Prove that the following equality holds in a group

Question

Let $G$ be a group (not necessary abelian) generated by $a, b$ that has a group presentation: $$\langle a, b \mid a^m = 1, b^n = 1, ba = a^rb\rangle. $$Then how can I prove that $(a^sb^t)(a^ub^v) = a^xb^y$, where $x$ is the remainder of $s + u(r^t)$ when divided by $m$, and $y$ is the remainder of $t + v$ when divided by $n$? It seems that it can be solved by changing $b^{...}a^{...}$ form into $a^{...}b^{...}$ form successively using the condition $ba = a^rb$, but I don't understand precisely how to do this.

Answer 1

First note that $$ ba^{k+1}=baa^k=a^rba^k$$ so that by induction on $k$, $$ ba^k=a^{rk}b.$$ Next, $$ b^{l+1}a^k=b^nba^k=b^la^{rk}b$$ so that by induction on $l$, $$ b^la^k=a^{r^lk}b^l.$$ Apply this to $a^sb^ta^ub^v$ and note that you may reduce exponents of $a$ modulo $m$ and exponents of $b$ modulo $n$.

Answer 2

It is enough to show $b^t a^u=a^{u(r^t)} b^t$, which is equivalent to $b^t a^u b^{-t}=a^{u(r^t)}$. After we show $b^t a b^{-t}=a^{r^t}$, we take $u$ power of both sides and finish the proof.
Now enough to show $b^t a=a^{r^t} b^t$, and it comes easily from repeated use of $ba=a^r b$.

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We get the last part basically from the idea you have mentioned. We start from the left hand side and switch $a,b$ so that $b$ move right each step. I think this is the most intuitive one.
However, I think induction on $t$ would be clearer. For $t=1$ it is obvious. To proceed from $t$ to $t+1$, we only have to show $b a^{r^{t}}=a^{r^{t+1}} b$. But it is equivalent to $a^{r^{t}}=b^{-1} a^{r^{t+1}} b$, and we obtain it by taking $r^t$ power of given relation $a=b^{-1} a^{r} b$.