Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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...)

share|improve this question

1 Answer 1

up vote 6 down vote accepted

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.

share|improve this answer
    
Hi,Oscar Randal-Williams, what is transfer? –  Hao Sep 13 '10 at 16:25
3  
The transfer is the map that comes under the heading "The Transfer" in your textbook on group homology. –  Oscar Randal-Williams Sep 13 '10 at 18:24

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.