Presentation of the fundamental group of an octagon with identified borders

Question

I need a little bit of clarification about the presentation of the fundamental group. My exercise is to consider all the possible identification of the opposite sides of an octagon and calculate their fundamental group, so I begun with $$G=\langle a,b,c,d|abcdabcd\rangle\;\;\;\;\;\operatorname{and}\;\;\;\;\;G'=\langle a,b,c,d|abcda^{-1}b^{-1}c^{-1}d^{-1}\rangle$$ The second one is equivalent to $\langle x,y|xyx^{-1}y^{-1}\rangle$ so the same as a torus, while the first one I thought that it was equivalent to the projective space and I thought that its representations should be $\langle x,y|xyxy\rangle$. Here I found out that the correct representation of the fundamental group is $\langle z|z^2\rangle$.
So my question here is why the representation of the torus can't be $\langle z|zz^{-1}\rangle$?

Answer 1

Regarding the question in the final paragraph of your post, the group presented by $\langle z \mid z z^{-1} \rangle$ is infinite cyclic, and the fundamental group of the torus is not infinite cyclic, so that cannot be a presentation of the fundamental group of the torus.

Regarding the part of your post asserting the presentations $\langle a,b,c,d \mid abcda^{-1}b^{-1}c^{-1}d^{-1} \rangle$ and $\langle xy \mid xyx^{-1}y^{-1} \rangle$ are the same group, the first has abelianization $\mathbb Z^4$ and the second has abelianization $\mathbb Z^2$, and so those cannot be isomorphic either.

Based on your comment, it looks like you are making mistakes in carrying out Tietze transformations.