Let $G\neq \{e\}$ be a finite non-abelian group. We say $G$ is quasi-simple if $G=[G,G]$ and $G/Z(G)$ is simple.
Let $G$ be a quasi-simple group. It is not so difficult to show that if $N$ is a non-trivial normal subgroup of $G$, then $N\leq Z(G)$. Indeed if $N$ is a normal subgroup of $G$ and $N$ is not contained in $Z(G)$, then necessarily $[G,G]\leq N$, whence $N=G$.
Now my question is about its converse. Suppose $G$ is a finite non-abelian group such that all its non-trivial normal subgroups are central. Can we conclude that $G$ is quasi-simple?
My attempt : Given the hypothesis it is immediate that $G/Z(G)$ is a finite simple group. We want to show $[G,G]=G$. If not then once again by hypothesis $[G,G]\leq Z(G)$, which implies that $G/Z(G)$ is abelian. Since it is also simple, we conclude that $G/Z(G)$ is cyclic, whence $G$ is abelian, a contradiction. Is this correct or am I saying something silly here!
Thanks in advance for any kind of comments or suggestions.