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Show that $M = \langle r,s \mid r^m = e, s^n = e, srs^{-1} = r^j \rangle$, where $j$ is a natural number satisfying $\operatorname{gcd}(j,m) = 1$ and $j^n \equiv 1 \pmod{m}$, has $mn$ elements.

I'm not sure how to start to show this rigorously, but I do have some ideas. If we can show that every element can be written in the form $r^as^b$ then clearly we would have $m$ choices for $a$ and $n$ choices for $b$, and hence by multiplying there would be $mn$ elements. But I'm not sure how to show that every product can be written in this form. I thought that maybe looking at the dihedral group and doing something analogous would help me, but all of the proofs that the dihedral group has $2n$ elements that I've seen use geometrical reasoning, which I don't seem to be able to do here.

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    $\begingroup$ Surely the relation $sr=r^{j}s$ will let you push all the $s$'s in a product along to the right? But then the difficult bit: your "clearly" is not so clear, why do you think there can't be coincidences in the $r^{a}s^{b}$? That will need proof. $\endgroup$ Commented Oct 31, 2018 at 7:58
  • $\begingroup$ @ancientmathematician How can I go about proving that? Like I said before, all of the similar proofs I've seen involve the dihedral group and just use geometric reasoning. $\endgroup$ Commented Oct 31, 2018 at 15:14

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You should be able to prove that any string of $r$'s and $s$'s can be reduced to some $r^{a}s^{b}$ where $0\leqslant a <m$ and $0\leqslant b <n$. As in the comments, you just need to push each of the $s$'s to the right using the relation $sr=r^{j}s$.

The group is therefore of order at most $mn$. It cannot be smaller, because there is an example of a group of order $mn$ satisfying these relations.

Here is such a group of $n\times n$ matrices over $\mathbb{C}$. This construction is just a generalisation of the usual dihedral group, which is the case $n=2$.

Let $\omega=e^{2\pi i/m}$ be a primitive $m$-th root of unity. Let $R,S$ be the $n\times n$ matrices

$$ R:=\begin{bmatrix} \omega & 0 & 0 & \dots & 0\\ 0 &\omega^{j} &0 & \dots & 0\\ 0 & 0 &\omega^{j^2} & \dots & 0\\ \vdots & \vdots &\vdots & \ddots &\vdots\\ 0&0&0&\dots&\omega^{j^{n-1}} \end{bmatrix}, \text{ and } S:=\begin{bmatrix} 0 & 1 & 0 & \dots &0\\ 0 & 0 & 1& \dots &0\\ 0 & 0 & 0 & \dots &0\\ \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \dots &1\\ 1 & 0 & 0 & \dots &0\\ \end{bmatrix}. $$ It is straightforward to check that $R$ and $S$ satisfy the three laws in the presentation. It is (fairly) straightforward the check that the $R^a S^b$ with $0\leqslant a <m$ and $0\leqslant b <n$ are distinct, so that the order of the group $\langle R, S\rangle$ is indeed $mn$.

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