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Hao
Hao
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If a group G has a subgroup H of finite index which is torsion free, then does it satisfy $H_\ast (G,Q) = H_ \ast (H,Q)$? (probably it is very well known...)

one more is: If the cohomological dimension agree with the homological dimension of a group? ((co)homological dimension is the biggest integer $n$ such that there is a $G$-module $M$ satisfies the (co)homology of $G$ with coefficient $M$ is nonzero). If $G$ is a duality group (of dimension $d$) (i.e., there exists a $G$-module $I$ and an element $e$ in $I$ such that $H^{k}(G,M)=H_{d-k}(G,M\otimes I)$), if we want to deduce the homological dimension of $G$ is $d$ then we need an $I^{-1}$ such that $I\otimes I^{-1}=1$ ,right?Or how can we deduce the homological dimension is $d$?

If a group G has a subgroup H of finite index which is torsion free, then does it satisfy $H_\ast (G,Q) = H_ \ast (H,Q)$? (probably it is very well known...)

one more is: If the cohomological dimension agree with the homological dimension of a group? ((co)homological dimension is the biggest integer $n$ such that there is a $G$-module $M$ satisfies the (co)homology of $G$ with coefficient $M$ is nonzero). If $G$ is a duality group (of dimension $d$) (i.e., there exists a $G$-module $I$ and an element $e$ in $I$ such that $H^{k}(G,M)=H_{d-k}(G,M\otimes I)$), if we want to deduce the homological dimension of $G$ is $d$ then we need an $I^{-1}$ such that $I\otimes I^{-1}=1$ ,right?Or how can we deduce the homological dimension is $d$?

If a group G has a subgroup H of finite index which is torsion free, then does it satisfy $H_\ast (G,Q) = H_ \ast (H,Q)$? (probably it is very well known...)

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Source Link
Hao
Hao
  • 113
  • 5

If a group G has a subgroup H of finite index which is torsion free, then does it satisfy $H_\ast (G,Q) = H_ \ast (H,Q)$? (probably it is very well known...)

one more is: If the cohomological dimension agree with the homological dimension of a group? ((co)homological dimension is the biggest integer $n$ such that there is a $G$-module $M$ satisfies the (co)homology of $G$ with coefficient $M$ is nonzero). If $G$ is a duality group (of dimension $d$) (i.e., there exists a $G$-module $I$ and an element $e$ in $I$ such that $H^{k}(G,M)=H_{d-k}(G,M\otimes I)$), if we want to deduce the homological dimension of $G$ is $d$ then we need an $I^{-1}$ such that $I\otimes I^{-1}=1$ ,right?Or how can we deduce the homological dimension is $d$?

If a group G has a subgroup H of finite index which is torsion free, then does it satisfy $H_\ast (G,Q) = H_ \ast (H,Q)$? (probably it is very well known...)

If a group G has a subgroup H of finite index which is torsion free, then does it satisfy $H_\ast (G,Q) = H_ \ast (H,Q)$? (probably it is very well known...)

one more is: If the cohomological dimension agree with the homological dimension of a group? ((co)homological dimension is the biggest integer $n$ such that there is a $G$-module $M$ satisfies the (co)homology of $G$ with coefficient $M$ is nonzero). If $G$ is a duality group (of dimension $d$) (i.e., there exists a $G$-module $I$ and an element $e$ in $I$ such that $H^{k}(G,M)=H_{d-k}(G,M\otimes I)$), if we want to deduce the homological dimension of $G$ is $d$ then we need an $I^{-1}$ such that $I\otimes I^{-1}=1$ ,right?Or how can we deduce the homological dimension is $d$?

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Hao
Hao
  • 113
  • 5

a small question about group homology

If a group G has a subgroup H of finite index which is torsion free, then does it satisfy $H_\ast (G,Q) = H_ \ast (H,Q)$? (probably it is very well known...)