Proof, [Proposition 1.3.10, AP]. For each $g \in H \wr_{I} \Gamma$ and subgroup $K \subset H \wr_{I} \Gamma$, let $g^{K} = \bracsn{kgk^{-1}|k \in K}$.

Let $g = (\xi, \gamma) \in H \wr_{I} \Gamma$. First suppose that $\xi \ne e_{H^{(I)}}$, then the support of $\xi$ is non-empty. By (a), the support of $\xi$ has infinite orbit under $\Gamma$, so $g^{G} \supset g^{\Gamma}$ is infinite.

Now suppose that $\xi = e_{H^{(I)}}$ and $\gamma \ne e_{\Gamma}$. If $\gamma \in \Gamma \setminus \Gamma_{f}$, then the conjugacy class of $\gamma$ is infinite, and the conjugacy class of $g$ is also infinite. Now suppose that $\gamma \in \Gamma_{f}$. By (b), there exists $i_{0} \in I$ with $\gamma i_{0} \ne i_{0}$. As $H$ is non-trivial, there exists $\xi_{0} \in H^{(I)}\setminus \bracsn{e_{H^{(I)}}}$ with support $\bracsn{i_0}$. In which case, if $g_{0} = (\xi_{0}, e_{\Gamma})$, then

and hence $g$ has infinite conjugacy class because $\xi_{0}^{-1}\gamma\xi_{0} \ne e_{H^{(I)}}$. Therefore $g^{G}$ is also infinite.$\square$