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I computed the fundamental group of the Klein bottle in two different ways and obtained two seemingly different answers: $$ \langle a,b \mid abab^{-1}\rangle $$ and $$ \langle c,d \mid c^2d^2\rangle. $$ I then tried to show that these two groups are in fact the same: \begin{align*} \langle a,b \mid abab^{-1}\rangle =\langle ab,b^{-1} \mid abab^{-1}\rangle &=\langle ab,b^{-1} \mid (ab)(ab)b^{-1}b^{-1}\rangle\\ &=\langle c,d \mid c^2d^2\rangle \end{align*} Is my method for doing so valid?

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Sure. The second step doesn't really make sense as a presentation, but if you skip it then everything is fine. It might be clearest that these groups are isomorphic if you added the intermediate group $$\langle a,b,c,d| abab^{-1},ab=c,d=b^{-1},c^2d^2\rangle$$ which admits obvious maps from each of the two desired groups which one readily proves to be surjective (since $c$ and $d$ are in the subgroup generated by $a$ and $b$) and injective (since the first and last relations are each implied by the other three.)

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