# Infinite group for which it is still unknown if it is simple

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?

• 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.) – YCor Apr 15 at 7:23
• 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. – YCor Apr 15 at 7:26
• 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.) – ADL Apr 15 at 8:35
• @ADL sure; this also applies to "trivial groups" — the input here is a finite group presentation. – YCor Apr 15 at 8:56
• @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. – YCor Apr 15 at 9:50

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.

• 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... – HJRW Apr 15 at 10:49
• @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. – YCor Apr 15 at 11:29
• @YCor: you're right, sorry. My question is really "what's the structure of this as a triangle of groups"? – HJRW Apr 15 at 11:45
• 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. – Uri Bader Apr 15 at 12:17
• @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. – Uri Bader Apr 15 at 12:43