It is well known that any group is a quotient a free group by a normal subgroup that is free. More precisely if $G$ is a group the exists a short exact sequence of groups $$1\rightarrow F^{'}\rightarrow F\rightarrow G\rightarrow 1 $$ where $F^{'}$ and $F$ are free groups.

Q1: Is any profinite group a quotient of a free profinite group by a normal subgroup that is free profinite?

Assuming that the answer to the first question is negative

Q2: what can we say about a profinite group if initially we know that it is a quotient of a free profinite group by a normal subgroup that is free profinite?

By the second question I do mean if such profinite group has some cohomological properties.

  • 1
    $\begingroup$ What do you mean by two free groups? $\endgroup$ – Yiftach Barnea Nov 15 '17 at 10:34
  • $\begingroup$ @YiftachBarnea I edited my question. $\endgroup$ – symmetry Nov 15 '17 at 10:39
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    $\begingroup$ Possibly relevant: Melnikov, Normal subgroup of free profinite groups 1978 AMS, Mathematics of the USSR-Izvestiya, 12(1) (iopscience.iop.org/article/10.1070/IM1978v012n01ABEH002289). It addresses the problem of understanding which normal subgroups of free profinite groups are free profinite. $\endgroup$ – YCor Nov 15 '17 at 13:17
  • $\begingroup$ @YCor the title looks promising! thanks $\endgroup$ – symmetry Nov 16 '17 at 7:52

We answer Q1 in affirmative, using the mentioned theory of Oleg V. Melnikov. We refer to Fried, Jarden, Field Arithmetic, 3rd edition as [FJ].

First some notation. For a group $F$ we denote by $|F|$ its cardinality.

Let $F$ be a free profinite group of an infinite rank $m\ge|G|$. There is an epimorphism $F \to G$. Let $N$ be its kernel, then $G \cong F/N$.

Let $S$ be a finite simple group. Then $r_F(S)=m$, that is, the largest quotient of $F$ which is the product of copies of $S$, is isomorphic to $S^m$ [FJ, Lemma25.7.1]. Notice that $|S^m|=2^m$. The quotient map $\pi\colon F \to S^m$ maps $N$ onto a closed normal subgroup of $S^m$, hence $\pi(N)\cong S^\kappa$ for some cardinality $\kappa\le m$.

Now, $|\pi(F)/\pi(N)|\cdot|\pi(N)|=|\pi(F)|$, that is, $|\pi(F)/\pi(N)|\cdot 2^\kappa=2^m$, but $|\pi(F)/\pi(N)|\le |F/N|=|G|\le m$, so $\kappa= m$. Hence, by [FJ, Theorem 25.7.3(b)], $N\cong F$.

Another version of the proof: Instead of $m\ge |G|$ assume just $m\ge\text{rank}\;G$. Then there is still an epimorphism $F \to G$. In fact, we can take it to be the composition of epimorphisms $F\to F\times G \to G$; then its kernel $N$ has $F$ as a quotient. Then [FJ, Lemma 24.9.2(a)] gives $r_N(S)\ge r_F(S) = m$, so $r_N(S)=m$, for every finite simple group $S$. Hence, by [FJ, Theorem 25.7.3(b)], $N\cong F$.

  • $\begingroup$ Maybe it would help the reader if you say before "Let $S$..." that you're going to prove that $N$ is free. $\endgroup$ – YCor Dec 16 '18 at 14:52
  • $\begingroup$ A next question is, when $G$ is 2nd-countable, whether one can choose $F$ 2nd-countable (i.e., of at most countable rank). Maybe it follows, just by choosing a suitable large enough subgroup of your $F$? $\endgroup$ – YCor Dec 16 '18 at 14:54
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    $\begingroup$ @YCor: Concerning your first comment: I did not want to kill the tension at the beginning of the story... Concerning your second comment: I edited the answer to answer your question. $\endgroup$ – Dan Haran Dec 17 '18 at 4:45

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