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I am trying to learn results related to the presentation complex of a group and I am new to this subject. So I apologize if the questions are silly.

Given a finite group $G$, and a presentation $P$ of $G$, consider the presentation complex $X_P$. I have computed using the Mayer-Vietoris sequence for the group $S_3$ (the symmetric group of order $6$) and get $H_2(X_P,\mathbb{Z})\neq 0$ (in fact it is torsion-free). Also, by using a similar method I was able to show that for finite cyclic groups $H_2(X_P,\mathbb{Z})=0$. I have the following questions:

  1. Is $H_2(X_P,\mathbb{Z})$ always torsion-free for a finite group $G$?

  2. Can we put some restriction on $G$ (here $G$ is a finite group) so that $H_2(X_P,\mathbb{Z})\neq 0$?

Any suggestions will be really helpful.

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    $\begingroup$ Maybe I'm being silly, but isn't this just the Schur multiplier for $G$? In this case, the answer to 1 is no -- the Schur multiplier of $C_2 \times C_2$ is $C_2$ (as is the Schur multiplier for $S_n$ with $n>3$, I believe). $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Sep 22, 2022 at 9:20
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    $\begingroup$ The presentation complex has dimension at most 2 and the 2-dimensional homology of a 2-dimensional complex with coefficients in the integers is always free. $\endgroup$
    – Fernando Muro
    Commented Sep 22, 2022 at 9:29
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    $\begingroup$ To get the Schur multiplier you need at least the 3-skeleton of a classifying space $\endgroup$
    – Benjamin Steinberg
    Commented Sep 22, 2022 at 10:23
  • $\begingroup$ @BenjaminSteinberg Ah, thanks! That's what I was missing. $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Sep 22, 2022 at 10:44
  • $\begingroup$ You can read a lot about presentation complexes from the book "Combinatorial group theory" by Lyndon&Schupp. (That text is a great starting point for anyone trying to learn low-dimensional topology from algebraic point of view!) $\endgroup$
    – Denis T
    Commented Sep 22, 2022 at 15:59

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