$\newcommand{\sym}[1]{{\rm Sym}\,#1}$
Statement (i) of Lemma 3.1
The problem was reported by Joris Buloron.
The statement claims that $$N_G(H)\lambda_H=C_{{\rm Sym}[G:H]}(G\varrho_H).$$
The statement is true, but the proof is incomplete for infinite groups.
The proof of the containment $N_G(H)\lambda_H\leq C_{\sym[G:H]}(G\varrho_H)$ is correct. We only show here that $C_{\sym[G:H]}(G\varrho_H)\leq N_G(H)\lambda_H$.
Let $c\in C_{\sym[G:H]}(G\varrho_H)$ and define $g_0\in G$ by $Hc = Hg_0^{-1}$. Note, for $g\in G$, that $$(Hg)c = H(g \varrho_H)c = Hc (g \varrho_H) = Hg_0^{-1} (g \varrho_H) = Hg_0^{-1}g.$$ Hence $(Hg)c = Hg_0^{-1}g$ holds for each $g\in G$. Thus the inverse $c^{-1}$ maps $H g_0^{-1}g$ to $Hg$ for all $g$. Setting $x = g_0^{-1}g$, we obtain that $c^{-1}$ maps $Hx$ to $Hg_0 x$ for all $x \in G$. If $x=1$, then this gives that $H c^{-1} = Hg_0$.
We claim that $H^{g_0}\leq H$. Let $h\in H$. Then $c(h\varrho_H)=(h\varrho_H) c$, and so $$Hg_0^{-1}h=Hc(h\varrho_H)=H(h\varrho_H)c=Hhc=Hc=Hg_0^{-1}.$$ That is, $Hg_0^{-1}hg_0=H$, and hence $g_0^{-1}hg_0\in H$ holds for all $h\in H$.
Our conclusion that $g_0 \in N_G(H)$ is not necessarily valid when $H$ is infinite. Our proof to this point showed only that $H^{g_0}\leq H$. We need also to show the reverse inclusion, or equivalently, we need to show that $H^{g_0^{-1}}\leq H$. To complete the proof note that $c\in C_{\sym[G:H]}(G\varrho_H)$ implies that $c^{-1}\in C_{\sym[G:H]}(G\varrho_H)$ and $Hc^{-1}=Hg_0$. Applying the argument above for $c^{-1}$, we obtain that $H^{g_0^{-1}} \leq H$; that is, $H\leq H^{g_0}$, and so $H^{g_0}= H$. Therefore $g_0\in N_G(H)$.
On line 4 of page 47, we should have written $H(h \lambda_H) = h^{-1}H$ instead of $Hh^{-1}$.