The group of Higman:
$ \langle \ a_0, a_1, a_2, a_3 \ | \ a_0 a_1 a_0^{-1}=a_1^2, \ a_1 a_2 a_1^{-1}=a_2^2, \ a_2 a_3 a_2^{-1}=a_3^2, \ a_3 a_0 a_3^{-1}=a_0^2 \ \rangle . $
Is it simple? What is actually known about it except the fact that it does not have non-trivial homomorphims into a finite group?
Higman's group is not simple. Indeed, if you look at Higman's paper, you will see that his group is an amalgamated product of two groups $K_{1,2}=\langle a_1, a_2, b_2\mid a_1^{-1}a_2a_1=a_2^2, a_2^{-1}b_2a_2=b_2^2\rangle$ and $K_{3,4}=\langle a_3, a_4, b_4\mid a_3^{-1}a_4a_3=a_4^2, a_4^{-1}b_4a_4=b_4^2\rangle$ with free amalgamated subgroups $\langle a_1, b_2\rangle$ and $\langle a_3, b_4\rangle$ with $a_1=b_4, b_2=a_3$. Now take $K_{1,2}$ and add the relations $w(a_1,b_2)=1, w(b_2,a_1)=1$ for some "complicated" word $w$. You get a non-trivial group $K$. Similarly imposing $w(a_3,b_4)=1=w(b_4,a_3)$ on $K_{3,4}$ you get a non-trivial group $K'$. Then there is a homomorphism from the Higman group to the amalgamated product of $K$ and $K'$ with a non-trivial kernel. The fact that $K$ (and, equally, $K'$) is not trivial is not obvious but it is not very difficult to prove.
A simplification. Instead of adding two relations, in fact it is enough to add a relation $w(a_1,b_2)=w(b_2,a_1)$ and similarly for $a_3,b_4$. For example, $(a_1b_2)^3=(b_2a_1)^3$.
Update 1. Probably the easiest way to be convinced that $K$ ($=K'$) is non-trivial is to input its presentation into GAP or Magma (you do not need 3 in the simplified relation above, 2 is enough).
Update 2. An even easier way is to impose further relations $a_2=b_2=1$ on $K$ (as in the simplification above). The factor-group is the infinite cyclic group. Hence $K$ is infinite (and $K'$ is infinite too).
Update 3. I forgot to mention that one also needs that after we impose the relation as in the simplification, we should make sure that the subgroup of $K$ generated by $a_1,b_2$ has an automorphism switching $a_1$ and $b_2$ (otherwise we cannot form the proper amalgamated product of $K$ and $K'$). That is why we cannot replace 3 in the Simplification above by $1$.
Update 4. Gilbert Baumslag told me that Higman did not know whether his group is simple, but Paul Schupp proved it. And indeed, Schupp, Paul E. Small cancellation theory over free products with amalgamation. Math. Ann. 193 (1971), 255–264 proved much more: Higman's group is SQ-universal, so every countable group embeds into one of its quotients.
Reference:
Higman, Graham, A finitely generated infinite simple group, J. Lond. Math. Soc. 26, 61-64 (1951). ZBL0042.02201.
It is shown that $G$ is infinite and has no proper normal subgroups of finite index, except $G$.
It is easy to see that this group is perfect: it has trivial abelianization.
I have heard through the grapevine that the space $X$ with four one-cells and four two-cells (corresponding, respectively, to generators and relations) is the classifying space of the group (I don't have a reference).
