of ${\hat{G}}$ on ${\mathcal{O}}_\infty$ whose crossed product is purely infinite. Our method is close to that of [@mathOA14061208].

Pick a finite sequence of elements $S := (g_1, g_2, \ldots g_n)$ in $G$. Then this induces a finite dimensional representation $$\pi_S : C(G) \to \mathbb{M}_n, \ \pi_S(f) = {\rm diag}(f(g_i))$$ and the corresponding representation $U_S := {\rm diag}(\lambda_{g_i}) \in M(C^*_\lambda(G) \otimes \mathbb{M}_n)$ of ${\hat{G}}$. Observe that $$U_S \otimes U_T := (U_T)_{13} (U_S)_{12} \simeq U_{S \otimes T}$$ where $S \otimes T := \{hg \mid g \in S, h \in T\}$. Hereafter, we fix symmetric $S = (g_1,g_2, \ldots, g_n)$ which topologically generates $G$ and contains $1$.

We prepare the following elementary lemma which plays an essential role in the proof of purely infiniteness of the crossed product.

\[Lemma\] For any $\varepsilon > 0$ and any finite subset $1 \not \in F \subset {\operatorname{\rm Irr}}(G)$, there exists $k \in {\mathbb Z}_+$ and a nonzero projection $p \in \mathbb{M}_n^{\otimes k}$ such that $$\| (\pi \otimes {\mathrm{id}})((1_{C^*_\lambda(G)} \otimes p)(U_S^{\otimes k})^* (1_{C^*_\lambda(G)} \otimes p)U_S^{\otimes k}) \| _{{\mathbb{B}}({\mathscr{H}}_\pi) \otimes \mathbb{M}_n^{\otimes k}} < \varepsilon$$ for any $\pi \in F$.

Since $\displaystyle \bigcup_k S^{\otimes k}$ is dense in $G$, we have a net of vectors $\xi_i \in \ell^2(S^{\otimes {k_i}})$ 