$ commutator subgroup. Now we have that $G^n$ is characteristic and hence normal.
So I am convinced that I want to let $K$ be a minimal normal subgroup of $G^{n}$, since $
Best Answer
So here is a full proof to complete the question.
Let $M$ be a minimal normal subgroup of $G$. Then $M$ is the direct product of isomorphic simple groups $S$. If $S$ is cyclic then $M$ is a $p$-subgroup such that $N_G(M)=G$, so $G$ is soluble. Thus $S$ is non-abelian and $M$ is the direct product of $n$ copies of $S$. If $Q$ is a Sylow $p$-subgroup of $S$ for some $p\mid |S|$ then $n-1$ copies of $S$ appear in $N_G(Q)$. Since $N_G(Q)$ is soluble though, this means $n=1$. Thus $M$ is non-abelian simple.
Finally, by a Frattini argument $G=M\cdot N_G(Q)$, where $Q\in\mathrm{Syl}_p(M)$. Since $N_G(Q)$ is soluble by assumption, and $G/M\cong N_G(Q)/N_M(Q)$, we see that $G/M$ is soluble.