most recent 30 from http://mathoverflow.net 2013-06-20T12:38:42Z http://mathoverflow.net/feeds/question/95495 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/95495/open-subgroups-of-free-pro-c-groups Lior Bary-Soroker 2012-04-29T09:19:51Z 2012-04-29T11:01:05Z

<p>This question is related to <a href="http://mathoverflow.net/questions/95384/open-subgroups-of-free-profinite-groups" rel="nofollow">this</a> mathoverflow question that I've asked recently. </p> <p>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. </p> <p>In the literature I find this theorem:</p> <blockquote> <p><strong>Theorem.</strong> 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-$\mathcal{C}$. </p> </blockquote> <p>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 $\mathcal{C}$. </p> <p>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. </p> <p>My question is </p> <blockquote> <p>For which $\mathcal{C}$ the theorem above holds true? Does it suffice that $\mathcal{C}$ is closed to quotients, fiber products, and subgroups? </p> </blockquote>

http://mathoverflow.net/questions/95495/open-subgroups-of-free-pro-c-groups/95498#95498 Pavel Zalesskii 2012-04-29T11:01:05Z 2012-04-29T11:01:05Z

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