Group isomorphic to all its proper quotients, and not simple

Question

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 $\{\pm 1\}$.

How does $G/H$ trivial not imply $G=H$? If $g\in G\setminus H$, then $gH\ne 1H$ in $G/H$. — Hagen von Eitzen, Jun 06 '18 at 13:46
You write non-trivial, but I think it's proper you really mean. — Andreas Caranti, Jun 06 '18 at 13:53

Andreas Caranti · Accepted Answer · 2018-06-06T13:59:52.070

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$\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*}

edited Jun 06 '18 at 13:59

answered Jun 06 '18 at 13:57

Andreas Caranti

68,873

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@Alphonse the wikipedia reference basically has it all, in particular it lists all the subgroups, which are normal, $G$ being abelian. – Andreas Caranti Jun 06 '18 at 14:02
A possibly hard question is whether these are the only examples. – YCor Jun 06 '18 at 23:52
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@YCor You have previously solved your possibly hard question: https://math.stackexchange.com/a/1985569/10513 – user1729 Jun 07 '18 at 09:17
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@user1729 oops, thanks very much, I forgot this! – YCor Jun 07 '18 at 09:52
@YCor, brilliant, thanks! – Andreas Caranti Jun 07 '18 at 13:20

Group isomorphic to all its proper quotients, and not simple

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