This question is related to this 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:

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-$\mathcal{C}$.

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

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

  • $\begingroup$ @Lior, as soon as you lose closure under extensions open subgroups of free discrete groups don't inherit their pro-C topology from the ambient. This is a problem. $\endgroup$ Apr 29, 2012 at 16:02

1 Answer 1


No, it does not hold if $C$ is a class of finite nilpotent groups.


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