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Let $G$ be a geometrically finite group, i.e. there exists a finite CW complex of type $K(G,1)$. By Serre's Theorem, every finite-index subgroup $H$ of $G$ satisfies $cd(H)=cd(G)$, but what about the opposite implication?

If $H$ is a subgroup of $G$ with $cd(H)=cd(G)$, does it already follow that $H$ has finite index? If not, are there additional conditions on $G$ under which this would be true?

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    $\begingroup$ Think of some simple examples such as free groups of rank at least 2 or, more generally, free products. For your 2nd question, Think about Poincare duality groups. $\endgroup$
    – Moishe Kohan
    Commented Sep 14, 2023 at 14:21
  • $\begingroup$ Thank you, Moishe! May I ask where I can find a reference for the statement for Poincare duality groups? $\endgroup$
    – Stephan Mescher
    Commented Sep 14, 2023 at 14:30
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    $\begingroup$ Probably in Brown's book. $\endgroup$
    – Moishe Kohan
    Commented Sep 14, 2023 at 14:49
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    $\begingroup$ The result for Poincaré duality groups is due to Strebel. $\endgroup$
    – HJRW
    Commented Sep 14, 2023 at 18:33

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For the sake of completeness:

  1. Is false, just take the free product $G=H\star H$, where $H$ is a (geometrically finite) group of cohomological dimension $n$, $0<n<\infty$.

  2. This is true when $G$ is a Poincare Duality group, you can find it in exercise 6 (with a hint), section 10 of chapter 8 of Ken Brown's book "Cohomology of Groups." The fact is purely cohomological, you do not need geometric finiteness.

A complete proof can be found in Strebel, R., A remark on subgroups of infinite index in Poincaré duality groups, Comment. Math. Helv. 52, 317-324 (1977). ZBL0365.20040.

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