Property (T) implies compact generation

Let $\mathcal{C}$ be the set of open compactly generated subgroups of $G$; it is a set because it is a subcollection of $\mathcal{P}(G)$. It covers $G$: for $g\in G$, choose a compact neighborhood $K$ of $g$ and an open $U$ with $g\in U\subseteq K$ by [F2]. The subgroup $H=⟨K⟩$ contains the nonempty open set $U$; if $u\in U$, then $u^{-1}U$ is an open identity neighborhood contained in $H$, and $H={\bigcup }_{h\in H}h(u^{-1}U)$ is open. It is locally compact Hausdorff by [F4], and $K$ is a compact generator, so $H\in \mathcal{C}$ and $g\in H$.

For any compact $Q\subseteq G$, the compact set $Q\cup \{e\}$ is covered by the open subgroups in $\mathcal{C}$ by step 1.1. Take a finite subcover $H_1,\dots ,H_n$ with $n\ge 1$ by [F5], and for each $i$ take a compact generating set $K_i$ for $H_i$ by [F3]. The finite union $K={\bigcup }_{i=1}^n{K}_i$ is compact, and $H_0=⟨K⟩$ contains every $H_i$; because it contains the open subgroup $H_1$, it is open and locally compact Hausdorff by [F4]. Thus $H_0\in \mathcal{C}$ and $Q\subseteq H_0$.

Suppose for contradiction that $G$ is not compactly generated. Then every $H\in \mathcal{C}$ has infinite index: if $G\mathrm{/}H$ were finite, adjoining finitely many left coset representatives to a compact generating set of $H$ would give a compact set by [F5] generating $G$ by [F3]. By [F6], each quasi-regular representation ${\lambda }_{G\mathrm{/}H}$ has no nonzero $G$-invariant vector. Form the Hilbert direct sum $\pi ={\widehat{⨁}}_{H\in \mathcal{C}}{\lambda }_{G\mathrm{/}H}$; this is a strongly continuous unitary representation by [F7]. Its invariant vectors are coordinatewise invariant, so $\pi$ has no nonzero invariant vector.

For every compact $Q\subseteq G$, choose $H_0\in \mathcal{C}$ containing $Q$ by step 2.1. The vector ${\delta }_{H_0}$ in its coordinate of $\pi$ is a unit vector fixed by every $q\in Q$ by [F6]. Therefore $\pi$ has almost invariant vectors.

By property (T) and [F1], $\pi$ has a nonzero invariant vector. At least one coordinate of this vector is nonzero, and that coordinate is $G$-invariant in some ${\lambda }_{G\mathrm{/}H}$; [F6] then says that $G\mathrm{/}H$ is finite. The finite-index argument in step 2.2 makes $G$ compactly generated, contradicting the assumption there. Hence $G$ is compactly generated.

If $G$ is discrete, every compact subset is finite: the cover of a compact subset by its open singletons has a finite subcover by [F5]. A compact generating subset supplied by step 4.1 is therefore finite, so $G$ is finitely generated.