Let $G$ be a group (not necessary abelian) generated by $a, b$ that has a group presentation: $$\langle a, b \mid a^m = 1, b^n = 1, ba = a^rb\rangle. $$ I have to prove that $G$ is of order $mn$ if and only if $r^n = 1 \pmod{m}$. I was able to prove the $\leftarrow$ side, but I'm stuck in proving the $\rightarrow$ side.
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From $b^n = 1$ and $bab^{-1} = a^r$ it follows, that $a = b^{n}ab^{-n} = a^{r^n}$, which means $a^{r^n - 1} = 1$.
Because we also have $a^m = 1$ we can conclude that $a^{GCD(m, r^n - 1)} = 1$. And because it is not hard to see, that $a \mapsto a^r$ is an automorphism of $\langle a \rangle_{GCD(m, r^n - 1)}$ we can see that
$$\langle a, b | a^m = 1, b^n = 1, bab^{-1} = a^r\rangle \cong \langle a \rangle_{GCD(m, r^n - 1)} \rtimes_{a \mapsto a^r} \langle b \rangle_{n}$$
which is a group of order $nGCD(m, r^n - 1)$
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\pmod{m}to produce the spacing, parentheses, and "mod" text in roman typeface.\modproduces the wrong spacing. – Arturo Magidin Jan 07 '21 at 19:42