Explicit Construction of the Alternating Subgroup of a Coxeter Group

Question

Given a Coxeter group $(W,S)$ with $S=\{s_1,\dots,s_n\}$, I want to show that the alternating subgroup $H$ (containing all elements with even length w.r.t. $S$) is generated by $\{s_is_n\}_{i=1}^{n-1}$.

If we pick any element $w\in H$, we can expressed $w=s_{\alpha_1}\cdots s_{\alpha_{2k}}$ for some $k=0,1,\dots$ with $l(w)=2k$
But then I have no idea how to "insert" those $s_n$ in between. I have tried using the order,
i.e. $\left(s_{\alpha_i}s_n\right)^{m(s_{\alpha_i},s_n)}=e\implies s_{\alpha_i}=s_n\left(s_{\alpha_i},s_n\right)^{m(s_{\alpha_i}s_n)-1}$
But I don't know how to deal with the $s_n$ in front, and I can't even handle the cases where $m(s_{\alpha_i},s_n)=\infty$.

Any help is much appreciated!

Answer 1

Remember that the generating elements are all involutions. For all $i,j<n$ we see that $$ s_is_j=(s_is_n)(s_n^{-1}s_j)=(s_is_n)(s_n^{-1}s_j^{-1})=(s_is_n)(s_js_n)^{-1} $$ is an element of the subgroup $H$. The claim follows from this.