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?
 A: 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.
