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...)
1 Answer
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No, it is not true. For example, let $\mathbf{Z}/2$ act on $\mathbf{Z}$ by inversion, and $G$ be the semidirect product. Then $\mathbf{Z}$ is a torsion-free finite index normal subgroup of $G$, but one easily computes that the rational cohomology of $G$ is trivial, by the Hochschild-Leray-Serre spectral sequence for the extension.
In general, if $H$ is a normal subgroup of finite index, then $H^*(G;\mathbb{Q}) \cong H^*(H;\mathbb{Q})^{G/H}$, for the action of $G/H$ on $H$ by outer automorphisms.
If the subgroup is of finite index but not normal, the most one can say is that $H^*(G;\mathbb{Q}) \to H^*(H;\mathbb{Q})$ is split injective, which is proved using the transfer.
answered Sep 13, 2010 at 14:33
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$\begingroup$ Hi,Oscar Randal-Williams, what is transfer? $\endgroup$– HaoCommented Sep 13, 2010 at 16:25
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3$\begingroup$ The transfer is the map that comes under the heading "The Transfer" in your textbook on group homology. $\endgroup$ Commented Sep 13, 2010 at 18:24