# Do cyclically presented groups of positive word length four relators satisfy the Tits Alternative?

I finished an MPhil a year ago that focused on the following question. I've moved on to a different area of group theory now, so I thought I'd ask it here.

Definition: Let $$w\in F_n$$ for the free group $$F_n$$ on $$\{x_i\mid i\in\overline{0,n-1}\}$$ and $$n\in \Bbb N$$. A cyclically presented group is a group $$G_n(w)$$ with presentation

$$P_n(w):=\langle x_0, \dots, x_{n-1}\mid w, \theta(w), \dots, \theta^{n-1}(w)\rangle,$$

where \begin{align}\theta: F_n&\to F_n,\\ x_i&\mapsto x_{i+1\pmod{n}}.\end{align}

Let $$w=x_0x_jx_kx_l$$. This is a positive word of length four.

Definition: A class $$\mathcal{G}$$ of groups satisfies the Tits Alternative if for any $$G$$ in $$\mathcal G$$ either $$G$$ has a free, non-abelian subgroup or $$G$$ has a solvable subgroup of finite index.

## The Question:

Do cyclically presented groups of positive word length four relators satisfy the Tits Alternative?

My dissertation is available here and here. A paper by myself and my supervisor at the time, Prof. G. Williams, on the topic can be found here and here.

We found that, using mainly Bogley & Parker's paper [1], for

$$(A)=\left\{\begin{array}{rlc} 2k &\equiv 0& \text{or}\\ 2j &\equiv 2l& \end{array}\right.$$

$$(B)=\left\{\begin{array}{rlc} k &\equiv 2j& \text{or}\\ k &\equiv 2l& \text{or}\\ 2k&\equiv j+l& \text{or}\\ 0&\equiv j+l& \end{array}\right.$$

$$(C)=\left\{\begin{array}{rlc} l&\equiv j+k& \text{or}\\ j &\equiv l+k& \end{array}\right.$$

all modulo $$n$$, in all cases but when $$(A)$$ and $$(C)$$ are false while $$(B)$$ is true, the Tits Alternative holds. We gathered evidence in GAP for the remaining case up to $$n=20$$.

References:

[1] Bogley, W., A., Parker, F., W. (2018) Cyclically Presented Groups with Length Four Positive Relators. J. Group Theory 21(5) (2018).

When $$j,|k-j|, |l-j|, |n-l|$$ are all distinct, then the presentation complex is a $$CAT(0)$$ square complex, since the length of any path in the link of the vertex (which is a bipartite graph because of the positivity assumption) is at least 4. In this case, the Tits alternative follows from a result of Sageev-Wise. There are no finite subgroups in this case since the complex is $$CAT(0)$$ and hence aspherical, so their theorem applies.