I just came across this statement in Bowditch's notes on geometric group theory that $\langle a,b\ \ aba^{1}b^{2},a^{2}b^{1}ab \rangle$ is a presentation of the trivial group. Does anyone know if all presentations of the form $\langle a,b\ \ a^{i_1}b^{j_1}\cdots a^{i_n}b^{j_n},a^{j_1}b^{i_1}\cdots a^{j_n}b^{i_n} \rangle$ generally present the trivial group? We can realize the presented group as the fundamental group of $$\text{glue two disks to $S^1\vee S^1$ along the relations}$$ and it seems like this construction is homotopy equivalent to $S^2$.

2$\begingroup$ If $n=1$ and $i_1=j_1$, the group is not trivial. $\endgroup$– Mariano SuárezÁlvarezNov 12, 2010 at 7:03

1$\begingroup$ Put $I=i_1+\dotsb+i_n$ and $J=j_1+\dotsb+j_n$. Then the abelianisation has order $I^2J^2$ (or is infinite if $I^2=J^2$). $\endgroup$– Neil StricklandNov 12, 2010 at 7:50
2 Answers
If $n$ is big enough comparing to $i$'s and $j$'s, and $i$'s and $j$'s are sufficiently different, then this group satisfies the small cancellation condition $C'(\lambda)$ with $\lambda\le 1/6$ (that is these words do not contain common subwords of length $\gt 1/6$ of their length). This implies that the group is infinite, hyperbolic, and not virtually cyclic. This means that "generically" this group is infinite. By the way, the group in Bowditch's notes is a particular case of complexes of BaumslagSolitar groups studied here.
For any $n$, setting all $i$'s and $j$'s to $k$ gives a group which certainly surjects onto $C_k\times C_k$, for any $k$, which means it's definitely nontrivial.