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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).

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1 Answer 1

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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.

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