Making a profinite group free

Question

Let $F$ be a free profinite group, $G$ a profinite group. Suppose that the free profinite product $F \amalg G$ is a free profinite group. Must $G$ be a free profinite group?

For abstract groups the answer is positive in view of the Nielsen-Schreier theorem.

Answer 1

Following Steinberg comment: If $F$ is free profinite group of infinite rank, and $G$ is projective of rank at most the rank of $F$, then the free product is free. Indeed, it is quasi-free (in the sense that every finite split embedding problem has the rank many solutions) and projective, so free.

Edit: Here are some more details: Let $H$ be the free product of a free profinite group $F$ of infinite rank $m$ and a profinite group $G$ of rank at most $m$.

Claim: $H$ is quasi-free of rank $m$.

Proof: If $\alpha \colon H\to A$ and $\beta \colon B\to A$ are two epimorphisms of profinite groups with $B,A$ finite and $\beta$ not an isomorphism, and if $\beta$ has a group theoretic section $\gamma\colon A\to B$, then we can construct $m$ distinct solutions as follows: Since $F$ is free we have $m$ distinct solutions to the restricted EP: $\alpha|_{F}\colon F\to \alpha(F)$ and $\beta'\colon \beta^{-1}(\alpha(F))\to A$, denote them by $\psi_i\colon F\to B$. Then for each $i$, $\psi_i$ and $\gamma\circ \alpha|_{G}\colon G\to B$ define a homomorphism $\Phi_i\colon H\to B$, by the universal property of free products.

Clearly the $\Phi_i$ are distinct (since there restrictions to $F$ are $\psi_i$). Since $\psi_i$ is surjective, $\psi_i(F)$ contains the kernel of $\beta$, hence also $\Phi_i(H)$. But $\beta(\Phi_i(H))=\alpha(H)=A$, so $\Phi_i(H)=B$.

This finishes the proof of the claim.

Now if in addition $G$ is projective, then clearly $H$ is projective, hence by the theorem Benjamin Steinberg mentioned above that says quasi-free+projective=free, $H$ is free.