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.



Making a profinite group free
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. 


