# show trivial group

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.

• Are you sure of that second relation? Because it makes things really easy: $$aba=baba\iff e=b$$and then use the first relation and voila... – DonAntonio Apr 27 '14 at 4:30
• Forgive me, thanks a lot! That was a typo, and a very bad typo indeed... Now fixed, my apologies. – TLR Apr 27 '14 at 4:31
• true @sea, that was a typo. Thanks...and the result is the same, of course. – DonAntonio Apr 27 '14 at 4:32

$aba=bab\Leftrightarrow bab^{-1}=a^{-1}ba$, 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!).
• Thank you so much, in fact I was trying to do some of the conjugation before. However, if you don't mind, may I ask a very silly question, say I realise conjugate $b$ with some group element end up still equal to $b$, but what does that mean? I just cannot remember what conclusion I can say from that? Sorry, I know this is a very silly basic group theory question. – TLR Apr 27 '14 at 13:57
• @TLR If $xbx^{-1}=b$ then $x$ and $b$ commute, and are in each other's centralizers. – blue Apr 27 '14 at 14:51