For $m\in\mathbb{N}$, consider the group $G_m=\langle s_1,\dots,s_{n-1}\rangle$ generated by the relations \begin{align*} s_i^m&=1\\ s_is_j&=s_js_i &|i-j|>1 \\ s_is_js_i&=s_js_is_j & |i-j|=1 \end{align*}

If $m=1$, $G_m$ is trivial. If $m=2$, $G_m$ is the symmetric group $\mathrm{Sym}_n$. If $m=0$, the first relation is trivial and we get the braid group $B_n$.

Is there a name for this group for general $m$? Does it have any interesting properties (for example, its irreducible representations in the case $m=3$)?

Any comments or references are appreciated!

  • $\begingroup$ Was $s_{n-1}$ what you meant? Something seems mismatched between “$s_{n-1}$” and "$\text{Sym}_n$”. $\endgroup$
    – MJD
    Dec 6 '21 at 9:27
  • 2
    $\begingroup$ Yes, I mean $\mathrm{Sym}_n$ to be the permutations of the set $\{1,\dots,n\}$ and $s_i$ to be the transposition $(i,i+1)$. Hence $s_{n-1}=(n-1,n)$ is the 'largest' swap you need to generate the relevant symmetric group. $\endgroup$
    – GossipM
    Dec 6 '21 at 11:46
  • 2
    $\begingroup$ Those are Coxeter relations. Check the theory: en.wikipedia.org/wiki/Coxeter_group $\endgroup$
    – Phicar
    Dec 6 '21 at 12:39
  • 3
    $\begingroup$ The definition of $G_m$ depends on both $m$ and $n$, right? $\endgroup$
    – user1729
    Dec 6 '21 at 14:02

There are two extremely important classes of groups related to your question: Artin groups and Coxeter groups. If $m=0$ we have an Artin group, and if $m=2$ we have a Coxeter group. If $m$ is arbitrary, we get a "Pride group".

Let's introduce these all formally. We can define Artin and Coxeter groups using graphs: Let $\Gamma$ be a simplicial graph with vertex set $\{1, \ldots, n\}$ and with edges $[i, j]$ labelled by natural numbers $p_{i, j}$.

The Artin group $A_{\Gamma}$ is the group with presentation $\langle x_1, \ldots, x_n\mid (x_ix_j)^{p_{i, j}}=(x_jx_i)^{p_{i, j}}\rangle$. Braid groups are Artin groups.

The Coxeter group $W_{\Gamma}$ is the group with presentation $\langle x_1, \ldots, x_n\mid x_i^2, (x_ix_j)^{p_{i, j}}=(x_jx_i)^{p_{i, j}}\rangle$, i.e. is a quotient of an Artin group by adding the relations $x^2=1$. Symmetric groups are Coxeter groups.

We can define Pride groups using graphs too, but more data is required. The motivation here is to prove very general results, so the definition is deliberately pretty general. In particular, it uses "free products". So, let $\Gamma$ be a simplicial graph with vertex set $\{1, \ldots, n\}$. Associate with each vertex a group $G_i$, and label the edge $[i, j]$ with a set of elements $\mathbf{t}_{i, j}\subset G_i\ast G_j$ (this is our free product - basically $\mathbf{t}_{i, j}$ consists of words $W=u_1v_1\cdots u_kv_k$ where $u_r\in G_i$ and $v_s\in G_j$). Then the Pride group $P_{\Gamma}$ is the group with relative presentation $$\langle G_1, \ldots, G_n\mid\mathbf{t}_{i, j}\rangle.$$

Therefore, we see that Artin and Coxeter groups are Pride groups, where the groups $G_i=\langle x_i\rangle$ are infinite cyclic/cyclic of order $2$, respectively, and the set $\mathbf{t}_{i, j}$ consists of the word $(x_ix_j)^{p_{i, j}}(x_jx_i)^{-p_{i, j}}$. In your case, the groups $G_i=\langle x_i\rangle$ are cyclic of order $m$, and you have the same sets of words but with $p_{i, j}\in\{2, 3\}$ according to certain rules.

Pride groups are named after Stephen J. Pride, who introduced them in the article Groups with presentations in which each defining relator involves exactly two generators. J. London Math. Soc. (2) 36 (1987), no. 2, 245–256 (doi). A natural question for your groups is whether they have decidable word problem, and this was addressed in a paper of one of Pride's students: Peter Davidson, The word problem for Pride groups, Comm. Algebra 42 (2014), no. 4, 1448–1459 (doi, Pure). However, the results of Davidson are not be applicable to the groups in the question as the relations $s_is_j=s_js_i$ do not satisfy his "asphericity" condition, but possibly his techniques can be adapted.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.