Something I thought of while studying for qualifying exams...

Given a group and a choice of presentation $<S|R>$ (here, I'm not sure if we must assume finitely generated) we can build a 2-complex as follows. First, take a wedge of $|S|=n$ circles. Then, attach one 2-cell for each relation $r \in R$ whose attaching map reads off the word $r$.

Now, given an arbitrary group $G$, there may be more than one distinct group presentation for $G$, say $<S_1|R_1>$ and $<S_2|R_2>$. Let these presentations build the 2-complexes $X_1$ and $X_2$, respectively. My question is: what is the relationship between $X_1$ and $X_2$?

My thoughts are the following. Well, obviously they have isomorphic fundamental groups. Is that the most we can say? Perhaps they are homotopic, or at least of the same homotopy type? If so, this would be very surprising to me. Going backwards, given two arbitrary group presentations, it seems to me that the task of deciding whether or not those presentations generate groups which are isomorphic is (usually) impossible. However, if given that $X_1$ and $X_2$ are not of the same homotopy type implies that $<S_1|R_1>$ and $<S_2|R_2>$ generate different groups, then we may have a valuable reformulation of the problem. Although, deciding whether or not two 2-complexes are homotopic might seem like an even bigger mess.

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    $\begingroup$ The presentation $\langle z\mid\rangle$ gives a circle; the presentation $\langle a,b\mid [a,b], a\rangle$ gives a torus with a solid disk inside, homotopic to $S^2\vee S^1$. $\endgroup$ – Steve D Jul 22 '17 at 4:32

Pathological example: take $\mathbb{Z}_2 = <a | a^2> = <a | a^2, a^{-2}>$, and denote corresponding complexes by $C_1, C_2$. Space $C_2$ is homotopy equivalent to $C_1 \vee S^2$, hence has different $H_2$, so $C_1$ and $C_2$ cannot be homotopy equivalent.

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