2

I'm trying to prove that Higman's group is not trivial.

In order to do that, first of all I have to define the following groups:

$ \langle h_{i},h_{i+1}| h_{i+1}h_{i}h_{i+1}^{-1}=h_{i}^2\rangle$ for $i\in \mathbb{Z}/4\mathbb{Z}$

and I have to prove that each group has two disjoint cyclic subgroups $\langle h_{i}\rangle$ and $\langle h_{i+1} \rangle$. The idea is to prove that every element of each group can be written uniquely as $h_{i}^nh_{i+1}^m$ for some $n$, $m\in \mathbb{Z}$.

My idea is to find the expressions of $h_{i+1}h_{i}$, $h_{i+1}h_{i}^{-1}$, $h_{i+1}^{-1}h_{i}$ and $h_{i+1}^{-1}h_{i}^{-1}$ as $h_{i}^nh_{i+1}^m$ and it would be done, but I have only been able to see that $h_{i+1}h_{i} = h_{i}^2h_{i+1}$.

Can anyone help me or give me an alternative way? Thank you.

  • 3
    It is not true that each element can be written as $h_i^nh_{i+1}^m$. For example (with $i=1$), the element $h_2^{-1}h_1h_2$ cannot be written in that form. These 2-generator groups are the Baumslag-Solitar groups ${\rm BS}(1,2)$ and they have been much studied. For example they are isomorphic to groups of $2 \times 2$ matrices over the rationals. – Derek Holt Mar 05 '17 at 21:37

1 Answers1

2

One possibility is following: Taking the amalgamated product $$\langle a,b \mid a^b=a^2 \rangle \underset{a=y}{\ast} \langle d,y \mid d^y=d^2 \rangle$$ of two copies of $BS(1,2)$ produces the group $$\langle a,b,d \mid a^b=a^2,d^a=d^2 \rangle.$$ Now, by amalgamated two copies of this group $$\langle a,b,d \mid a^b=a^2,d^a=d^2 \rangle \underset{b=p,d=r}{\ast} \langle p,c,r \mid p^c=p^2,c^r=q^2 \rangle,$$ we find the Higman's group $$\langle a,b,c,d \mid a^b=a^2,b^c=b^2,c^d=c^2,d^a=d^2 \rangle.$$ The only thing to verify is that the amalgamations are well defined. The first one is clear, and for the second one it is sufficient to notice that the subgroup $\langle b,d \rangle$ of our first amalgamation is a non abelian free group.

In particular, the generators of $a,b,c,d$ have infinite order in Higman's group.

Seirios
  • 34,083