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Let $G$ be a finitely presentable group. If we assume $H_2(G,Z/pZ) =0$, $p$ is a prime, then can we always find a finite presentation $\mathcal{P}$ of $G$ so that its presentation complex $K_{\mathcal{P}}$ satisfies $H_2(K_{\mathcal{P}},Z/pZ)=0$?

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Group-cohomology encompasses group homology; the previous tag is more abundant. – Chris Gerig Apr 5 '13 at 8:18
Hi Li: Just a remark. By Hopf exact sequence, $H_2(K_P,Z/pZ)=0$ if and only if the Hurewicz map $\pi_2(K_p)\tensor Z/p \rightarrow H_2(K_P,Z/pZ)$ is trivial. But $\pi_2(K_p)$ is the same as the second homology of its universal covering space. If the (mod $p$) cohomological dimension of $G$ is large and the group homomorphism between mod $p$ homology groups induced by covering map is not trivial, then $H_2(K_P,Z/pZ)$ is not zero. – yeshengkui Apr 5 '13 at 12:56
HI: yeshengkui: $\pi_2(K_{\mathcal{P}})$ is not an invariant of $G$. So when we change the presentation $\mathcal{P}$ of $G$, the size of the second homotopy group of $K_{\mathcal{P}}$ may be reduced. – Li Yu Apr 5 '13 at 13:48

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