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I've read in many places that every compactly generated group $G$ satisfying Kazdhan's property (T) is not virtually indicable (there is no subgroup $H\leq G$ of finite index which surjects onto $\mathbb{Z}$). However, it is claimed as an obvious fact and I can't find a proof of this fact.

Any help or references will be highly appreciated.

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    $\begingroup$ Well, it's not hard but this depends on the definition of Property T. If a compactly generated group has a finite index open subgroup with a continuous nonzero homomorphism to $\mathbf{Z}$, then it actually admits a fixed-point-free continuous isometric action on some Euclidean space. $\endgroup$
    – YCor
    Nov 30, 2022 at 15:51
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    $\begingroup$ Also the book "Kazhdan's Property T" by Bekka-Harpe-Valette contains everything to prove this (Property T passes to finite index open subgroups, to quotients, and $\mathbf{Z}$ doesn't have Property T). $\endgroup$
    – YCor
    Nov 30, 2022 at 15:52
  • $\begingroup$ @YCor Where is the proof of property $T$ passing to finite index open subgroups? I've seen this to be true in this book just for discrete groups. $\endgroup$
    – Marcos
    Nov 30, 2022 at 15:59
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    $\begingroup$ No, they do something much more general, see their Theorem 1.7.1: it passes to arbitrary closed subgroups of finite covolume (this includes open subgroups of finite index!). $\endgroup$
    – YCor
    Nov 30, 2022 at 16:03
  • $\begingroup$ I think we have established that this does indeed follow easily from the results in Bekka--de la Harpe--Valette. I suggest someone (@YCor?) posts an answer; if not, I'm going to vote to close. $\endgroup$
    – HJRW
    Dec 1, 2022 at 10:19

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