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 proper 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 $\text{SL}_3(\mathbb{F}_2((x)))$, 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.

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.