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 $H_2(K_{\mathcal{P}},Z/pZ)=0$?

Group-cohomology encompasses group homology; the previous tag is more abundant.
– Chris GerigApr 5 '13 at 8:18

1

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.
– yeshengkuiApr 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 YuApr 5 '13 at 13:48