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Given a finite group $G$, consider a presentation $P$ of $G$ and consider $X_P$, the presentation complex. Now let $Y$ be any $2$-dimensional CW-complex with $\pi_1(Y)=G$. Is there any relation between $Y$ and $X_P$ for some presentation $P$ of $G$?

If we consider $X_P\vee S^2$ then the fundamental group for $X_P$ and $X_P\vee S^2$ are the same. However, for an arbitrary finite $2$-dimensional CW-complex $Y$ with $\pi_1(Y)=G$, can one say that such $Y$ is always obtained (up to homotopy) by taking the wedge product of $2$-spheres with some $X_P$? or $Y$ is obtained by some modification of some $X_P$?

I apologize for the sloppy writing!

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  • $\begingroup$ I'm not sure I understood your question correctly: is $P$ allowed to depend on $Y$, or is $P$ fixed in advance? $\endgroup$
    – HJRW
    Commented Oct 4, 2022 at 13:40
  • $\begingroup$ Anyway, if $P$ is allowed to depend on $Y$ (as your wording suggests), then all you need to do is to contract a maximal tree in $Y$. This gives some $Y'\simeq Y$ with exactly one vertex. But 2-complexes with one vertex are presentation complexes! $\endgroup$
    – HJRW
    Commented Oct 4, 2022 at 13:46
  • $\begingroup$ I am considering $Y$ an arbitrary finite $2$-complex with $\pi_1(Y)=G$. For such $G$, I am choosing presentation complex $X_P$, where $P$ varies over all presentations of $G$. $Y$ is not depending on $P$. But for such $Y$ can I get an $P$, so that $Y$ is homotopic to wedge of $S^2$'s and $X_P$? $\endgroup$
    – gola vat
    Commented Oct 4, 2022 at 14:08
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    $\begingroup$ You don't even need to wedge an $S^2$! As I explained above, every connected 2-complex is homotopy equivalent to a 2-complex with a single vertex. This is a presentation complex: the generators are given by the 1-cells; the attaching maps of the 2-cells give (up to conjugacy) elements of $\pi_1$ of the 1-skeleton, which can be written as words in your generators; these are your relations. $\endgroup$
    – HJRW
    Commented Oct 4, 2022 at 15:03
  • $\begingroup$ There are two presentation complexes of a group related to braid group in three strands. They are wedge-indecomposable, not homotopy equivalent, and become homotopy equivalent after wedging with 2-sphere. This is explained by existence of stably free non-free modules over group ring. (If somebody remembers exact reference, please post an answer; afair this example is by Ratcliffe) $\endgroup$
    – Denis T
    Commented Oct 21, 2022 at 18:06

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