# Making a profinite group free

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.

• Have you tried using the characterization free=quasi-free+projective and then using embedding problems? Jul 9, 2014 at 19:27
• I am not able to solve embedding problems here... A counterexample may be to take a closed normal nonfree subgroup $G$ of a free f.g profinite group $L$. Now taking free profinite product with a free infinitely generated profinite group $F$ will make a free profinite group of $G$ since the new quotient group $L \amalg F / G \amalg F$ is f.g. Is this true? Jul 9, 2014 at 19:54
• Should your H in the last line be N? I don't follow it. Jul 9, 2014 at 19:55
• I have edited it now... yes. Jul 9, 2014 at 19:56
• I still can't parse your comment because it seems G has two meanings. Jul 9, 2014 at 20:06

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.

• Do you use the splitness of the embedding problems somehow? Can you simply solve any (finite) embedding problem many times? Jul 10, 2014 at 15:51
• The point is that if you can solve split EPs many times and Frattini EPs once, then you can solve any finite EP many times. Jul 10, 2014 at 17:37
• You don't give a proof here so I don't know if the way you solve the embedding problem uses its splitness in some way, or you can handle a general embedding problem directly. Do you use the wreath product approach? Jul 10, 2014 at 18:33
• I added more details to the answer Jul 11, 2014 at 4:52
• You could have avoided the use of split EPs or finite ones since you have only used it to solve weakly for $G$ which is automatic if $G$ is projective. In fact, this has been carried out in Proposition 9.1.11 of Ribes-Zalesskii 2nd ed. What we get here is in fact an equivalent characterization for projectivity ($G$ is projective iff $G \amalg F$ is free). Furthermore, we get an abundance of examples of quasi-free groups which are not free (groups of the form $F \amalg G$ with $F$ free and $G$ not projective). It would be interesting to know if all quasi-free groups arise in this way... Jul 14, 2014 at 16:45