This question is related to [this](https://mathoverflow.net/questions/95384/open-subgroups-of-free-profinite-groups) mathoverflow question that I've asked recently. 


The question rose while I prepared my lectures on Profinite Groups in an advance course in Tel Aviv University. Let $\mathcal{C}$ be a family of finite groups, and $F$ a free pro-$\mathcal{C}$ group on a basis $X$. I always assume that $\mathcal{C}$ is closed to taking quotients and to taking fiber products (the latter can be reformulated as if $N\_1,N\_2$ are normal subgroups of some group $G$ and if $G/N\_1,G/N\_2\in \mathcal{C}$, then $G/N\_1\cap N\_2\in \mathcal{C}$). Also, to avoid trivialities, we assume that $\mathcal{C}$ contains a group generated by at most $|X|$ elements. 


In the literature I find this theorem:



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> **Theorem.** In the above notation and assumption, if $\mathcal{C}$ is closed to taking normal subgroups and extensions, then every open pro-$\mathcal{C}$ subgroup $H$ of $F$ is free pro-$\mathcal{C}$. 
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> 
> 


In the proof the extra closeness conditions are crucially used, so it seems that the proof can't be (easily) modified to relax the conditions on $\mathcal{C}$. 


On the other hand if $\mathcal{C}$ is, for example, the family of finite abelian groups, then $\mathcal{C}$ is not closed to extensions, but still satisfies the theorem. 


My question is 



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> For which $\mathcal{C}$ the theorem above holds true? Does it suffice that $\mathcal{C}$ is closed to quotients, fiber products, and subgroups? 
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> 
> 



Open subgroups of free pro-C groups
No, it does not hold if $C$ is a class of finite nilpotent groups.


