If $Z(G)$ is the only nontrivial normal subgroup of $G$, show $G/Z(G)$ is simple

Question

I was hoping to approach by saying if you mod out the only nontrivial normal subgroup, then the quotient group has no nontrivial normal subgroups. But maybe that's an oversimplification.

Thanks

Answer 1

Let $N$ be a normal subgroup of $G/Z(G)$.
By Correspondence Theorem, $N=H/Z(G)$ where $Z(G)\lhd H \lhd G$.
By hypothesis, $H=Z(G)$ or $H=G$.
Hence $N=\{Z(G)\}$ or $N=G/Z(G)$.

In general,

$N$ is a maximal normal subgroup of $G$ iff $G/N$ is simple.

Answer 2

Let $p:G\rightarrow G/Z(G)$ the canonical projection and $N$ a proper normal subgroup of $G/Z(G)$, $p^{-1}(N)$ is a proper normal subgroup of $G$, it is $Z(G)$ by hypothesis this implies that $N=e$.