Suppose $G$ is a discrete group and $H \leq G$ a subgroup of finite index. If $H$ has Kazhdan property (T), does it follow that $G$ has property (T)? (I've read somewhere that (T) is preserved by exact sequences, so if $N$ is normal and $G/N$ is finite, then the fact above holds ; here, however, we do not assume $H$ to be normal)
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3 Answers
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Yes. For groups $H\subset G$, with H a lattice, H has (T) iff G has (T). When both groups are discrete being a lattice is the same as being finite index.
Almost every thing you ever need to know about Property (T) can be found here
answered Oct 25, 2010 at 16:31
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I think even more is true. See Proposition 2.5.5 in "the book":
Bekka, Bachir; de la Harpe, Pierre; Valette, Alain (2008), Kazhdan's property (T), New Mathematical Monographs, 11, Cambridge University Press, MR2415834, ISBN 978-0-521-88720-5, http://perso.univ-rennes1.fr/bachir.bekka/KazhdanTotal.pdf
for (I think) a stronger result about property (FH)
answered Oct 25, 2010 at 16:32
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Since $H$ contains a subgroup of finite index which is normal in $G$, what you are asking also follows from Theorem 1.7.1 of the book of Bekka, de la Harpe and Valette (mentioned in Jon Bannon's comment.)
