# Every profinite group is a quotient of a profinite free group by a normal subgroup that is free profinite?

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.

• What do you mean by two free groups? – Yiftach Barnea Nov 15 '17 at 10:34
• @YiftachBarnea I edited my question. – symmetry Nov 15 '17 at 10:39
• 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. – YCor Nov 15 '17 at 13:17
• @YCor the title looks promising! thanks – 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)], $$N\cong F$$.

• Maybe it would help the reader if you say before "Let $S$..." that you're going to prove that $N$ is free. – YCor Dec 16 '18 at 14:52
• 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 $F$? – YCor Dec 16 '18 at 14:54
• @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. – Dan Haran Dec 17 '18 at 4:45