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