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.

  • 3
    $\begingroup$ 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/…. $\endgroup$ – Igor Belegradek Aug 13 '19 at 0:22
  • 2
    $\begingroup$ And there are many more examples, see en.wikipedia.org/wiki/Simple_group#Infinite_simple_groups. $\endgroup$ – Igor Belegradek Aug 13 '19 at 0:29
  • 1
    $\begingroup$ 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 '19 at 6:07
  • 6
    $\begingroup$ 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 '19 at 6:09
  • 5
    $\begingroup$ So you're looking for a simple simple group? $\endgroup$ – Ian Agol Aug 13 '19 at 6:57

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.

| cite | improve this answer | |
  • $\begingroup$ This is perfect. Thanks! $\endgroup$ – Isaac Aug 13 '19 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 '19 at 16:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.