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.

  • 1
    $\begingroup$ Have you tried using the characterization free=quasi-free+projective and then using embedding problems? $\endgroup$ Jul 9, 2014 at 19:27
  • $\begingroup$ 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? $\endgroup$
    – Pablo
    Jul 9, 2014 at 19:54
  • $\begingroup$ Should your H in the last line be N? I don't follow it. $\endgroup$ Jul 9, 2014 at 19:55
  • $\begingroup$ I have edited it now... yes. $\endgroup$
    – Pablo
    Jul 9, 2014 at 19:56
  • 1
    $\begingroup$ I still can't parse your comment because it seems G has two meanings. $\endgroup$ Jul 9, 2014 at 20:06

1 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.

  • $\begingroup$ Do you use the splitness of the embedding problems somehow? Can you simply solve any (finite) embedding problem many times? $\endgroup$
    – Pablo
    Jul 10, 2014 at 15:51
  • $\begingroup$ 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. $\endgroup$ Jul 10, 2014 at 17:37
  • $\begingroup$ 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? $\endgroup$
    – Pablo
    Jul 10, 2014 at 18:33
  • $\begingroup$ I added more details to the answer $\endgroup$ Jul 11, 2014 at 4:52
  • $\begingroup$ 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... $\endgroup$
    – Pablo
    Jul 14, 2014 at 16:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.