Semi-free subgroups of a profinite surface group

Theorem 1.1.

Let F be a nonabelian finitely generated free profinite group, and let H≤cF be a closed subgroup of infinite index in F. Then there exists a free profinite subgroup H≤L≤cF of rank ℵ0.

Theorem 1.2.

Let N⊲cΓg be a normal subgroup of infinite index in a profinite surface group of genus g≥2. Then there exists a semi-free profinite subgroup M⊲cΓg that contains N.

Theorem 1.3.

Weakly maximal subgroups of profinite surface groups are semi-free profinite.

Definition 2.1.

An infinite profinite group G is called just infinite if for every {1}≠N⊲cG the quotient G/N is finite. Equivalently, every nontrivial normal subgroup of G is open. We say that G is hereditarily just infinite if every open normal subgroup of it is just infinite.

Definition 2.2.

Let H be a closed subgroup of infinite index in a profinite group G. We say that H is weakly maximal in G if every H⪇K≤cG is open.

Definition 2.3.

Given groups G,A,B and surjective homomorphisms

we define the embedding problemℰ⁢(G,A,B,α,β) as the problem of finding a homomorphism φ:G→A such that β=α∘φ. Such a homomorphism φ is called a solution to the problem. If moreover φ is surjective, then it is called a proper solution. An embedding problem ℰ⁢(G,A,B,α,β) is called finite if A is finite. Note that since α is surjective, B is also finite.

Definition 2.4.

An embedding problem ℰ⁢(G,A,B,α,β) is called split if there is a homomorphism γ:B→A such that α∘γ=idB. In such a case we have A≅Ker⁢(α)⋊B.

Definition 2.5.

A profinite group G of rank ℵ0 is called semi-free if every finite split embedding problem ℰ⁢(G,A,B,α,β) has a proper solution.

Definition 2.6.

The profinite surface group of genus g is the group given by the profinite presentation

The set x1,…,xg,y1,…,yg from such a presentation will be called a surface basis.

Fact 2.7.

An open subgroup H of a genus g profinite surface group Γ is a profinite surface group of genus [Γ:H](g-1)+1.

Claim 2.8.

Let E⁢(G,A,B,α,β) be an embedding problem and let φ:G→A be a solution of E. If Ker⁢(α)⊆Im⁢(φ), then φ is a proper solution.

Proof.

Let a∈A. Denote b=α⁢(a) and let g∈G be such that β⁢(g)=b. Then

therefore

As Im⁢(φ) is a subgroup, we conclude that a∈Im⁢(φ) so Im⁢(φ)=A. ∎

Lemma 3.1.

Let

be a profinite surface group, and let N≤cΓ. Consider the diagram

where A,B are finite groups, α is a surjection, and β¯|N=β is a surjection as well. Denote by

the two finite embedding problems in the above diagram. Let φ:Γ→A be a solution to E¯, and set

Suppose that g≥s⁢n+s and that for all 1≤i,j≤s⁢n+s,

Then E admits a proper solution.

Proof.

Choose a set of generators B:={k1,…,ks} of K. Define η⁢(x1)=φ⁢(x1) and

Let η coincide with φ for all other generators of Γ, that is,

Since

we get

Therefore

Thus η extends to a homomorphism. Since x1-1⁢xi∈N for 1≤i≤s⁢n+s, we conclude that kj+1=η⁢(x1-1⁢xj⁢n+2)∈η⁢(N) for 0≤j≤s-1, hence η⁢(N) contains K, so the result follows by invoking Claim 2.8. ∎

Lemma 3.2.

Let Γg be a profinite surface group of genus g, let

be a finite split embedding problem, and suppose that g≥2⁢|A|2⁢|B|. Then E has a proper solution, and Γg has a surface basis x1,…,xg,y1,…,yg such that

Furthermore, if N≤cΓg is such that β⁢(N)=B and

then the embedding problem E¯=E⁢(N,A,B,α,β|N) is properly solvable.

Proof.

Let z1,…,zg,w1,…,wg be a surface basis of Γg. Let γ be a section of α and set φ=γ∘β. As before, put K=Ker⁢(α), s=d⁢(K), and n=|K⁢φ⁢(Γg)|=|A|. Note that

and thus

Each of the pairs (φ⁢(zi),φ⁢(wi)) can attain at most |B|2 values, whence by the pigeonhole principle (since g≥m⁢|B|2) there are

such that

Suppose j1≠1. Then

and so we can replace {zi,wi}i=1g with a new surface basis

By repeating this process with j2,…,jm, we obtain a surface basis x1,…,xg, y1,…,yg of Γg such that xi=zji,yi=wji for 1≤i≤m, and so

Since s⁢n+s≤m, we can apply Lemma 3.1 (with N=Γg if necessary) and the result follows. ∎

Corollary 3.3.

The finite split embedding problem

has a proper solution once g≥2⁢|K|2⁢|H|3, Γg is a profinite surface group of genus g, and G is any profinite group.

Proof.

Write H0 for β⁢(Γg). According to Lemma 3.2, the finite split embedding problem

has a proper solution φ1. Let φ2:G→K⋊H be the map defined by

for every f∈G. There exists a unique homomorphism

such that

By the universal property of free products, φ is a solution to the original embedding problem. Since K is contained in Im⁢(φ1), it is also contained in Im⁢(φ) so by Claim 2.8, φ is a proper solution. ∎

Theorem 3.4.

Let N⊲cΓg be a normal subgroup of the profinite surface group of genus g≥2 such that Γg/N is hereditarily just infinite. Then N is semi-free.

Proof.

Let

be a finite split embedding problem for N. We shall prove it has a proper solution.

Using [3, Lemma 1.2.5 (c)], we can extend our embedding problem to a subgroup

such that

where m=2⁢|K|2⁢|H|. By Fact 2.7, F is a profinite surface group of genus

Note that

Applying Lemma 3.2 to F, the extended embedding problem and m, we obtain a proper solution φ, and a surface basis x1,…,xh,y1,…,yh of F such that

If xi⁢x1-1∈N for all 1≤i≤m, then by Lemma 3.2, our original embedding problem has a proper solution. Assume henceforth that xi⁢x1-1∉N for some i with 1≤i≤m.

The homomorphism β factors modulo

Note that

and that

so β even factors through

where G=〈x1,y1∣[x1,y1]m=1〉.

Let

be the quotient map, and set M=ψ⁢(N). Since Γg/N is hereditarily just infinite, F/N is just infinite, so ψ-1⁢(M) either equals N or is open in F. The former is impossible as there exists 1≤i≤m such that

Hence ψ-1⁢(M) is an open subgroup of F, so M can be seen as an open subgroup of Γh-m∐G.

We can now write the original embedding problem as

so it is sufficient to properly solve it for M. Applying the Kurosh theorem (see [6, Theorem D.3.1]), we find that M≅Γt∐G¯, where

and so by Corollary 3.3 the desired proper solution exists. ∎

Theorem 3.5.

Let Γg be a surface group and N≤cΓg weakly maximal. Then N is semi-free profinite.