Free groups and Kazhdan's property (T) Showing non-amenability of a (non-abelian) free group  is somewhat easy and one can do this immediately after the definition of amenability.
Is there an easy proof of the fact that free groups do not have property (T)?
 A: Remark: For those unacquainted with Property $(T)$, the standard reference is the freely available book by Bekka-de la Harpe-Valette.
Since groups with property $(T)$ are finitely generated, we can assume that the  rank of the free group $F$ is finite. If a group $G$ has property $(T)$ then so does every quotient(1), recall also my answer to your earlier question. However, for a free group $F$ of rank $k$ we have $\mathbb{Z}^k = F/[F,F]$ as abelianization, and it is not difficult to prove that $\mathbb{Z}^k$ doesn't have property $(T)$ directly from the definition(2) or from the fact that an infinite amenable group doesn't have property $(T)$.
Added: I thought that adding a few details wouldn't hurt:
(1) This follows from the observation that if $\rho: G/N \to U(\mathcal{H})$ has almost invariant vectors then so does $\rho\pi$, where $\pi: G \to G/N$ is the quotient map. If $G$ has property $(T)$ then there is a $\rho\pi$-invariant unit vector, but this vector is also $\rho$-invariant, and hence $G/N$ has property $(T)$. 
(2) To see that $\mathbb{Z}^k$ doesn't have property $(T)$ note that for every finite set $K \subset \mathbb{Z}^k$ the normalized characteristic function of $[-n,n]^k$ with $n$ large enough is an almost $K$-invariant vector for the left regular representation of $\mathbb{Z}^k$ on $\ell^2(\mathbb{Z}^k)$. But there are no non-zero invariant functions in $\mathcal{H}$ because $\mathbb{Z}^k$ is infinite.
Exercise: If $H$ is an amenable group use Følner sets to prove that the left regular representation $\ell^2(H)$ has almost invariant vectors. Conclude that:


*

*an infinite amenable group does not have property $(T)$.

*if $G$ has property $(T)$ then $G/[G,G]$ is finite.
