Remark

Consider a dihedral group $W$ of order $2m$ generated by two distinct elements $s$ and $s'$ of order 2. Denote by $s_q$ (resp. $s'_q$) the sequence of length $q$ whose odd (resp. even) numbered terms are equal to $s$ and whose even (resp. odd) numbered terms are equal to $s'$, and let $w_q$ (resp. $w'_q$) be the product of the sequence $s_q$ (resp. $s'_q$). We have

If $s = (s_1, \ldots, s_q)$ is a reduced decomposition (with respect to $\{ s, s' \}$) of an element $w$ of $W$, then clearly $s_i \neq s_{i+1}$ for $1 \leq i \leq q-1$. Hence, $s = s_q$ or $s = s'_q$.

If $m = \infty$, the elements $(ss')^n$ and $(ss')^n s$ for $n \in \mathbf{Z}$ are distinct. Hence, the elements $w_q$ ($q \geq 0$) and $w'_q$ ($q > 0$) are distinct, and if $s$ is a reduced decomposition of $w_q$ (resp. $w'_q$) we necessarily have $s = s_q$ (resp. $s = s'_q$). It follows from this that $l(w_q) = l(w'_q) = q$ and that the set of reduced decompositions of the elements of $W$ consists of the $s_q$ and the $s'_q$. Moreover, every element of $W$ has a unique reduced decomposition.

Suppose now that $m$ is finite. If $q \geq 2m$, we have $w_q = w_{q-2m}$ and $w'_q = w'_{q-2m}$; if $m \leq q \leq 2m$, we have $w_q = w'_{2m-q}$, $w'_q = w_{2m-q}$. Hence, neither $s_q$ nor $s'_q$ are reduced decompositions if $q > m$. It follows that each of the $2m$ elements of $W$ is one of the $2m$ elements $w_0 = w'_0$, $w_q$ and $w'_q$ for $1 \leq q \leq m-1$, and $w_m = w'_m$. These $2m$ elements are thus distinct and it follows as above that $l(w_q) = l(w'_q) = q$ for $q \leq m$ and that the set of reduced decompositions of elements of $W$ consists of the $s_q$ and the $s'_q$ for $0 \leq q \leq m$. Every element of $W$ except $w_m$ has a unique reduced decomposition; $w_m$ has two.