The symmetric group, in terms of presentation, is given by a group with generators $x_1,x_2,\cdots,x_n$ with following types of relations:

(R1) $x_1x_{i+1}x_i=x_{i+1}x_ix_{i+1}$

(R2) $x_ix_j=x_jx_i$ for $|i-j|\geq 2$

(R3) $x_i^2=1$

There is a class of groups which are closely related with this group. The Braid group $B_n$ has presentation:

$$B_n=\langle x_1,x_2,\cdots,x_n : R1, R2\rangle.$$ In other words, dropping relations (R3) from above presentation of symmetric group, we get Braid group. Then my question is

Question: If we drop relations (R1) or (R2) from presentation of $S_n$, what kind of groups we get? Are they finite? Are these groups studied well?

  • $\begingroup$ Well, if you replace (R1) by something suitable you get Coxeter groups of other types than $A$ (with possibly some small adjustments to (R2) for some types). $\endgroup$ – Tobias Kildetoft Jun 7 '16 at 8:44
  • $\begingroup$ You mean: the relations (R1) (using (R3)) can be written as $(x_ix_{i+1})^3=1$; instead of this, we write $(x_ix_{i+1})^n_i=1$, then we get Coxeter group. $\endgroup$ – p Groups Jun 7 '16 at 8:46
  • $\begingroup$ Yes, precisely (or we can remove the condition by letting the element have infinite order, which obviously gives an infinite group). $\endgroup$ – Tobias Kildetoft Jun 7 '16 at 8:50
  • $\begingroup$ Also note that is may be more suggestive to write all the relations here in the form $(xy)^n = 1$ for some $n$. $\endgroup$ – Tobias Kildetoft Jun 7 '16 at 9:11

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