Proposition 3

Assume that (W, S) is a Coxeter system. Then, two elements s and s' of S are conjugate$^2$ in W if and only if the following condition is satisfied:

(I) There exists a finite sequence $(s_1, \ldots, s_q)$ of elements of S such that $s_1 = s, s_q = s'$ and $s_j s_{j+1}$ is of finite odd order for $1 \leq j < q$.

Let s and s' in S be such that $p = ss'$ is of finite order $2n + 1$. By (9), $sp^{-n} = p^n s$ hence

and $s'$ is conjugate to s.

For all s in S, let $A_s$ be the set of $s' \in S$ satisfying (I). With the hypotheses in (I), the elements $s_j$ and $s_{j+1}$ are conjugate for $1 \leq j < q$ by the above, hence every element $s'$ of $A_s$ is conjugate to s.

Let $f$ be the map from S to $M = \{1, -1\}$ equal to 1 on $A_s$ and to -1 on $S - A_s$. Let $s'$ and $s''$ in S be such that $s's''$ is of finite order m. Then $f(s') f(s'') = 1$ if $s'$ and $s''$ are both in $A_s$ or both in $S - A_s$. In the other case, $f(s') f(s'') = -1$, but m is even. Thus $(f(s') f(s''))^m = 1$ in all cases. Since (W, S) is a Coxeter system, there exists a homomorphism g from W to M inducing f on S. Let $s'$ be a conjugate of s. Since s belongs to the kernel of g, so does $s'$, hence $f(s') = g(s') = 1$ and finally $s' \in A_s$. Q.E.D.