I have a group presentation here, which is $\langle a,b|a^n = b^{n+1}, aba=bab\rangle$, $n$ is any fixed integer. I want to show this presentation is in fact the trivial group.
So far I have been trying many ways to rearrange relations, but still not working. For example, I got $a^n b^{-n} = b$, and $b^{-1} = b^n a^{-1}$, then I substituted into the relator $abab^{-1}a^{-1}b^{-1}=1$, but still nothing nice came out.
Can anyone please give me a hand on this? Thanks a lot.
$aba=bab\Leftrightarrow bab^{-1}=a^{-1}ba$, so try conjugating $a^n$ by $b$ and $b^{n+1}$ by $a$ and see what happens!
Remember conjugations are automorphisms (so commute with taking powers), and powers of $g$ are invariant under conjugation by $g$ for any group element $g$, so in particular $a^n$ and $b^{n+1}$ are invariant under conjugation by $a$ and $b$ (not just respectively; read that in any order!).