Maybe I am missing something obvious here. But in this series of lectures the following is proved to be equivalent under the assumption that the group $W$ has a (finite) set of generators $S$ whose elements are all of order 2:
- $(W,S)$ is a Coxeter system
- $(W,S)$ has the exchange property
- $(W,S)$ has the deletion property
I am puzzled why the assumption is necessary because it seems to imply that there is a group $W$ generated by the elements of a set $S$ with $s\in S\implies s^2=e$ which is not a Coxeter group. But is that property not enough to conclude that $W$ is a Coxeter group? In other words, can there be anything else than the relations $(s_{i}s_{j})^n=e$ with $n\in\mathbb{N}\cup \{ \infty \}$?
Edit: I should have waited for the next (the seventh) video where it was (AFAICT) mentioned for the first time, that a Coxeter group can have only such relations that are consequences of the given orders of the product of all the pairs of generators. I can see now that my question looks a bit silly.