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enter image description here (4) shows the difficulty of proving a finite presentation group is trivial or not. Moishe Cohen mensioned the first group is called Higman group. I can't see why the Higman group is infinite but the second one is trivial.

I guess for the second part there is an elementary combinatorics proof, and for the first part we need to deal with some classical groups as a surjective homomorphic image of the shown group. Both parts I don't know either now. Help!

Minghui Ouyang
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  • First of all, you should type the questions (including the group presentations) to make your question searchable. Second, consider reading Serre's book "Trees", the first part is proven there (section 1.4); this group even has its own wikipedia page. – Moishe Kohan Aug 15 '17 at 22:42
  • For the second part see https://math.stackexchange.com/questions/1348065/exercise-from-serres-trees-prove-that-a-given-group-is-trivial – Colin McQuillan Aug 16 '17 at 04:00
  • The first group is a free product with amalgamtion of two (isomorphic) $3$-generator groups, both of which are HNN-extensions, and its infiniteness follows from the general theory of these things. – Derek Holt Aug 16 '17 at 07:46
  • @DerekHolt Wow. So we can't prove it from scratch? – Minghui Ouyang Aug 16 '17 at 08:04
  • I don't know any other proof. It is not hard to prove that this group has no nontrivial finite quotient groups. That suggests that proving it to be infinite (or even proving it nontrivial) will not be completely straightforward. – Derek Holt Aug 16 '17 at 08:28
  • @DerekHolt You still need to prove it is non-trivial, which is the hardest part I think. – Minghui Ouyang Aug 16 '17 at 08:29

2 Answers2

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I will give a few more details of my comment on proving the Higman group infinite, but you need to know a bit about HNN-extensions and amalgamated free products to understand it.

The group $G_2 := \langle a,b \mid a^b=a^2 \rangle$ is an HNN-extension of one infinite cyclic group by another, and so is infinite. In fact this is the Baumslag-Solitar group ${\rm BS}(1,2)$ and has been much-studied.

The group $G_3 := \langle a,b,c \mid a^b=a^2, b^c=b^2 \rangle$ is an HNN-extension of $G_2$ by $\langle c \rangle$, and is infinite. In fact it is easy to see from Britton's Lamma on HNN-extensyions that no nontrivial reduced word in the subgroup $\langle a,c \rangle$ is trivial in $G_3$, so that subgroup is free of rank $2$.

Now $\langle c,d, a \mid c^d=c^2, d^a=d^2 \rangle$ is isomorphic to $G_3$ and again the subgroup $\langle a,c \rangle$ is free.

So the Higman group $G_4 := \langle a,b,c,d \mid a^b=a^2, b^c=b^2, c^d=c^2, d^a=d^2 \rangle$ is a free product with amalgamation of the two copies of $G_3$ amalgmated on the subgroup $\langle a,c \rangle$. By the theory of free products with amalgamation, both subgroups embed into $G_4$, so it is infinite, and in fact the subgroups $\langle a,c \rangle$ and $\langle b,d \rangle$ are free.

Derek Holt
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That the $4$-generator group (Higman's group) is infinite was shown in this answer of Serios (Derek Holt's answer here is similar).

That the $3$-generator group is trivial was shown in this answer of Jim Belk. Derek Holt has some interesting, related comments here.

user1729
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