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Let $G_1 \le G_2 \le \cdots $ be countable groups with Kazhdan property (T). Let $G = \bigcup_i G_i$. Does it necessarily follow that $G$ has (T)?

This seems false but I cannot find a counterexample.

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    $\begingroup$ Hint: Property T discrete groups are finitely generated. $\endgroup$ Jun 5, 2023 at 3:14
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    $\begingroup$ For a specific example take the union of $SL(n,\mathbb Z)$ over $n\ge 3$. The chain of standard inclusions $SL(n,\mathbb Z)\to SL(n+1,\mathbb Z)$ does not stabilize so the union isn't finitely generated. $\endgroup$ Jun 5, 2023 at 13:06
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    $\begingroup$ Examples are abundant: they include all infinite countable locally finite groups. $\endgroup$
    – YCor
    Jun 6, 2023 at 6:57
  • $\begingroup$ @YCor is it as easy when each $G_i$ is taken to have properties FH or FW? Does $SL_{\infty}(Z)$ have property FW? $\endgroup$
    – jonan
    Aug 19, 2023 at 4:01
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    $\begingroup$ @jonan every countable group with Property FW is finitely generated $\endgroup$
    – YCor
    Aug 19, 2023 at 6:32

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