Find the order of a presentation

[Mr Davis 97]

Homework Statement

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,

Homework Equations

The Attempt at a Solution

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 geomretical reasoning, which I don't seem to be able to do here.

 

[fresh_42]

Homework Statement

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,

Homework Equations

The Attempt at a Solution

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 geomretical reasoning, which I don't seem to be able to do here.

That every product can be written in this form is trivial: With $sr=r^js$ we can change every word until all $s$ are on the right and all $r$ are on the left. You can do this by induction over the number of changes $(s^qr^p)$ in words. Therefore $M=\{\,r^ps^q\,|\,1\leq p \leq m\, , \,1\leq q \leq n\,\}$ What is left to show is, that we do not count the same element multiple times, i.e. that $r^ps^q=r^{p'}s^{q'} \Longrightarrow p=p' \,\wedge \, q=q'$