If $F$ is a prosoluble subgroup of a free profinite product $\amalg G_i$ and $F \cap G_i^g$ is pro-$p$, is also $F$ pro-$p$?

Question

There is a 1995 paper (Manusc. Math., DOI link) of Florian Pop where he proves the following:

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Theorem. Let $G$ be the free product of profinite groups $G_i$ and $F$ a closed prosoluble subgroup of $G$. One of the following statements holds:

(i) There is a prime $p$ such that $F \cap G_i^g$ is pro-$p$ for every $i \in I$ and $g \in G$.

(ii) All non-trivial intersections $F \cap G_i^g$ for $i \in I$ and $g \in G$ are finite and conjugate in $F$.

(iii) Some conjugate of $F$ is a subgroup of $G_i$ for some $i \in I$

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It arises a natural question to me:

Is there some non pro-$p$ subgroup $F$ satisfying the item (i) (considering at least one non-trivial intersection)?

I'm working on it for a few days but I cannot conclude anything. So, I would like to know if someone knows an example of such group, an argument showing that such groups does not exists or even if there is no anwser.

Thanks in the advance.

Answer 1

Let $I = \{1,2\}$ and let $G_1 = G_2$ be groups of order $p=2$. Their free product is $G = \langle \delta, \varepsilon |\, \delta^2= \varepsilon^2=1\rangle$, which is $\langle \tau, \varepsilon |\, \varepsilon^2=1, \tau^\varepsilon = \tau^{-1} \rangle$, where $\tau = \varepsilon \delta$. So $G$ is a profinite dihedral group, the semi-direct product of $\langle \varepsilon \rangle$ and $\langle \tau \rangle$, with $\langle \varepsilon \rangle$ acting on $\langle \tau \rangle$. This presentation shows that $G$ is prosoluble, but not pro-$2$.

Put $F = G$. Then the intersections in (i) are $G_i^g$ of order $2$, but $F$ is not pro-$2$.