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Is there an example of an infinite group for which it is still unknown whether or not it is simple?

Is there a criterion for determining the simplicity of infinite groups?

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    $\begingroup$ The question is very vague, but let me mention that if A=2^N/fin is the quotient Boolean algebra of the power set of N by finite subsets, then it's undecidable in ZFC whether Aut(A) is simple. (But it's a theorem of ZFC, due to Rubin, that its derived subgroup is simple, and it's a theorem of ZFC+CH that Aut(A) is simple.) $\endgroup$ – YCor Apr 15 at 7:23
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    $\begingroup$ I also think I remember there are some diffeomorphism groups ($\mathrm{Diff}^k_0(M)$ for suitable $k$ and manifold $M$, index $0$ meaning isotopic to identity), whose simplicity is unknown — and certainly unrelated to set-theoretic subtleties as in my previous example. $\endgroup$ – YCor Apr 15 at 7:26
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    $\begingroup$ Finitely presented simple groups have decidable word problem, and so being simple is a Markov property, and so it is undecidable if a given presentation defines a simple group. (In his survey article "Decision problems for groups: survey and reflections", Miller attributes this to Kuznetsov. He also provides an easy, half-page proof.) $\endgroup$ – ADL Apr 15 at 8:35
  • $\begingroup$ @ADL sure; this also applies to "trivial groups" — the input here is a finite group presentation. $\endgroup$ – YCor Apr 15 at 8:56
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    $\begingroup$ @ADL this is indeed imprecise but the way the group comes into play is also "to the appreciation of the reader": as group presentation? as group explicitly described as acting by automorphisms on some structure. In any case, I quite like Uri's answer. $\endgroup$ – YCor Apr 15 at 9:50
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Below I will present an example of a finitely presented group which is conjecturally simple. Thank you for the opportunity to communicate this example to a wider community.


Consider the group $$\Gamma=\langle s,t,u \mid s^7=t^7=u^7=1, u=s^3t^3, u^3=st \rangle. $$ Then $\Gamma$ itself is not simple: it has a non-trivial homomorphism to $\mathbb{Z}/7\mathbb{Z}$, defined by $s\mapsto 1$, $t\mapsto -1$, $u\mapsto 0$. However, the kernel $\Gamma_0$ of this homomorphism, is conjecturally simple. Clearly, $\Gamma_0$ is finitely presented, as it is of finite index in $\Gamma$. Much is known about the groups $\Gamma$ and $\Gamma_0$. For example:

  • These groups are infinite groups and they have Kazhdan's property (T).
  • Every nontrivial normal subgroup in either group is of finite index.
  • These groups are non-linear over any field.

These groups act cocompactly on an exotic $\tilde{A}_2$-building of thickness 3, so they are somehow related to the Fano plane, the projective plane over the field with two elements, which hints about the role of the power 3 and 7 in the presentation of $\Gamma$. The group $\Gamma$ is briefly discussed in section 10.4 here. It is closely related to the group $$\langle s,t,u \mid s^7=t^7=u^7=1, u=st, u^3=s^3t^3 \rangle $$ which happens to be an index 3 subgroup of an arithmetic lattice in $\mathrm{SL}_3(\mathbb{F}_2(\!(x)\!))$, thus residually finite.


The above example is in fact a part of the infinite family of groups acting cocompactly on affine buildings. It is conjectured that if the corresponding building is not a Bruhat-Tits building then the acting group contains a simple subgroup of finite index.

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  • $\begingroup$ Is there by any chance a slightly more symmetrical presentation of $\Gamma$, where each relator uses at most two generators? To put it another way, I guess this is a triangle of groups... $\endgroup$ – HJRW Apr 15 at 10:49
  • $\begingroup$ @HJRW of course $u$ is redundant so there's an obvious presentation on generators $s,t$. Technically this satisfies your requirement "each relator uses at most two generators", but at the same time I don't think this is what you're asking. $\endgroup$ – YCor Apr 15 at 11:29
  • $\begingroup$ @YCor: you're right, sorry. My question is really "what's the structure of this as a triangle of groups"? $\endgroup$ – HJRW Apr 15 at 11:45
  • $\begingroup$ YCor, thanks for editing. @HJRW, as YCor said, $\Gamma$ is clearly generated by $s$ and $t$. Now, one easily observes that $s\mapsto s^2$, $t\mapsto t^2$ extends to an automorphism of $\Gamma$ which is of order 3. Constructing the corresponding semi-direct product, we get a supgroup in which $\Gamma$ is of index 3. This supgroup is isomorphic to the group $G_4$ in Ronan's "Lectures on buildings", Theorem 2.5, and $G_4$ is known to act freely transitively on the chambers of the corresponding building. I suppose this should give you a precise presentation as a triangle of groups. $\endgroup$ – Uri Bader Apr 15 at 12:17
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    $\begingroup$ @HJRW I believe I knew it at some point, but it has been a while since I spent my time calculating stuff about this group. I'd be happy to see you (and others) spending some time on it too, Henry ;) In any case, possibly it is described in Ronan's book cited above, but currently I don't have an access. Also, Pierre-Emmanuel Caprace or Stefan Witzel might know. $\endgroup$ – Uri Bader Apr 15 at 12:43

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