Proposition 2

Assume that $S$ consists of two distinct elements $s$ and $s'$ of order 2.

(i) The subgroup $P$ of $W$ generated by $p = ss'$ is normal, and $W$ is the semi-direct product of the subgroup $T = \{1, s\}$ and $P$. Moreover, $(W : P) = 2$.

(ii) Let $m$ be the order (finite or infinite) of $p$. Then $m \geq 2$ and $W$ is of order $2m$. There is a unique isomorphism $\varphi$ from $D_m$ to $W$ such that $\varphi(\rho) = s$ and $\varphi(\rho') = s'$.

(i) We have $sps^{-1} = sss's = s's = p^{-1}$, and hence

for every integer $n$. Since $W$ is generated by $\{ s, s' \}$, and hence by $\{ s, p \}$, the subgroup $P$ is normal. It follows that $TP$ is a subgroup of $W$, and since $TP$ contains $s$ and $s' = sp$, we have $W = TP = P \cup sP$. To prove (i), it is therefore enough to show that $W \neq P$. If $W = P$, the group $W$ would be commutative, so $p^2 = s^2 {s'}^2 = 1$. But then the only elements of $W = P$ would be 1 and $p$, contradicting the hypothesis that $W$ contains at least three elements, namely 1, $s$ and $s'$.

(ii) Since $s \neq s'$, we have $p \neq 1$ and so $m \geq 2$. Since $P$ is of order $m$ and $(W : P) = 2$, the order of $W$ is $2m$. If $m$ is finite (resp. infinite), there exists an isomorphism $\varphi'$ from $\mathbf{Z}/m\mathbf{Z}$ (resp. $\mathbf{Z}$) to $P$ taking $\pi$ to $p$. Moreover, there exists an isomorphism $\varphi''$ from $M = \{ 1, -1 \}$ to $T$ taking $-1$ to $s$. The group $W$ is the semi-direct product of $T$ and $P$. In view of the formulas (9) and $\rho \pi^n \rho^{-1} = \pi^{-n}$, $\varphi'$ and $\varphi''$ induce an isomorphism $\varphi$ from $D_m$ to $W$ such that $\varphi(\rho) = s$ and $\varphi(\pi) = p$, and hence $\varphi(\rho') = s'$. The uniqueness of $\varphi$ follows from the fact that $D_m$ is generated by $\{ \rho, \rho' \}$.