I would like an easy example of an infinite simple group along with an embedding into a finitely presented group. I know that there are infinite simple finitely presented group such as Thompson’s group $V$ and I know that any group with solvable word problem embeds into a simple subgroup of a finitely presented group, but I am looking for an example of a more elementary nature.

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    $\begingroup$ One example is Burger-Mozes's group in "Finitely presented simple groups and products of trees", ethz.ch/content/dam/ethz/special-interest/math/department/…. $\endgroup$ Aug 13, 2019 at 0:22
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    $\begingroup$ And there are many more examples, see en.wikipedia.org/wiki/Simple_group#Infinite_simple_groups. $\endgroup$ Aug 13, 2019 at 0:29
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    $\begingroup$ That Burger-Mozes groups are finitely presented is somewhat immediate once constructed. But the construction, and the proof of simplicity are both subtle. $\endgroup$
    – YCor
    Aug 13, 2019 at 6:07
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    $\begingroup$ The group $A$ of alternating finitely supported permutations of an infinite countable set embeds into Houghton's group $H_3$. The latter is quite easy to construct, and that it is finitely generated and contains $A$ is easy, and that $A$ is simple is also easy. That $H_3$ is finitely presented is more technical with some combinatorial arguments. $\endgroup$
    – YCor
    Aug 13, 2019 at 6:09
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    $\begingroup$ So you're looking for a simple simple group? $\endgroup$
    – Ian Agol
    Aug 13, 2019 at 6:57

1 Answer 1


I think D. L. Johnson's article Embedding some recursively presented groups should answer your question. The abstract is:

We seek to illustrate the Higman Embedding Theorem by finding actual embeddings of various popular recursively presented groups in finitely presented ones, and are successful in at least one case.

In particular, Section 3 contains a short and elementary proof of the fact that the group $S_\infty$ of bijections of $\mathbb{N}$ with finite supports (which contains the infinite simple group $A_\infty$ of alternating bijections) embeds into a finitely presented group. A presentation for such a group, simplified in Section 4, is $$\langle a,b,x \mid a^2=1, (xaxa^{-1})^3=1, [x,a^2xa^{-2}]=1, x=[a,b], axa^{-1}=bxb^{-1} \rangle.$$ Actually, the group defined by the presentation coincides with Houghton's group $H_3$ already mentioned by Yves in the comments.

  • $\begingroup$ This is perfect. Thanks! $\endgroup$
    – Isaac
    Aug 13, 2019 at 16:07
  • $\begingroup$ To have a complete proof of an embedding, one needs to check that $[x,a^nxa^{-n}]=1$ in this group for all $n\ge 2$. $\endgroup$
    – YCor
    Aug 13, 2019 at 16:46

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