# Understanding groups that have a similar presentation as $S_n$

It is well known that the symmetric group $S_n$ can re presented using the generatos $\sigma_1, \dots, \sigma_{n-1}$ subject to the relations: $$\sigma_i^2=1,\\ \left[\sigma_i ,\sigma_j\right] = e \quad (|i-j| \neq 1), \\ \langle\sigma_i,\sigma_{i+1} \rangle = e.$$ Where $\langle a,b\rangle$ is the triple relation between $a$ and $b$: $$aba = bab.$$

I am interested in groups that have a very similar presntation, but slightly different: One (and only one) of the commutators is replaced with a triple relation involving the same generators (e.g. $[\sigma_1,\sigma_3]=e$ is replaced with $\langle \sigma_1, \sigma_3 \rangle = e$). Such groups arise as the fundumental group of some geometric objects that I am interested in.

Can something be said about these groups? In particular, I am interested in surjecting such groups on $S_n$ and finding the kernel of this surjection (note that the straightforward surejction does not work). What are the surjections from such groups onto $S_n$ and how can I find their Kernel?

• Such a group is still a Coxeter group, like $S_n$, so there is a lot that can be said. However, any group such as you describe will be infinite. The case where $\sigma_1$ and $\sigma_{n-1}$ get the triple relations has a nice combinatorial model, if that's of interest. – Hugh Thomas Apr 27 '18 at 15:29
• @HughThomas , this case is indeed of interest to me (just like the others). Can you elaborate? – Mike Apr 27 '18 at 15:32
• If you do this with the first and last generator, you get the affine Weyl group of type $A$. – Tobias Kildetoft Apr 27 '18 at 15:37
• The way of thinking about it is that you consider bijections from $\mathbb Z$ to $\mathbb Z$, with the property that $\pi(x+(n-1))=\pi(x)+n-1$. $\sigma_i$ acts by swapping the entries in positions $k(n-1)+i$ and $k(n-1)+i+1$ for all $k\in\mathbb Z$ simultaneously. – Hugh Thomas Apr 27 '18 at 15:38

The replacement of the relation that you are describing is equivalent to replacing $(\sigma_i\sigma_j)^2=1$ by $(\sigma_i\sigma_j)^3=1$, and I prefer to think of it that way because we are actually talking about Coxeter groups.
After a bit of experimentation on the computer, I am reasonably convinced that the answer to your question is that, if $j=i+2$ then there is still a unique (up to conjugation of the image in $S_n$) epimorphism onto $S_n$ (or two when $n=6$). For example, when $i=1$ and $j=3$, we can map $\sigma_1,\sigma_2,\sigma_3$ to $(1,2),(2,3),(2,4)$, and $\sigma_i$ to $(i,i+1)$ when $i>3$.
But when $j>i+2$ then there are no epimorphisms onto $S_n$. I expect that this could proved but I don't have the time or energy to try and prove it now!
• Is this map (for $i=j+2$) an isomorphism? I think not, but can not see why. What is the kernel of this map? As for the case $i > j+2$, I think I can prove that there is no epimorphism that sends $\sigma_i$ to transopisitions, but is that the only possible case? – Mike Apr 28 '18 at 13:19
• How do you see that it is infinite? Finding an element of an inifinite order seems rather hard to me. Also, what about the case $i > j+2$ that I have mentioned above? – Mike Apr 28 '18 at 20:01