Theorem.

There are two subgroups H1 and H2 of a free group F such that H1 and H2 satisfy conditions (C1)–(C2) and r¯⁢(H1)=3, r¯⁢(H2)=5, r¯⁢(H1∩H2)=8, and r¯⁢(〈H1,H2〉)=2. In particular, inequality (1) fails for H1,H2.

Proof.

Let 𝒜 be a set, called an alphabet, and let 𝒜±1:=𝒜∪𝒜-1, where 𝒜-1 is the set of formal inverses of elements of 𝒜, 𝒜-1 is disjoint from 𝒜. Let F⁢(𝒜) denote the free group whose free generators are elements of 𝒜. A graph Γ is called labeled if Γ is equipped with a function φ:E⁢Γ→𝒜±1, where E⁢Γ is the set of oriented edges of Γ, such that φ⁢(e-1)=φ⁢(e)-1 for every e∈E⁢Γ. If p=e1⁢…⁢eℓ is a path in a labeled graph Γ, where e1,…,eℓ∈E⁢Γ, then the label φ⁢(p) of p is the word φ⁢(p):=φ⁢(e1)⁢…⁢φ⁢(eℓ) over 𝒜±1. The initial vertex of a path p is denoted p- and the terminal vertex of p is denoted p+.

We now construct some subgroups of a free group by means of their Stallings graphs, for definitions see [1, 12, 17].

Let k,ℓ,m,n be some positive integers such that m≥max⁡(ℓ,n). Consider three closed labeled paths qa,qb,qc such that φ⁢(qa)=ak⁢ℓ, φ⁢(qb)=bk⁢m and φ⁢(qc)=ck⁢n. A vertex of qx, x∈{a,b,c}, at a distance from (qx)-=(qx)+ that is divisible by k, is called a phase vertex of qx. We connect some ℓ phase vertices of qb to the ℓ phase vertices of qa with edges f1,…,fℓ so that (fi)-∈qb, (fi)+∈qa are all distinct and φ⁢(fi)=d, i=1,…,ℓ. We also connect some n phase vertices of qb to the n phase vertices of qc with edges g1,…,gn so that (gi)-∈qb, (gi)+∈qc are all distinct and φ⁢(gi)=e for all i=1,…,n. Let Γ⁢(k,ℓ,m,n) denote a labeled graph constructed this way, where

see Figure 1 where the graph Γ⁢(1,1,1,1), also denoted Γ⁢(1¯), is depicted.

Note that Γ⁢(k,ℓ,m,n) is not uniquely defined because there are choices for (fj)-,(gj′)-∈qb. Let H⁢(Γ⁢(k,ℓ,m,n)) denote the subgroup of the free group F⁢(𝒜) which consists of all the words φ⁢(p), where p is a closed path in a fixed graph Γ⁢(k,ℓ,m,n) such that p-=p+=(qb)-. It is immediate from the definitions that H⁢(Γ⁢(1¯))=〈b,d⁢a⁢d-1,e⁢c⁢e-1〉 and, for every Γ⁢(k,ℓ,m,n), the subgroup H⁢(Γ⁢(k,ℓ,m,n)) is a subgroup of H⁢(Γ⁢(1¯)). It is easy to check that

Consider two such graphs Γ⁢(7,1,3,2), Γ⁢(11,2,5,3) and associated subgroups

Denote Δ1:=Γ⁢(7,1,3,2) and Δ2:=Γ⁢(11,2,5,3). Since mi=ℓi+ni for the graph Δi, i=1,2, we may assume that every phase vertex of Δ1 and Δ2 has degree 3. Note that the parameters are chosen so that gcd⁡(|qx,1|,|qx,2|)=1 for every x∈{a,b,c}. Indeed, |qa,1|=7, |qa,2|=22, |qb,1|=21, |qb,2|=55, |qc,1|=14, |qc,2|=33. Now it is easy to see that the entire pullback Δ1×Γ⁢(1¯)Δ2 is a graph of the form Γ⁢(77,2,15,6). In particular, this pullback is connected and r¯⁢(H1∩H2)=-χ⁢(Γ⁢(77,2,15,6))=8. On the other hand, it is immediate that 〈H1,H2〉=H⁢(Γ⁢(1¯)), hence r¯⁢(〈H1,H2〉)=2. It remains to show that the subgroups H1,H2 satisfy conditions (C1)–(C2).

Suppose that, for i=1,2, Ki is a finitely generated subgroup of Hi and K1,K2 satisfy K1∩K2=H1∩H2. Let Δ⁢(Ki) denote the Stallings graph of Ki, i=1,2. Then there is a canonical locally injective graph map Δ⁢(Ki)→Γ⁢(1¯), i=1,2, that factors through Δi. Since K1∩K2=H1∩H2, it follows that

is equal to core⁢(Δ⁢(K1)×Γ⁢(1¯)Δ⁢(K2)) and the projection map Ψ→Δi factors through Δ⁢(Ki), i=1,2. Observe that for every vertex v∈V⁢Δi of degree 3, there exists a vertex w∈V⁢Ψ of degree 3 that maps to v. Hence, there is a vertex u∈V⁢Δ⁢(Ki) of degree 3 that maps to v. This means that -χ⁢(Δ⁢(Ki))≥-χ⁢(Δi), whence r¯⁢(Ki)≥r¯⁢(Hi), i=1,2. If, in addition, Ki is a free factor of Hi, we also have r¯⁢(Ki)≤r¯⁢(Hi). Therefore, r¯⁢(Ki)=r¯⁢(Hi) and Ki=Hi as desired. Thus condition (C1) is proven.

Observe that if pi is a closed reduced path in Δi, then the sum of exponents ε on edges xε of p, where φ⁢(x)∈{a,b,c}, is divisible by 7 if i=1 and is divisible by 11 if i=2. Since 〈H1,H2〉=〈d⁢a⁢d-1,b,e⁢c⁢e-1〉, we see that no path p∈Hi, where Hi=π1⁢(Δi,(qb,i)-) is the fundamental group of Δi at (qb,i)-, can be a free generator of H⁢(Γ⁢(1¯))=〈d⁢a⁢d-1,b,e⁢c⁢e-1〉. Hence condition (C2) is also satisfied for H1, H2, and the theorem is proved. ∎

Problem.

Does inequality (1) hold true under the assumption

Equivalently, does this assumption imply that r¯⁢(〈H1,H2〉)=1?

Example.

Suppose that H0 and H1 are proper normal subgroups of finite index of a free group F⁢(a,b) and F⁢(a,b)=H0⁢H1. For instance, one can take normal subgroups H0 and H1 of F⁢(a,b) so that the quotient groups F⁢(a,b)/H0, F⁢(a,b)/H1 have coprime orders. Then

and

Next, let

be a subgroup of the free group F⁢(a,b,c1,…,ck-1). Then it is not difficult to check that

and

which implies that

Hence, the reduced rank r¯(〈H1,H2, S(H1,H2)〉) could be arbitrarily high in the situation when r¯⁢(H1,H2)=r¯⁢(H1)⁢r¯⁢(H2)>0.