Sort-of-simple non-Hopfian groups

Question

A finite simple group is one which has no homomorphic images apart from itself and the trivial group. However, the simple-groups tag does not include the condition "finite". My question is the following.

Is the following true?

Claim: A simple group is one where, up to isomorphism, the only homomorphic images are itself and the trivial group.

However, I am asking this question not because of the tag wiki, but because I think any counter-example would be interesting, especially if it was finitely generated (and especially especially if it was finitely presented). That is,

Does there exists a (finitely presented) group $G$ which is not simple but where the only homomorphic images, up to isomorphism, are itself and the trivial group?

Such a counter-example would be non-Hopfian, and would be sort-of-simple, by the definition of a simple group that we want to exist. Thus the title of the question.

Answer 1

An infinitely generated example is the Prüfer group $\mathbb{Z}[\frac{1}{p}]/\mathbb{Z}$.

But there are no finitely generated examples. For if $G$ is such a group, and $S$ a generating set of minimal size, then no proper normal subgroup $N$ can contain any elements of $S$, or the remaining elements of $S$ would give a smaller generating set of $G/N\cong G$. So the union of any chain of proper normal subgroups contains no element of $S$ and is therefore a proper normal subgroup. By Zorn's Lemma, there is a maximal proper normal subgroup $N$. But then $G\cong G/N$ is simple.

Answer 2

Here is a nonabelian (countable) example:

Let $S$ be a permutation group on a set $X$ with distinguished point. By the wreath product $G\wr S$ I thus mean $G^X\rtimes S$, where $S$ permutes the factors, and I endow it with the embedding of $G$ to the factor of $G^X$ corresponding to the distinguished point.

Now assume $S$ to be the alternating group on a set $X$ with at least 5 points. Define $S_0=1$, $S_1=S$, and $S_{n+1}=S_n\wr S$. (This is a subgroup of the automorphism group of a regular rooted tree of depth $n$: if we were considering the full symmetric group we'd get the whole automorphism group.)

Define $S_\infty$ as the inductive limit (at this point it's important we were precise how $S_n$ is embedded into $S_{n+1}$).

Then $S_\infty$ is the desired example.

To prove it, we first need to describe the normal subgroups of $S_n$ for all $n$: they form an increasing chain $N_{n,k}$, $0\le k\le n$, where $N_{n,n}=S_n$, and for $k\le n-1$, $N_{n,k}$ is the subgroup $N_{n-1,k}^X$ of $S_{n-1}\wr S$ (this is the kernel of the projection onto $S_{n-1}\wr S\to (S_{n-1}/N_{n-1,k})\wr S$). A simple induction (using that $S$ is simple nonabelian) shows that these are the only normal subgroups of $S_n$ for all $n$.

Let $f_n$ be the embedding $S_n\to S_\infty$. Then for fixed $k$, the subgroups $f_n(N_{n,k})$ form an increasing chain and its union $N_k$ is a normal subgroup of $S_\infty$. Then it is easy to see that the only normal subgroups of $S_\infty$ are the $N_k$ and $S_\infty$ itself (strategy: if $N$ is a proper normal subgroup, choose $n$ such that $N$ does not contain $S_n$; then $N\cap S_n=N_{n,k}$ for some $k<n$; then it is immediate that $N\cap S_m=N_{m,k}$ for all $m\ge n$).

Next, we have, for all $n\ge 1$, a surjective homomorphism $p_n:S_n\to S_{n-1}$, defined by induction: for $n=1$ it's the map to the trivial group; if $n\ge 2$, it's defined by considering $p_{n-1}\times\dots\times p_{n-1}:S_{n-1}^X\to S_{n-2}^X$ and extend it to the wreath products as the identity on $S$. The kernel of $p_n$ is $N_{n,1}$ (and iterations have kernel $N_{N,2}$, etc). Then since the the choice of embedding, $p_n$ extends $p_{n-1}$, it extends to a surjective endomorphism $p$ of $S_\infty$ whose kernel is $N_1$, and more generally the kernel of $p^i$ is $N_i$. So $S_\infty/N_i$ is isomorphic to $S_\infty$ for all $i$.