n_k f_{\eta_k}=0$ with $n_k\in {{\mathbb Z}}$. Choose then $\theta \in [\alpha]^\omega$ with $\eta_1,\ldots , \eta_{m}, \in
    \theta$. Since $\pi_{\mathcal{D}_\theta}^G(f_{\eta_k})=g_{{\eta_k},\theta}$ and the set $\{g_{\eta,\theta} \colon \eta \in \theta\}$ is independent, the independence of $F$ follows. Since $\pi_{\mathcal{D}_\theta}(\chi^\sigma(f_{\eta}))=\chi^{\mathcal{D}_\theta}(g_{\eta,\theta})$, $\chi^\sigma(F)$ is also independent. It is easy to see, now, that $\langle F\rangle$ and $\langle\chi^\sigma(F)\rangle$ are isomorphic.

2.  *The subgroup $\langle \chi^\sigma(F)\rangle$ has property ${{\mathbf \sharp}}$.* Let $N$ be a countable subgroup of ${\langle\,}\chi^\sigma(F){\,\rangle}$. Let $\theta\in [\alpha]^\omega$ be such that $N\subseteq {\langle\,}\chi^\sigma(f_\eta) \colon \eta \in \theta{\,\rangle}$ and define $N_\theta:={\langle\,}f_\eta \colon \eta \in \theta {\,\rangle}$.

    Observe finally that $\pi_{\mathcal{D}_\theta}^{{{\mathbb T\,}}}(N)=
    \chi^{\mathcal{D}_\theta}(\pi_{\mathcal{D}_\theta}^G(N_\theta))$. This last subgroup is just $j_\theta\left(\left\langle \chi^\sigma(
    d_{\eta}) \colon \eta \in \theta \right\rangle\right)$ and the latter carries by construction its maximal totally bounded topology, since the restriction of $\pi_{\mathcal{D}_\theta}^{{{\mathbb T\,}}}\colon
    {{\mathbb T\,}}^{\sigma}\to{{\mathbb T\,}}^{\mathcal{D}_\theta}$ to $N$ is a group isomorphism onto $\pi_{\mathcal{D}_\theta}^{{{\mathbb T\,}}}(N)
    =\chi^{\mathcal{D}_\theta}(\pi_{\mathcal{D}_\theta}^G(N_\theta))$, Lemma \[lem:quotients\] applies.

3.  *$\langle F\rangle $ has property ${{\mathbf \sharp}}$.* Take $\pi=\chi^{\sigma}$, $K=G^\sigma$ and $ L={{\mathbb T\,}}^\sigma$. Bearing in mind that the restriction to ${\langle\,}F{\,\rangle}$ is an isomorphism because $F$ and $\chi^{\sigma}(F)$ are independent sets, Lemma \[lem:quotients\] applies again.

4.  *$F$ is a $G_\delta$-dense subset of $G^\sigma$.* Observe that, for every $\eta<\alpha$, $f_\eta$ coincides with $d_\eta$ on the set of 