TITLE: Group isomorphic to all its proper quotients, and not simple
QUESTION [5 upvotes]: Let $G$ be a group such that for any proper normal subgroup $H \subset G$, we have $G/H \cong G$ (of course this isomorphism may not be given by the projection $G \to G/H$). Does it follow that $G$ is a simple group? What can we say about such a group $G$ in general?
Notice that $G/H \cong G$ doesn't imply $H = \{e\}$, for instance $z \mapsto z^2$ is a surjective morphism $\Bbb C^{\times} \to \Bbb  C^{\times}$ of kernel $\{\pm 1\}$.

REPLY [8 votes]: $\newcommand{\C}{\mathbb{C}}$$\newcommand{\Set}[1]{\left\{ #1 \right\}}$The Prüfer $p$-group is non-simple, and satisfies your hypothesis.
Given a prime $p$, you realize it as the subgroup of the multiplicative group $\C^{*}$ given by
\begin{equation*}
  G = \Set { z \in \C^{*} : z^{p^{k}} = 1 \ \text{for some $k \ge 0$}}.
\end{equation*}