On $P=\operatorname{span}\{\alpha,\beta\}$, each $s_\alpha$ is reflection in the line $\alpha^\perp\cap P$. The product of two plane reflections is a rotation. If $\theta$ is the oriented angle from $\beta$ to $\alpha$, then
$$
s_\alpha s_\beta
$$
rotates through $2\theta$ modulo $2\pi$.

If this rotation has order $m$, conjugation by either reflection inverts it. Hence
$$
\langle s_\alpha,s_\beta\rangle
=\langle r,s:r^m=s^2=1, srs=r^{-1}\rangle
$$
is a <dihedral group>, with rotational subgroup $\langle r\rangle=\langle s_\alpha s_\beta\rangle$.

For the simple-root angles from part ii, the rotation orders and Weyl groups are
$$
\begin{array}{c|c|c}
\text{type}&m&|W|\\ \hline
A_1\times A_1&2&4\\
A_2&3&6\\
B_2&4&8\\
G_2&6&12.
\end{array}
$$
These are the <Weyl group of a rank-two root system>[rank-two Weyl groups].

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 102 / 4 / iv / Solution

The two simple reflections generate a <dihedral group>. For $A_1\times A_1$, $A_2$, $B_2$, and $G_2$, the rotation has order $2,3,4,6$, so the Weyl groups have orders $4,6,8,12$ respectively.

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