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Let $F$ be a finitely generated nonabelian free profinite group, $p$ a prime number, $L \lhd_o F$ with $[F : L]$ coprime to $p$, $N \lhd_c^\infty F$ contained in $L$ with $L/N$ pro-$p$, and $N \leq H \leq_c^\infty F$. Suppose that there exists a discrete $H$-module structure on $M = (\mathbb{Z}/p\mathbb{Z})^m$ for some $m \in \mathbb{N}$ such that $H^1(H,M)$ is finite.

Must $L/N$ be free pro-$p$?

The case when one can take $M = \mathbb{Z}/p\mathbb{Z}$ and trivial as an $H$-module (every element of $H$ acts as an identity) and $H^1(H,M) = \{0\}$ is already of interest.

In this special case, I want to show that $L/N$ is free pro-$p$ given that $\text{Hom}(H,\mathbb{Z}/p\mathbb{Z}) = \{0\}$.

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  • $\begingroup$ Why should it be? $\endgroup$
    – HJRW
    Commented Sep 18, 2014 at 18:54
  • $\begingroup$ I forgot to say that I assume $[F : N] = \infty$ $\endgroup$
    – Pablo
    Commented Sep 18, 2014 at 19:49
  • $\begingroup$ @HJRW: To get a feeling try to take $L = F$ and $H = N$. In this case I am quite sure this works. $\endgroup$
    – Pablo
    Commented Sep 18, 2014 at 20:02
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    $\begingroup$ But why can't you take $H=F$? $\endgroup$
    – HJRW
    Commented Sep 18, 2014 at 20:08
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    $\begingroup$ I see why you're interested in Greenberg's theorem! $\endgroup$
    – HJRW
    Commented Sep 18, 2014 at 20:18

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