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?

  • $\begingroup$ Do you have a reference for where this is constructed? $\endgroup$ – Igor Rivin Feb 2 '12 at 19:46
  • $\begingroup$ I would had expected the last equation to be $a_3a_0a_3^{-1}=a_0^2$. $\endgroup$ – Ralph Feb 2 '12 at 20:01
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    $\begingroup$ @Richard: it is still not clear (to me) if it is simple... $\endgroup$ – Kate Juschenko Feb 2 '12 at 22:53
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    $\begingroup$ @Kate: Oh yes, I agree that it is not clear that the group is simple. Higman does not claim that it is, but merely that once you mod out by the maximal normal subgroup, you get an infinite simple group. I would love to know the answer to your question. +1. $\endgroup$ – Autumn Kent Feb 2 '12 at 23:01
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    $\begingroup$ The infinite simple group is not finitely presented, but I don't have a reference at hand. I will look a little bit later tonight. $\endgroup$ – Steve D Feb 2 '12 at 23:40

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.

  • $\begingroup$ @Kate: I am sure that Higman knew that his group is not simple. But I am curious to know how and if he proved it back then. If I see Gilbert Baumslag in New York today, I will ask him. $\endgroup$ – Mark Sapir Feb 3 '12 at 10:37
  • $\begingroup$ As for other properties, you have the same access to Google as everyone else. And Google does not give much info about it. $\endgroup$ – Mark Sapir Feb 3 '12 at 10:38


Higman, Graham 

A finitely generated infinite simple group. J. London Math. Soc. 26, (1951). 61--64

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).

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    $\begingroup$ About the last statement: it is correct. The group is an amalgamated product of two aspherical groups with free associated subgroups. So the group is (combinatorially aspherical), hence its presentation complex is a $K(.,1)$ $\endgroup$ – Mark Sapir Feb 7 '12 at 4:51

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