Is Kazhdan's property (T) stable under extensions? i.e. if $G$ is an extension of a group with property (T) by a group with property (T), does it follow that $G$ has property (T)?
Extensions and Kazhdan's Property (T)
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0Yes I am. I have edited the question. – 2011-12-25
2 Answers
The answer to your question is yes:
See Bekka–de la Harpe–Valette, Kazhdan's property $(T)$, page 69 (the standard reference on property (T), freely available on Bekka's homepage):
The proof is not very difficult, and it is given in a clear fashion in the book, so it doesn't make much sense to reproduce it here.
Note also that Property (T) is closed under taking quotients, but it is not closed under passing to normal subgroups. The standard example is the semi-direct product $\mathrm{SL}_{n}(\mathbb{R}) \ltimes \mathbb{R}^n$ which has property (T) for $n \geq 3$ but whose normal subgroup $\mathbb{R}^n$ doesn't.
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0Sure, no problem (i$f$ someone knows how to add that to Wikipedia this would certainly be a worthwhile thing to mention). It's a fine book and now you have access :) – 2011-12-25
If I understood your question correctly, the answer is NO. For instance, take property T to be being cyclic. The Klein-4 group is not cyclic, but it has three subgroups of order 2, all of them cyclic.
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0Then that does it then. – 2011-12-25