Proof, [Theorem 3.5, Fol16]. (1): $(H, \pi)$ is irreducible if and only if $\pi(G)'$ contains no non-trivial projections. Since $\pi(G)'$ is a von Neumann algebra, Theorem 39.6.2 implies that the span of projections in $\pi(G)'$ is dense in $\pi(G)'$. Therefore $(H, \pi)$ is irreducible if and only if $\pi(G)' = \complex I$.

(2): For any $T \in \mathcal{C}(\pi, \tau)$, $T^{*} \in \mathcal{C}(\tau, \pi)$, so $T^{*}T \in \mathcal{C}(\pi, \pi) = \pi(G)'$, and $TT^{*} \in \mathcal{C}(\tau, \tau) = \tau(G)'$. Given that $(H, \pi)$ and $(K, \tau)$ are irreducible, there exist $\lambda, \mu \in \complex$ such that $T^{*}T = \lambda I_{H}$ and $TT^{*} = \mu I_{K}$. As such, either $T = 0$ or $\lambda^{-1/2}T$^[1] is unitary, so $\mathcal{C}(\pi, \tau)$ is non-zero if and only if $(H, \pi)$ and $(K, \tau)$ are unitarily equivalent.

If $\mathcal{C}(\pi, \tau) \ne \bracsn{0}$, then $\mathcal{C}(\pi, \tau)$ consists of scalar multiples of unitary operators. In which case, for any unitary operators $S, T \in \mathcal{C}(\pi, \tau)$, $S^{-1}T = S^{*}T \in \mathcal{C}(\pi, \pi)$. Thus there exists $\lambda \in \complex$ such that $S^{-1}T = \lambda I_{H}$, so $T = \lambda S$, and $\mathcal{C}(\pi, \tau)$ is one-dimensional.$\square$