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Suppose I am looking at the group $W=\langle s_\alpha, t_\beta \rangle$ where $s_\alpha$ and $t_\beta$ are reflections in $\mathbb{R}^2$ coming from two vectors $\alpha$ and $\beta$ making an angle of $\pi/m$ between their perpendicular hyperplanes $H_\alpha$ and $H_\beta$, where $m \geq 2$. I need to show that $s_\alpha \cdot t_\beta$ has order $m$ and $W$ has order $2m$. It is easy enough to show for $m=2$, but I am not quite comfortable dealing with presentations of groups yet. However $W$ ought to be the dihedral group somehow, and therefore have order $2m$ by definition.

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Your intuition is correct: what you get is indeed the dihedral group of order $2m$. One way of showing that $s_{\alpha} t_{\beta}$ has order $m$ is to represent it by a suitable matrix. Your element should correspond (in the expected dihedral group) to a rotation of angle $2\pi /m$. Hence it remains to find a suitable basis of your two-dimensional real space such that the matrix of $s_{\alpha} t_{\beta}$ with respect to that basis is the matrix of a rotation of order $m$.

For simplicity, set $s=s_\alpha$, $t=t_\beta$. Once you have proven that the order of $st$ is $m$, since $s$ and $t$ have order $2$ you get the so-called "braid relation" $$s t s \cdots = t s t \cdots,$$ where there are $m$ factors on each side. But since your elements $s$ and $t$ have order $2$, any relation involving the generators $s$ and $t$ is of the form $$s t s \cdots = t s t \cdots,$$ where is there are $n_1$ factors on the left and $n_2$ on the right, $n_2+n_1$ is even (otherwise both end with the same letter $s$ or $t$ which you can simplify). But you can then rewrite your relation as $(st)^{\frac{n_1+n_2}{2}}=1$, hence $\frac{n_1+n_2}{2}$ is a multiple of $m$ since $st$ has order $m$, and you get no new relation. This means that your group has two generators $s$, $t$ and relations $s^2=1$, $t^2=1$ and the braid relation $sts\cdots = tst\cdots$, which is exactly the presentation of a dihedral group of order $2m$.

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