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

Is the following true?

Claim: A simple group is one where 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 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.

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  • $\begingroup$ Apparently these exist but are difficult to construct. The Wikipedia article Simple group references an example by Graham Higman (1951) and some other examples. $\endgroup$ – hardmath Apr 25 '14 at 10:07
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    $\begingroup$ I am not sure what you mean by "counter-example". Counter-example to what? Why would you imagine that the definition of a simple group applies only to finite groups? Also, I don't understand your comment about non-Hopfian. Simple groups are clearly Hopfian. You could also search for *Thompson groups" for interesting examples of finitely presented infinite simple groups. $\endgroup$ – Derek Holt Apr 25 '14 at 10:11
  • $\begingroup$ @hardmath I cannot find what you are talking about in this article. The only reference to Higman I can find is his construction of an infinite simple group (and that paper is the only cited paper in the only references). $\endgroup$ – user1729 Apr 25 '14 at 10:12
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    $\begingroup$ OK, sorry. I've got it now. I'll think about it. If you get no answer within a day or so then you could ask it on MathOverflow. $\endgroup$ – Derek Holt Apr 25 '14 at 10:20
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    $\begingroup$ @hardmath Does there exist a non-simple group $G$ such that every proper, non-trivial normal subgroup $N$ is such that $G/N\cong G$. (The point of my question is that you take a statement of normal subgroups and naively translate it into terms of homomorphic images, and my question is "is this naive translation actually correct?". So translating the claim back to normal subgroups would be...difficult...translating the possible counter-example is easier, which is what my first sentence is doing.) $\endgroup$ – user1729 Apr 25 '14 at 11:44
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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.

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  • $\begingroup$ Yeah, okay, thanks. I'd love a finitely presented example though :-) $\endgroup$ – user1729 Apr 25 '14 at 12:53
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    $\begingroup$ @user1729: I've edited my answer to add a proof that there are no fin.gen. examples. $\endgroup$ – Jeremy Rickard Apr 25 '14 at 14:22
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    $\begingroup$ It would also be interesting to know whether there are any nonabelian examples. $\endgroup$ – Derek Holt Apr 25 '14 at 14:26
  • $\begingroup$ This is really very nice! Would I be pushing my luck if I asked you to not assume AoC?... $\endgroup$ – user1729 Apr 25 '14 at 16:05
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    $\begingroup$ @user1729 you don't need AC to prove that nontrivial finitely generated groups admit simple quotients (see my comment to math.stackexchange.com/questions/1324763) $\endgroup$ – YCor Oct 26 '16 at 8:17
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Here is a nonabelian (countable) example:

Let $S$ be a permutation group on a set $X$ with distinguished point. By $G\wr S$ I thus mean $G^X\rtimes S$, where $S$ permutes the factor, 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 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 $N_{n-1,k}\wr S$ of $S_{n-1}\wr S$. 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 $i$.

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