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 $H_2(K_{\mathcal{P}},Z/pZ)=0$?
$\begingroup$
$\endgroup$
3
-
$\begingroup$ Group-cohomology encompasses group homology; the previous tag is more abundant. $\endgroup$– Chris GerigCommented Apr 5, 2013 at 8:18
-
1$\begingroup$ 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. $\endgroup$– yeshengkuiCommented Apr 5, 2013 at 12:56
-
$\begingroup$ 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. $\endgroup$– Li YuCommented Apr 5, 2013 at 13:48
Add a comment
|