On the lattice of subgroups of a free group: complements and rank
jjournal of Groups, Complexity, CryptologyVolume 12, Issue 1, 2020, pp. 1:1–1:24https://gcc.episciences.org/ Submitted Sept. 11, 2019Published Feb. 29, 2020
ON THE LATTICE OF SUBGROUPS OF A FREE GROUP:COMPLEMENTS AND RANK
JORDI DELGADO AND PEDRO V. SILVACentro de Matemática, Universidade do Porto, Portugal e-mail address : [email protected] de Matemática, Universidade do Porto, Portugal e-mail address : [email protected]
Abstract. A ∨ -complement of a subgroup H (cid:54) F n is a subgroup K (cid:54) F n such that H ∨ K = F n . If we also ask K to have trivial intersection with H , then we say that K is a ⊕ -complement of H . The minimum possible rank of a ∨ -complement (resp., ⊕ -complement)of H is called the ∨ -corank (resp., ⊕ -corank) of H . We use Stallings automata to study thesenotions and the relations between them. In particular, we characterize when complementsexist, compute the ∨ -corank, and provide language-theoretical descriptions of the sets ofcyclic complements. Finally, we prove that the two notions of corank coincide on subgroupsthat admit cyclic complements of both kinds. Introduction
Subgroups of free groups are complicated. Of course not the structure of the subgroupsthemselves (which are always free, a classic result by Nielsen and Schreier) but the relationsbetween them, or more precisely, the lattice they constitute. A first hint in this direction isthe fact that (free) subgroups of any countable rank appear as subgroups of the free groupof rank 2 (and hence of any of its noncyclic subgroups) giving rise to a self-similar structure.This scenario quickly provides challenging questions involving ranks. Classical examplesinclude intersections of finitely generated subgroups, and subgroups of fixed points ofautomorphisms; both proved to be finitely generated in the second half of last century(see [9] and [7] respectively), and both having a long and rich subsequent history in thequest for bounds for those finite ranks (see [6, 14] and [4] repectively).In this paper we shall be mainly concerned with a kind of dual of the previous problem:given a finitely generated subgroup H of F n , what is the minimum number of generatorsthat must be added to H in order to generate the full group F n ? What happens if the addedsubgroup is also required to intersect trivially with H ? These numbers, called respectivelythe join ( ∨ ) and direct ( ⊕ ) coranks of H , shall be investigated using Stallings automata.Although previously studied using other techniques, maybe the most enlighteningapproach to subgroups of the free group F n was their geometric interpretation as covering Key words and phrases: free-group, subgroups, lattice of subgroups, Stallings automata,complement, rank, corank, join . Journal of GROUPS,COMPLEXITY, CRYPTOLOGY c (cid:13)
J. Delgado and P. V. Silva CC (cid:13) Creative Commons :2 J. Delgado and P. V. Silva
Vol. 12:1 spaces of the bouquet of n circles. It soon became clear that the (mainly topological)original viewpoint by Serre and Stallings (see [19, 22]) admitted an appealing restatementin terms of automata (see [11, 1]). We briefly summarize this modern approach in Section 3.Furthermore, similarly to the strategy followed in [21] and [15] to deal with free factors, wehighlight the role played by identifications of vertices in the Stallings automaton.In Section 4, we introduce the various notions of complement and corank studied in thispaper.Section 5 is devoted to join complements and join corank. We discuss the possiblecombinations between rank and join corank, show that the latter is always computable (aresult previously proved by Puder in [15]), and prove that the set of join cocycles is alwaysrational. We note that, from a language theoretical viewpoint, proving that a set of solutionsis rational is highly desirable: in view of all the closure properties satisfied by rationallanguages, this allows for an efficient search for solutions of particular types.In Section 6, the simpler case of meet complements is discussed. Existence is essentiallydetermined by the subgroup having finite or infinite index, and this will be useful for thediscussion of direct complements.Section 7 contains the main results of the paper, devoted to the harder case of directcomplements and coranks. Existence is shown to depend on the index of the subgroup only,and the possible combinations between rank and direct corank turn out to be the same asin the join case. The set of cyclically reduced direct cocycles is also rational, but the setof direct cocycles needs not to be: in general, it is only context-free (the next level aboverational, in the classical Chomsky’s hierarchy from language theory). We also prove thatthe concepts of join cocyclic and direct cocyclic coincide, and raise the natural question: dojoin corank and direct corank coincide in the general case?We finally point out that Stallings’ geometric interpretation converts the corank problemsinto problems about equations in automata (that, is equations between automata that includearcs labelled by variable strings). We note that this is a very appealing general problemthat, in particular, includes that of equations in the free group which has become one of themain topics in modern group theory (see [12, 16, 10]).2. Preliminaries
Throughout the paper we assume that A = { a , . . . , a n } is a finite set of letters that we callan alphabet , and we denote by A ∗ the free monoid on A (consisting of all finite words on A including the empty word , which is denoted by 1).The subsets of A ∗ are called languages (over A ), or A -languages. An A -language is saidto be rational if it can be obtained from the letters of A using the operators union, productand star (submonoid generated by a language) finitely many times. A direct consequenceof a fundamental theorem of Kleene (see for example [18, Theorem 2.1]) is that the set ofrational A -languages is closed under finite intersection and complement.We also denote by F n the free group with basis A ; that is, F n = F A = h A | −i . Moreprecisely, we denote by A − the set of formal inverses of A . Formally, A − can be defined asa set A equipotent and disjoint with A , together with a bijection ϕ : A → A ; then, for every a ∈ A , we call aϕ the formal inverse of a , and we write a − = aϕ . So A − = { a − : a ∈ A } ,and A ∩ A − = ∅ .The set A ± = A t A − , called the involutive closure of A , is then equipped with aninvolution − (where ( a − ) − = a ), which can be extended to ( A ± ) ∗ in the natural way: ol. 12:1 ON THE LATTICE OF SUBGROUPS OF A FREE GROUP: COMPLEMENTS AND RANK 1:3 ( a . . . a n ) − = a − n . . . a − for all a , . . . , a n ∈ A ± . An alphabet is called involutive if it isthe involutive closure of some other alphabet.A word in ( A ± ) ∗ is said to be (freely) reduced if it contains no consecutive mutuallyinverse letters (i.e., it has no factor of the form aa − , where a ∈ A ± ). It is well known thatthe word obtained from w ∈ ( A ± ) ∗ by successively removing pairs of consecutive inverseletters is unique; we call it the free reduction of w , and we denote it by w . Similarly, wewrite S = { w : w ∈ S } , for any subset S ⊆ ( A ± ) ∗ , and we denote by R A (or R n ) the set ofreduced words in ( A ± ) ∗ , that is the rational language R A = ( A ± ) ∗ = ( A ± ) ∗ (cid:114) S a ∈ A ± ( A ± ) ∗ aa − ( A ± ) ∗ . In a similar vein, a word in A ± is said to be cyclically reduced if all of its cyclic permutationsare reduced (that is, if it is reduced and its first and last letters are not inverses of eachother). The cyclic reduction of a word w ∈ F n , denoted by w , is obtained after iterativelyremoving from w the first and last letters if they are inverses of each other. We also extendthis notation to subsets, and denote by C A (or C n ) the set of all cyclically reduced words in( A ± ) ∗ , that is the rational language C A = ( A ± ) ∗ = ( A ± ) ∗ (cid:114) S a ∈ A ± (cid:0) ( A ± ) ∗ aa − ( A ± ) ∗ ∪ a ( A ± ) ∗ a − (cid:1) . The free group F A (with basis A ) can then be thought as the set of reduced words in A ± withthe operation consisting of “concatenation followed by reduction”. We recall that Benois’Theorem (see [2]) allows us to understand the rational subsets of F A as reductions of rational A ± -languages. Definition 2.1.
The rank of a group G , denoted by rk( G ), is the smallest cardinality of agenerating set for G .It is well known that the free group F n has rank n , and that every subgroup of F n isagain free and can have any countable rank if n ≥
2. We will see in Theorem 3.5 thatone can biunivocally assign to every subgroup H (cid:54) F n a geometric object — called the Stallings automaton St ( H ) of H — which provides a lot of useful information about thesubgroup. What is more, if H is given by a finite family of generators, then St ( H ) is fastlycomputable (see [24]), and many algorithmic results regarding subgroups of the free groupfollow smoothly from the Stallings construction.For example, we shall see that the rank of a subgroup H of F n is precisely the (graph)rank of St ( H ); hence one can always compute the rank of a finitely generated subgroup H of F n from a finite family of generators.One of the goals in this paper is to to obtain an analogous result for the join corank of H , that is, the minimum number of elements that must be added to H in order to generatethe whole group F n (see Theorem 5.4).2.1. Graphs, digraphs, and automata.
As we have already mentioned, we will useautomata (that is essentially labelled digraphs) to describe subgroups of the free group.Since parallel arcs and loops are allowed, we shall define digraphs in the sense of Serre.
Definition 2.2. A directed multigraph (or digraph for short) is a tuple ~ Γ = ( V , E , ι, τ ),where V is a nonempty set (called the set of vertices of ~ Γ), E is a set (called the set of arcs or directed edges of ~ Γ), and ι, τ : E → V are (resp. initial and final) incidence functions .Then, for each arc e ∈ E , we say that e is incident to ι ( e ) and τ ( e ), which are called the :4 J. Delgado and P. V. Silva
Vol. 12:1 origin (or initial vertex ), and end (or final vertex ) of e , respectively. Two vertices are saidto be adjacent if there exists an edge incident to both of them, and two edges are said to be incident if some vertex is incident to both of them.We denote by V ~ Γ and E ~ Γ the respective sets of vertices and arcs of a digraph ~ Γ. Adigraph ~ Γ is called finite if the cardinal V ~ Γ t E ~ Γ) is finite.Note that no incidence restrictions have been applied; in particular we are allowing boththe possibility of arcs having the same vertex as origin and end (called directed loops ), andof different arcs sharing the same origin and end (called parallel arcs ). Definition 2.3. A walk in a digraph ~ Γ is a finite alternating sequence γ = p e p . . . e n p n of successively incident vertices and arcs in ~ Γ (more precisely ι e i = p i − and τ e i = p i for i = 1 , . . . , n ). We then say that γ goes from p to p n — or that γ is a ( p , p n ) -walk — andwe write γ : p p n . If the first and last vertices of γ coincide then we say that γ is a closed walk . A closed walk from p to p is called a p -walk . The length of a walk is the numberof arcs in the sequence. The walks of length 0, called trivial walks , correspond precisely tothe vertices in ~ Γ. Definition 2.4.
Given an alphabet A , a (pointed) A -automaton Γ = ( P , E , ι, τ, ‘, ) is adigraph ~ Γ = ( P , E , ι, τ ) called the underlying digraph of Γ , together with a labelling on thearcs ‘ : E → A , and a distinguished vertex called the base vertex or basepoint of Γ . Remark 2.5.
In the general automata setting, pointed automata correspond to automatahaving a unique coincident initial and final state.A vertex p in an A -automaton is said to be complete if for every letter a ∈ A there is(at least) one a -arc with origin p . An automaton is complete if all its vertices are complete.Otherwise (if there exists a vertex p and a letter a such that there is no a -arc with origin p ),we say that both the vertex p and the automaton are a - deficient . A vertex is said to be an a -source if it is the origin of one single arc, and this arc is labelled by a . An automatonwhose basepoint is an a -source is also called an a -source.An A -automaton is said to be deterministic if no two arcs with the same label departfrom the same vertex. Hence, a deterministic automaton is complete if and only if everyletter induces a total transformation on the vertex set. Definition 2.6.
The label of a walk γ in an A -automaton Γ = ( V , E , ι, τ, ‘, ) is defined tobe ‘ ( γ ) = (cid:26) , if γ = p , (an empty walk in Γ ) ,‘ ( e ) · · · ‘ ( e k ) , if γ = p e p . . . e k p k (a nonempty walk in Γ ) .We then say that γ reads or spells the word ‘ ( γ ) ∈ A ∗ , and that the word ‘ ( γ ) labels thewalk γ . We also write ‘ ( Γ ) = { ‘ ( e ) : e ∈ E } ⊆ A (the subset of letters appearing as labels ofarcs in Γ ). If P and Q are subsets of vertices in a labelled digraph Γ , then we denote by L PQ ( Γ ) the set of words read by walks from vertices in P to vertices in Q . In particular, if p , q are vertices of Γ , then L pq ( Γ ) = { ‘ ( γ ) : γ : p q in Γ } . In view of the well-knownKleene’s Theorem the languages L PQ ( Γ ) are rational if Γ is finite.The set of words read by -walks in an A -automaton Γ is called the language recognized by Γ , and is denoted by L ( Γ ); that is L ( Γ ) = L ( Γ ). It is clear that L ( Γ ) is a submonoidof the free monoid A ∗ . Definition 2.7. An involutive A -automaton is an A ± -automaton together with an involution e e − on its arcs (called inversion of arcs ) such that: ol. 12:1 ON THE LATTICE OF SUBGROUPS OF A FREE GROUP: COMPLEMENTS AND RANK 1:5 (i) No arc is the inverse of itself (i.e., e − = e , for every e ∈ E ).(ii) Inverse arcs are reversed (i.e., ι e − = τ e , for every e ∈ E ).(iii) Arc inversion is compatible with label inversion (i.e., ‘ ( e − ) = ‘ ( e ) − , for every e ∈ E ).Thus, in an involutive automaton, for every labelled arc e ≡ p q reading a ∈ A , therealways exists a reversed arc e − ≡ p q reading a − (called the inverse of e ), so that arcsappear by pairs as shown in Figure 1. aa − Figure 1: Arcs in an involutive automatonA walk in an involutive automaton is said to be reduced if it has no two consecutiveinverse arcs.An arc in an involutive A -automaton is said to be positive (resp. negative ) if it islabelled with a letter in A (resp. A − ). We respectively denote by E + Γ and E − Γ the sets ofpositive and negative arcs in an involutive automaton Γ . The positive representation of an A -involutive automaton Γ is the A -automaton obtained after removing all the negative arcsfrom Γ . Remark 2.8.
An involutive automaton is fully characterized by its positive representation(with the tacit assumption that every positive arc, say reading a ∈ A , is allowed to be crossedbackwards reading the inverse label a − ). Definition 2.9.
The underlying graph of an involutive automaton Γ is the undirectedmultigraph, denoted by Γ , obtained by identifying all the pairs of respectively inverse arcsin the underlying digraph of Γ . Note that this is the same as ‘forgetting the labels anddirection’ in the positive representation of Γ . Remark 2.10.
Any undirected multigraph Γ can be seen as the underlying graph of someinvolutive automaton; an edge in Γ is then an unordered pair { e , e − } . We shall referto undirected multigraphs simply as graphs (in contraposition to digraphs, introduced inDefinition 2.2). Convention 2.11.
If not stated otherwise, the automata appearing in this paper hereinafterwill be assumed to be pointed and involutive, and we shall represent them by their positivepart. It is clear that the language recognized by such an automaton is the same as thelanguage recognized by the connected component containing the basepoint. With thisconvention, an (involutive) A -automaton Γ is complete if and only if (in the positiverepresentation of Γ ), for every vertex p and every positive letter a ∈ A , there is an a -arcstarting at p and an a -arc ending at p . Definition 2.12.
The (cycle) rank of a graph Γ, denoted by rk(Γ), is the minimum numberof edges that must be removed from Γ to obtain a forest (i.e., to break all the cycles in Γ).It is well known (see for example [3]) that, if Γ is finite, thenrk(Γ) = e − v + c , (2.1)where e , v , and c are respectively the number of edges, vertices, and connected componentsin Γ. :6 J. Delgado and P. V. Silva
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The previous considerations make it possible (and convenient) to extrapolate graph-theoretical notions to involutive automata from their underlying graphs. For example,the rank , connectivity , or vertex degree of an automaton Γ are defined in terms of thehomonymous notions in its underlying graph. In the same vein, we will call an involutiveautomaton a path , a cycle , a tree , or a spanning tree if their underlying graph is so. Remark 2.13.
If an (involutive) automaton Γ is deterministic, then a walk γ in Γ is reducedif and only if its label ‘ ( γ ) is reduced. Definition 2.14.
The reduced label of a walk γ in an involutive A -automaton is defined tobe ‘ ( γ ) = ‘ ( γ ) ∈ F A . We write L PQ ( Γ ) = L PQ ( Γ ). The set of reduced labels of -walks in Γ is a subgroup of F A called the subgroup recognized by Γ , and is denoted by h Γ i , so that h Γ i = L ( Γ ) (cid:54) F A . Definition 2.15.
An automaton Γ is said to be core if every vertex appears in some -walkwith reduced label.Note that, if Γ is deterministic, this is the same as Γ being connected and not having“hanging trees” not containing the basepoint (we speak of hanging trees when we have a treeadjoined to a vertex of degree strictly greater than 1 in a graph). Accordingly, we definethe core of a deterministic automaton Γ , denoted by core( Γ ), to be the automaton obtainedafter taking the basepoint component of Γ and removing from it all the hanging trees notcontaining the basepoint. Note that we then have that h core( Γ ) i = h Γ i . Definition 2.16.
An involutive pointed automaton is said to be reduced if it is deterministicand core (and hence connected).
Definition 2.17. A subautomaton of an automaton Γ is any automaton obtained from Γ through restriction (of vertices or arcs). We then write ∆ (cid:54) Γ (or ∆ < Γ , if ∆ = Γ , that isif ∆ is a strict subautomaton of Γ ).2.2. Operations on automata.
Throughout this paper, two main kinds of operations onautomata appear, namely vertex and arc identification.
Definition 2.18.
Let Γ be an A -automaton and let P be a nonempty subset of verticesof Γ . The quotient of Γ by P , denoted by Γ / P , is defined to be the automaton obtainedafter identifying in Γ the vertices in P (and inheriting the adjacencies, the labelling, and theinitial and terminal vertices from Γ ). Remark 2.19.
Note that V ( Γ / P ) = V Γ − P + 1, and E ( Γ / P ) = E Γ . Hence,rk Γ ≤ rk Γ / P ≤ rk Γ + P − , (2.2)where the lower (resp., upper) bound in (2.2) corresponds to the case where all the vertices in P belong to different (resp., the same) connected components of Γ . In particular, identificationof vertices can never decrease the rank of an automaton.If P is finite, the exact rank of Γ / P is immediately obtained from the case of two vertices(denoted by Γ / p = q = Γ / { p , q } ), which is detailed below. Lemma 2.20.
Let p , q be two different vertices in Γ , then: rk Γ / p = q = (cid:26) rk Γ + 1 if p and q are connected in Γ , rk Γ if p and q are not connected in Γ . ol. 12:1 ON THE LATTICE OF SUBGROUPS OF A FREE GROUP: COMPLEMENTS AND RANK 1:7 Hence, if Γ is finite, the identification of two vertices either increases the rank exactlyby one (if they are connected) or keeps the rank equal (if they are disconnected). Definition 2.21.
Let Γ and ∆ be A -automata. The sum of Γ and ∆ , denoted by Γ + ∆ ,is then the automaton obtained after identifying the basepoints of Γ and ∆ .It is clear that the sum of automata is associative and commutative. We write P ni =1 Γ i = Γ + . . . + Γ n . Then, it is also clear that rk( P i Γ i ) = P i rk( Γ i ), and that h P i Γ i i = h S i h Γ i i i = W i h Γ i i .Regarding arcs, we shall only be interested in a very specific kind of identification. Definition 2.22. A folding in an automaton Γ is the identification of two different arcs in Γ with the same origin and label (inheriting the adjacencies, the labelling, and the basepointfrom Γ ). A folding is said to be closed if the identified arcs are parallel and open otherwise.Recall that our automata are always involutive and we use the convention of onlyrepresenting its positive part, that is, we are always implicitly assuming that every time afolding is performed (in the positive part) the corresponding folding between the respectivelyinverse arcs is also performed. Hence every folding (of arcs) induces an identification betweenthe corresponding incident edges in the underlying graph. Remark 2.23.
Note that, since foldings do not change the number of connected componentsin an automaton, the following statements are equivalent:(a) a folding is closed (resp., open);(b) a folding does not produce (resp., produces) an identification of vertices;(c) a folding reduces by one (resp., does not change) the rank of the automaton.Therefore, if e Γ is obtained from Γ after a sequence of foldings (i.e., if e Γ is a reduction of Γ ),then(i) E e Γ < E Γ (precisely, E + e Γ = E + Γ − V e Γ ≤ V Γ (precisely, V e Γ = V Γ − e Γ ≤ rk Γ (precisely, rk e Γ = rk Γ − h e Γ i = h Γ i .3. Subgroups of free groups and Stallings automata
We have seen that one can naturally assign a subgroup of F n = h A | −i to every A -automaton Γ . We shall see that every subgroup of F n admits such a description, which can be madeunique after adding some natural restrictions to the used automata. Notation 3.1. If p and q are vertices in a tree T (cid:54) Γ , then we denote by p T q theunique reduced walk in T from p to q .At the core of the alluded unicity is the following well-known fact which we state withouta proof. Proposition 3.2.
Let Γ be a connected A -automaton and let T be a spanning tree of Γ .Then the set S T = (cid:8) ‘ ( T • e • T ) : e ∈ E + Γ (cid:114) ET (cid:9) ⊆ F A :8 J. Delgado and P. V. Silva
Vol. 12:1 is a generating set for h Γ i . Furthermore, if Γ is reduced then S T is a free basis for h Γ i . On the other hand, given a reduced word w ∈ F A , we can always consider the petalautomaton Fl ( w ), i.e., the cyclic A -automaton spelling w (or w − if read in the oppositedirection). Then, given a subset S = { w i } i ⊆ F A , we define the flower automaton of S ,denoted by Fl ( S ), to be the automaton obtained after identifying the basepoints of the petalsof the elements in S ; that is, Fl ( S ) = P i Fl ( w i ). w w w p Figure 2: The flower automaton Fl ( w , w , . . . , w p )It is clear that h Fl ( S ) i = h S i (cid:54) F A . Hence, every subgroup H of F A is recognized bysome (clearly not unique) A -automaton. Note however that, if the words in S are reduced,then Fl ( S ) is core, and deterministic except maybe at the basepoint . A key result due toJ. R. Stallings is that determinism is the only missing condition in order to make thisrepresentation unique. Definition 3.3.
The (right)
Schreier A -automaton of a subgroup H (cid:54) G = h A i , denotedby Sch ( H ), is the automaton with vertices the (right) cosets Hg ( g ∈ G ), arcs Hg a Hga for every a ∈ A ± , and basepoint H . Note that Sch ( H ) is always deterministic (but notnecessarily core). Definition 3.4.
The
Stallings A -automaton of a subgroup H of F A is the core of the rightSchreier automaton Sch ( H ). We denote it by St ( H, A ) (or simply by St ( H ) if the generatingset is clear).Note that St ( F A , A ) has one single vertex and arcs labelled by each a ∈ A . Such anautomaton is called a bouquet . We denote by B n a bouquet with n positive arcs.By construction, St ( H, A ) is deterministic, core (i.e., it is a reduced A -automaton) andrecognizes H . Below, we see that every reduced A -automaton is the Stallings automaton ofsome subgroup of F n , making “reduced” and “Stallings” automata equivalent notions. Theorem 3.5 ([22]) . Let F n be the free group on A = { a , . . . , a n } ; then the map { Subgroups of F n } → { Reduced A -automata } H St ( H ) h Γ i ← (cid:91) Γ (3.1) is a bijection. Furthermore, finitely generated subgroups correspond precisely to finite Stallingsautomata, and in this case the bijection is algorithmic. As stated, if we restrict our attention to finitely generated subgroups, then the abovebijection is algorithmic. Given a finite Stallings A -automata Γ , one can always computea maximal tree T of Γ , and then use Proposition 3.2 to compute a free basis for h Γ i . Inparticular, rk h Γ i = E + Γ − V Γ + 1 = rk Γ .For the other direction, suppose that we are given a finite set of generators S for H .We have seen that Fl ( S ) is a finite core A -automaton recognizing H , although it might be ol. 12:1 ON THE LATTICE OF SUBGROUPS OF A FREE GROUP: COMPLEMENTS AND RANK 1:9 not deterministic. To fix this, one can apply successive foldings on possible arcs breakingdeterminism. Of course, a folding can provide new opportunities for folding, but, since thenumber of arcs in the graph is finite and decreases with each folding, after a finite numberof steps, we will obtain an A -automaton with no available foldings (i.e., deterministic, andhence a reduced A -automaton). Theorem 3.5 states that the resulting automaton must beprecisely St ( H, A ). Note that the bijectivity of (3.1) implies that the result of the foldingprocess depends neither on the order in which we perform the foldings nor on the starting(finite) set of generators for H , but only on the subgroup H itself.The bijection (3.1) has proven to be extremely fruitful and has provided natural proofs formany results on the free group, especially from the algorithmic point of view (see [11, 13, 1]).In particular, the Nielsen-Schreier theorem, the solvability of the subgroup membershipproblem, and the computability of rank and basis for finitely generated subgroups canbe easily derived from this geometric interpretation. Concretely, given a finite subset S of F n , one can decide whether a given element w ∈ F n belongs to h S i just by checkingwhether w reads a -walk in St ( S ). In the same vein, one can use Theorem 3.5 togetherwith Proposition 3.2 to compute a basis, and hence the rank, of the finitely generatedsubgroup h S i .Many other algebraic properties of subgroups become transparent from this geometricviewpoint. For example the Stallings automata of the conjugacy classes of a subgroup H of F n correspond exactly to the automata obtained after changing the basepoint in Sch ( H )and taking the corresponding core automaton. Remark 3.6.
Note that St ( H ) is complete if and only if St ( H ) = Sch ( H ), and otherwise Sch ( H ) is necessarily infinite. Hence a subgroup H has finite index in F n if and only if St ( H ) is finite and complete.From this, one can easily decide finite index in F n , and obtain other classical resultsinvolving subgroups of finite index, such as the Schreier index formula for finite indexsubgroups (rk H = | F n : H | ( n −
1) + 1).With some extra work, one can also study intersections and extensions of subgroupsof F n (we say that G is an extension of H if H is a subgroup of G ). For example, a neatproof for an algorithmic version of the classical theorem of Takahasi on free extensions iseasily obtained using Stallings automata. Theorem 3.7 ([23]) . Every extension of a given finitely generated subgroup H of F n is afree multiple of an element of a computable finite family of extensions of H . The minimal such computable set of extensions is called the set of algebraic extensions of H , and is denoted by AE( H ). Note that H ∈ AE( H ) (see [13] for details).Finally, we extend the previous scheme to arbitrary A -automata, that is, we define the Stallings reduction of an A -automaton Γ to be St ( Γ ) = St ( h Γ i ). We usually abbreviate St ( Γ )to Γ . Note that we then have that Γ = ∆ if and only if h Γ i = h ∆ i . Lemma 3.8. If Γ is an involutive A -automaton, then L ( Γ ) = L ( Γ ) .Proof. It suffices to consider one folding at the time. More precisely, let Γ be obtainedfrom Γ by folding the arcs p a q and p a q . Clearly, L ( Γ ) ⊆ L ( Γ ). On the other hand,for every u ∈ L ( Γ ), we can find some u ∈ L ( Γ ) by inserting factors of the form a − a into u . It follows that u = u and so L ( Γ ) = L ( Γ ) = L ( Γ ) = L ( Γ ). By iterating this argumentfor a sequence of foldings, we get the desired claim. :10 J. Delgado and P. V. Silva
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In the remainder of the section we summarize several consequences (on the relationbetween the graphical and the algebraic rank) of applying the Stallings bijection (3.1) toprevious results.The first one is an immediate consequence of the fact that foldings never increase therank of the affected automata.
Remark 3.9.
Let Γ be an automaton recognizing H (cid:54) F n ; then, rk H = rk Γ ≤ rk Γ .In particular (of course), if H, K (cid:54) F n , then rk( H ∨ K ) ≤ rk H + rk K , but this boundis not necessarily tight. Below, we use Stallings theory to precisely describe the maximumand minimum ranks attainable by extending a given finitely generated subgroup of F n . Proposition 3.10.
Let H be a finitely generated subgroup of F n . Then, (i) the minimum rank of an extension of H is the minimum of the ranks of the algebraicextensions of H . (ii) the maximum rank of an extension of H is either infinite (if H is of infinite indexin F n ) or equal to the rank of H (if H is of finite index in F n ).In other words, for every K (cid:54) F n , min H i ∈ AE( H ) rk( H i ) ≤ rk( H ∨ K ) ≤ (cid:26) ∞ if | F n : H | = ∞ rk( H ) if | F n : H | < ∞ , and all the bounds are tight and computable.Proof. (i) Let L be an extension of H of minimum rank. Since the algebraic extensions of H are certainly extensions of H , it is clear that rk( L ) ≤ min { rk( H i ) : H i ∈ AE( H ) } . Onthe other hand, since every extension of H is a free factor of some element in AE( H ) (seeTheorem 3.7), from Grushko theorem we have that rk( L ) ≥ min { rk( H i ) : H i ∈ AE( H ) } .Hence, rk( L ) = min { rk( H i ) : H i ∈ AE( H ) } , as claimed.(ii) If the index of H in F n is infinite, then there exists a vertex p in St ( H ) and agenerator a ∈ A such that there is no a -arc leaving p . So, it is enough to attach to p an a -source of infinite rank (for example the one in Figure 3) to obtain an extension (indeed afree multiple) of H of infinite rank. · · · a b Figure 3: An a -source of infinite rankOn the other hand, if | F n : H | < ∞ then St ( H ) is complete, and therefore any additionto St ( H ) would produce an identification of vertices. Therefore, after folding, we get adecrease in the index (and hence in the rank) of H .If R ⊆ F n , then we usually abuse language and write Γ + R = Γ + Fl ( R ). Note that wethen have that h Γ + R i = h Γ i ∨ h R i . Lemma 3.11.
Let ∆ and Γ be automata. If ∆ (cid:54) Γ , then rk ∆ (cid:54) rk Γ . Moreover, if ∆ , Γ are finite and connected then there exists a subset R of F n of cardinal rk Γ − rk ∆ such that Γ = ∆ + R . ol. 12:1 ON THE LATTICE OF SUBGROUPS OF A FREE GROUP: COMPLEMENTS AND RANK 1:11 Proof.
The first claim is obvious if Γ has infinite rank, and otherwise it follows from astraightforward analysis on (2.1) corresponding to the possible situations after removingan edge or an isolated vertex from Γ . For the second claim, it is enough to consider anyspanning tree T of the subgraph ∆ and realize that it can be expanded (e.g. by usingbreadth-first search) to a spanning tree T of Γ . We can then take R to be S T (cid:114) S T , whichclearly has the required cardinal. Since h Γ i = h S T i = h S T ∪ R i = h ∆ + R i , the claimedresult follows from Theorem 3.5.Note that, in general, a strict subgraph ∆ < Γ can still have the same rank as Γ (e.g. if Γ has isolated vertices, or “hanging trees”). Below we prove that this is no longer true if Γ isfinite and reduced. Corollary 3.12.
The rank of a finite core automaton is strictly greater than the rank ofany of its connected strict subautomata.Proof.
Let Γ be a finite core automaton. Note that any connected subautomaton of Γ isobtained by successively removing arcs, and discarding the eventual connected componentsnot containing the basepoint that may appear. Since Γ is core, the first removed arc cannotproduce an isolated vertex; hence the rank decreases in view of Lemma 3.11. Lemma 3.13.
Let Γ be a finite Stallings automaton, and let R ⊆ F n be a finite subsetof size r . If V ( Γ + R ) < V Γ , then Γ + R = Γ / p = q + R , where p , q are two differentvertices in Γ , and R ⊆ F n is a finite subset of size at most r − .Proof. First note that the hypothesis V ( Γ + R ) < V Γ entails the identification of two(different) vertices in Γ during the reduction of Γ + R .Let us call Γ -free any folding that does not produce an identification of vertices in Γ .We then let Γ be an automaton (note that it may not be uniquely determined) obtainedafter successively performing Γ -free foldings on Γ + R until no more Γ -free foldings arepossible (in particular, Γ (cid:54) Γ and Γ = Γ + R ). Note that Γ + R is core and so is Γ .But Γ cannot be reduced (otherwise Γ + R = Γ (cid:62) Γ , contradicting the hypothesis that V ( Γ + R ) < V Γ ) and therefore strictly contains Γ as a subautomaton. Hence, fromCorollary 3.12 and Lemma 3.11, we have that rk Γ + 1 ≤ rk Γ ≤ rk Γ + r .Since Γ is not reduced and there are no Γ -free foldings remaining, there must be anavailable folding in Γ identifying two different vertices (say p , q ) in Γ . Note that (since Γ isreduced by hypothesis) at least one of the arcs involved in the folding must lie outside Γ .Thus the following conditions hold:(i) Γ / p = q is a subautomaton of the automaton Γ obtained after performing the folding in Γ .(ii) rk Γ = rk Γ . (This follows immediately from (2.1), because we lose exactly onevertex and one positive arc in Γ after the folding, keeping the number of connectedcomponents equal to 1.)Therefore: rk Γ − rk Γ / p = q = rk Γ − rk Γ / p = q = rk Γ − rk Γ − ≤ rk Γ + r − rk Γ − r − , (3.2)where we have used that rk Γ ≤ rk Γ + r . :12 J. Delgado and P. V. Silva
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Finally, since Γ / p = q (cid:54) Γ , it follows from (3.2) and Lemma 3.11 that there exists a subset R ⊆ F n of size at most r − Γ + R = Γ = Γ / p = q + R , which is what we wantedto prove. 4. Complements and coranks of subgroups
Given a bounded lattice ( L, ∨ , ∧ ) with maximum 1 and minimum 0, we say that two elements a, b ∈ L are ∨ -complementary (resp., ∧ -complementary) if a ∨ b = 1 (resp., a ∧ b = 0); then wealso say that each of the elements is a ∨ -complement (resp., ∧ -complement ) of the other. Twoelements a, b ∈ L are said to be directly complementary if they are both ∨ -complementaryand ∧ -complementary; that is if a ∨ b = 1 and a ∧ b = 0; then we also say that each is a directcomplement of the other.Let Sgp ( G ) denote the set of subgroups of a group G , and let H, K ∈ Sgp ( G ). Wedefine the join of H and K to be H ∨ K = h H ∪ K i . It is easy to see that ( Sgp ( G ) , ∨ , ∩ ) is abounded lattice with maximum G and minimum the trivial subgroup { } . This immediatelyprovides the corresponding notions of complement in this setting. If H ∩ K = { } , then wesay that H and K are in direct sum , and we write H ∨ K = H ⊕ K . In particular, H, K (cid:54) G are directly complementary if and only if H ⊕ K = G . Remark 4.1.
Note that, if we denote the free product of two subgroups by H ∗ K , then H ∗ K = G ⇒ H ⊕ K = G ⇒ H ∨ K = G , but both converses are false. So, we havethree natural increasingly restrictive ways of “adding” subgroups; namely joins ( ∨ ), directsums ( ⊕ ), and free products ( ∗ ). The corresponding notions of complement are summarizedbelow. Definition 4.2.
Let
H, K (cid:54) G . If H ∨ K = G (resp., H ⊕ K = G , H ∗ K = G ) then we saythat H and K are ∨ - complementary (resp., ⊕ - complementary , ∗ - complementary ) in G , andthat each of them is a ∨ -complement ( ⊕ -complement , ∗ -complement ) of the other. Remark 4.3.
We note that all three notions of complement are well behaved with respectto conjugation. More precisely, if
H, K (cid:54) G , then, for every g ∈ G :( H ∨ K ) g = H g ∨ K g , ( H ⊕ K ) g = H g ⊕ K g , and ( H ∗ K ) g = H g ∗ K g , understanding that, in every equation, each of the sides is well defined if and only if theother side is well defined.We are interested in the behaviour of the complements of finitely generated subgroupswithin the free group. Concretely, we investigate which subgroups of F n admit each kindof complement, and the properties of the respective sets of complements. We note that,from the last remark, these notions work modulo conjugation. The case of free products hasbeen extensively studied and is well understood. Namely, from Grushko Theorem, a freecomplement of a subgroup H of G necessarily has (minimum and maximum) rank equalto rk( G ) − rk( H ). Moreover, given a finite subset S of F n , one can decide whether h S i admits a free complement either using the classical Whitehead’s peak reduction argument(see [25]) or, more efficiently, using techniques based on the Stallings description of subgroups(see [17, 21, 15]). We aim to extend some of these results to our weakened notions ofcomplement. ol. 12:1 ON THE LATTICE OF SUBGROUPS OF A FREE GROUP: COMPLEMENTS AND RANK 1:13 Definition 4.4.
Let H be a subgroup of G ; then the ∨ -corank (resp., ⊕ -corank ) of H isthe minimum rank of a ∨ -complement (resp., ⊕ -complement) of H . These are denoted bycrk ∨ ( H ), and crk ⊕ ( H ) respectively. Remark 4.5.
Note that crk ∨ ( H ) ≤ crk ⊕ ( H ), whenever they are well defined. Moreover,if H is a free factor of G , then crk ∨ ( H ) = crk ⊕ ( H ) = rk( G ) − rk( H ), which immediatelyfollows from Grushko theorem. In particular crk ∨ (1) = crk ⊕ (1) = rk( G ).A natural approach to this kind of questions is through the simplest possible complements,namely cyclic complements. We use the term cocycle to refer to cyclic complements (ortheir generators). Below is the precise definition in terms of each kind of complement. Definition 4.6. A ∨ -cocycle (resp., ∩ -cocycle , ⊕ -cocycle ) of a subgroup H (cid:54) F n is a cyclic ∨ -complement (resp., ∩ -complement, ⊕ -complement) of H , or any of its generators. Wedenote the corresponding sets of cocyles by Coc ∨ ( H ), Coc ∩ ( H ), and Coc ⊕ ( H ) respectively.A subgroup is said to be ∨ -cocyclic (resp., ⊕ -cocyclic ) if it admits a cyclic complement ofthe corresponding kind.Note that there is no need for the concept of ∩ -cocyclic subgroup since admitting acyclic ∩ -complement is equivalent to admitting a (general) ∩ -complement.5. Join complements
Recall that a subgroup K of G is a ∨ -complement of H (cid:54) G (in G ) if H ∨ K = G , and wedefine the ∨ -corank of H to be the minimum possible rank for such a subgroup K . We notethat this concept has previously appeared in the literature under other names. Concretely, itappears in [15] under the name of distance between subgroups , where the author also providesbounds for it and proves its computability. We will present our own proofs here, obtainedindependently, for the sake of completeness.Obviously, every subgroup H of F n admits a ∨ -complement of rank at most n (namely F n ).It is easy to see that the same holds for proper ∨ -complements if the involved subgroup isnot trivial. Lemma 5.1.
Every nontrivial subgroup of F n admits a proper ∨ -complement of rank n .Proof. It is enough to prove the claim for nontrivial cyclic subgroups h u i (cid:54) F n , where u iscyclically reduced.If n = 1, then, given a nontrivial subgroup h a k i (cid:54) Z = h a | −i , it is enough to considerany proper subgroup h a l i (cid:54) Z , where k, l are coprime integers.If n ≥
2, we distinguish two cases:(a) if the first and last letters of u are equal (say to a ∈ A ± ), then consider a subgroup ofthe form K = h{ ua i u − : a i ∈ A (cid:114) { a }} ∪ { bab − }i , where b ∈ A (cid:114) { a } .(b) if the first and last letters in u are different (say equal to b, a ∈ A ± respectively), thenconsider the subgroup K = (cid:10) { ua i u − : a i ∈ A (cid:114) { a }} ∪ { a } (cid:11) .Now, from the Stallings representation of K it is clear that in both cases: K = F n , H ∨ K = F n ,and rk( K ) = n . We have that K is a proper complement of H of rank n , and the proof isconcluded. :14 J. Delgado and P. V. Silva
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Our next goal is to compute the ∨ -corank (i.e., the minimum possible rank of a ∨ -complement) of a given finitely generated subgroup H of F n . Note that, in the context offree groups, we can restate this notion in graphical terms in the following way:crk ∨ ( H ) = min { | S | : S ⊆ F n and St ( H ) + S = B n } . First, we describe the range of possible values for the ∨ -corank of a subgroup of F n . Lemma 5.2. If H (cid:54) F n , then max { , n − rk( H ) } ≤ crk ∨ ( H ) ≤ n . (5.1) Moreover, for n > , this is the only general restriction between the rank and the ∨ -corank ofa subset of F n . Namely, for every pair ( r, c ) ∈ [1 , ∞ ] × [1 , n ] satisfying n − r ≤ c ≤ n thereexists a subgroup of F n with rank r and ∨ -corank c .Proof. It is clear that 0 ≤ crk ∨ ( H ) ≤ n since any basis of F n has n elements and is enoughto generate F n . On the other hand, since one cannot generate F n with fewer than n elements,it is also clear that rk( H ) + crk ∨ ( H ) ≥ n . The condition in (5.1) follows.Let F n = h a , a , . . . , a n | −i . For each c ∈ [1 , n ] and each positive r ≥ n − c , considerthe subgroup H ( r, c ) generated by { a c +1 , . . . , a n }∪ { ( a a ) i a ( a − a − ) i : 2 i + 1 ∈ [1 , r − n + c ] }∪ { ( a a ) j a a a − ( a − a − ) j : 2 j + 2 ∈ [2 , r − n + c ] } . Its Stallings automaton has loops at the basepoint labelled by a c +1 , . . . , a n and a chainobtained by concatenating r − n + c alternated cycles labelled by a and a . For instance, if n = 3 then St ( H (4 , a a a a a a a It is clear that H ( r, c ) has rank r . To compute its ∨ -corank, we define the subgroup K ( r, c ) = h a , . . . , a c , a a , a ( a a ) r i . The given generating set is clearly a basis (every reduced word on the generators is actuallyreduced as a word on A ± ); hence K ( r, c ) has rank c .Now St ( H ( r, c )) + St ( K ( r, c )) has loops at the basepoint labelled by a , . . . , a n . Afterfolding, we also get loops labelled by a and a . Hence St ( H ( r, c )) ∨ St ( K ( r, c )) = F n andso crk ∨ ( H ( r, c )) ≤ c . On the other hand, if we project F n onto ( Z / Z ) n , then most of thegenerators of H ( n, c ) collapse and the image has rank n − c . Since ( Z / Z ) n has rank n , thenit is impossible to find a ∨ -complement of H ( r, c ) with rank strictly less than c . Thereforecrk ∨ ( H ( r, c )) = c and we are done. Lemma 5.3.
Let Γ be a nontrivial finite reduced automaton; then, for every r ≥ , the joincorank crk ∨ h Γ i ≤ r if and only if crk ∨ h Γ / p = q i ≤ r − , for some pair of distinct vertices p , q in Γ .Proof. [ ⇒ ] If crk ∨ ( h Γ i ) ≤ r then there exists a reduced automaton ∆ of rank at most r such that Γ + ∆ = B n . Then, from Lemma 3.13, there exists a reduced automaton ∆ ofrank at most r − Γ / p = q + ∆ = B n . Hence crk ∨ h Γ / p = q i ≤ r − ol. 12:1 ON THE LATTICE OF SUBGROUPS OF A FREE GROUP: COMPLEMENTS AND RANK 1:15 [ ⇐ ] Since Γ / p = q = Γ + w , for some element w ∈ F n , and crk ∨ h Γ / p = q i ≤ r −
1, there existsa reduced automaton ∆ of rank at most r − Γ + ( w + ∆ ) = ( Γ + w ) + ∆ = Γ / p = q + ∆ = B n . So, h w + ∆ i is a ∨ -complement of h Γ i of rank at most r . Hence crk ∨ ( h Γ i ) ≤ r . Theorem 5.4 ([15]) . There exists an algorithm that, given a finite subset S of F n , outputsthe ∨ -corank of h S i in F n .Proof. Let us write Γ = St ( S ). It is clear that it is enough to be able to decide, for every k ∈ [0 , n − ∨ ( Γ ) ≤ k . We proceed by induction on k . The case k = 0 beingtrivial, we assume that k > ∨ ( Γ ) ≤ k − Γ is a bouquet (i.e., it has a single vertex); then the ∨ -corank problemis trivial: crk ∨ Γ = n − rk( Γ ). Hence we may assume that Γ has at least two vertices.In view of Lemma 5.3, we have crk ∨ ( Γ ) ≤ k if and only if crk ∨ h Γ / p = q i ≤ k −
1, for somepair of different vertices p , q in Γ . By the induction hypothesis, we can check if this conditionholds for every pair of vertices. Corollary 5.5.
Let H and K be two finitely generated subgroups of F n (given by respectivefinite generating sets); then one can algorithmically decide whether H (cid:54) K , and, if so,compute the ∨ -corank of H in K .Proof. The first claim immediately follows from the solvability of the membership problemfor free groups. Namely, H (cid:54) K if and only if every one of the given generators for H belongsto K , which can be checked by trying to read them as -closed walks in St ( K ). In theaffirmative case, the previous procedure provides the expression of the given H -generators aswords in some free basis B of K . Hence, the ∨ -corank of H in K is exactly the ∨ -corank of the(subgroup generated by the) new words in F B , which is computable using Theorem 5.4.We show next that, for a fixed ambient F n (and its canonical basis), we can computethe corank of a finitely generated subgroup H of F n in polynomial time with respect to thesize of H (the number of vertices of St ( H )). Corollary 5.6.
Let n ≥ be fixed. There exists an algorithm which computes the ∨ -corankof a given H ≤ fg F n of size m in time O ( m n +1) ) .Proof. We assume that H is given through its Stallings automaton which, in turn, can beassumed to have more than one vertex. Let V denote the vertex set of Γ = St ( H ), andlet m = V . Given I ⊆ V × V , let Γ /I denote the involutive automaton obtained from Γ by identifying all pairs of vertices belonging to I . Let n be the number of letters fromthe canonical basis of F n labelling edges in St ( H ). In view of the proof of Theorem 5.4,and folding being confluent, in order to compute the corank of H it would suffice to do thefollowing: • to fold St ( H ) /I for every I ⊆ V × V with | I | ≤ n , checking whether we get a bouquet; • to register the smallest r = | I | yielding a bouquet.The corank of H is then n − n + r .Thus it is enough to consider O ( m n ) subsets of V × V . Indeed, | V × V | = m and (cid:0) m j (cid:1) (cid:54) m j for every j (cid:54) n , yielding P nj =0 (cid:0) m j (cid:1) (cid:54) P nj =0 m j (cid:54) m n . :16 J. Delgado and P. V. Silva
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By [24, Theorem 1.6], we know that we can fold an involutive automaton with v verticesand e arcs in time O ( e + ( v + e ) log ∗ ( v )), where log ∗ ( v ) is the smallest positive integer k such that the k -fold iteration of the base 2 logarithm satisfies log k ( m ) (cid:54) St ( H ) /I we have | V | (cid:54) m , | E | (cid:54) mn and, although m is a verycoarse upper bound for log ∗ ( m ), it suffices to show that folding each St ( H ) /I can beperformed in time O ( m ). Therefore we can compute the ∨ -corank of H in time O ( m n m ) = O ( m n +1) ). Lemma 5.7.
The set of ∨ -cocycles of a finitely generated subgroup H of F n is rational andwe can effectively compute a finite automaton recognizing it.Proof. Let F n = h A | −i . Assume first that St ( H ) has a single vertex, that is, that St ( H ) isthe bouquet St ( H ) for some B ⊆ A .We now assume that B < A −
1. It is immediate that we have no ∨ -cocycles. If B = A , then Coc ∨ ( H ) = F A and we are also done. Thus we may assume that B = A \ { a } for some a ∈ A . In this case, it is easy to see thatCoc ∨ ( H ) = F B { a, a − } F B , where R B denotes the set of reduced words on the alphabet B ± . Hence Coc ∨ ( H ) is rationalwhenever St ( H ) is a bouquet.Thus we may assume that Γ = St ( H ) has at least two vertices. Let P denote the set ofall pairs of vertices ( p , q ) in Γ such that Γ / p = q is the bouquet B n on n letters. We claim thatCoc ∨ ( H ) = [ ( p , q ) ∈ P ( L p ( Γ ) L q ( Γ )) ∩ R A . (5.2)Indeed, let u ∈ R A . To compute the Stallings automaton of H ∨ u , we glue a cycle labeledby u to the basepoint of Γ and fold. If some edge of this new cycle is not absorbed by Γ inthe folding process, then Γ will be a subautomaton of St ( H ∨ u ). Since Γ has at least twovertices, then we won’t get the desired bouquet. Hence we may assume that u admits areduced factorization u = u u such that: • there exist paths u p and q u in St ( H ); • Γ / p = q = B n .Therefore (5.2) holds and we are done.6. Meet complements
Again, it is obvious that every subgroup admits a ∩ -complement (namely, the trivialsubgroup). In addition, if a subgroup admits a nontrivial ∩ -complement then it admits anontrivial ∩ -cocycle as well. Lemma 6.1. If St ( H ) is incomplete then H admits a ∩ -complement of any rank.Proof. Let p be a vertex in St ( H ) with no outgoing arc labelled by (say) a ∈ A , andlet w ∈ F n be a reduced word reading a walk from the basepoint to p in St ( H ). Thenthe subgroup recognized by any automaton of the form wa ∆, where ∆ is a labelleddigraph ( of any rank ) sharing no vertex with the walk labelled by w , has trivial intersectionwith H . ol. 12:1 ON THE LATTICE OF SUBGROUPS OF A FREE GROUP: COMPLEMENTS AND RANK 1:17 The characterization of the finitely generated proper ∩ -complements in Sgp ( F n ) followsfrom the previous result and a well-known general property of finite index subgroups. Proposition 6.2.
A finitely generated subgroup of F n admits a nontrivial ∩ -complement(of any rank) if and only if it has infinite index in F n . Hence it is decidable whether a givenfinitely generated subgroup of F n admits a nontrivial ∩ -complement.Proof. [ ⇒ ] If H is a subgroup of finite index of F n then the set { Hu k : k ∈ N } must be finitefor every u ∈ F n . Therefore, for every u ∈ F n , there exists k ≥ u k ∈ H . Since F n is torsion-free, it follows that H admits no nontrivial ∩ -complement.[ ⇐ ] Since a subgroup H of F n has finite index if and only if its Stallings automaton isfinite and complete (see Remark 3.6), the claim follows from Lemma 6.1. Lemma 6.3.
Let H (cid:54) F n be finitely generated, let m be the number of vertices of Γ = St ( H ) ,and let u ∈ C n \ { } ; then, the following statements are equivalent: (a) h u i ∩ H = { } (i.e., h u i is a (nontrivial) ∩ -complement of H ); (b) u k / ∈ H for every k ≥ ; (c) u m ! / ∈ H ; (d) u k / ∈ H for every k = 1 , . . . , m ; (e) u m / ∈ L V Γ ( Γ ) (i.e., u m cannot be read from the basepoint within Γ ).Proof. Note that (a) ⇔ (b) by definition, whereas the implications (e) ⇒ (b) ⇒ (c) ⇒ (d)are clear (and only the first one needs the assumption of u being cyclically reduced).Finally, to see that (d) ⇒ (e), suppose (by contraposition) that u m is readable in St ( H )from the basepoint, that is, there exists a walk= p u p u · · · u p m (6.1)in St ( H ). Since there are only m vertices in St ( H ), there must be a repetition among thevertices p , p , . . . , p m in (6.1), and, since St ( H ) is reduced, the first repeated vertex — say p k , for some k ∈ [1 , m ] — must be the basepoint (otherwise there would be two differentwalks reading u arriving at the first repeated vertex). Hence there is a -walk in St ( H )reading u k , that is, there exists some k ∈ [1 , m ], such that u k ∈ H , contrary to the conditionin (d). Therefore, [(d) ⇒ (e)], and all the five statements are equivalent, as claimed.The previous characterization provides a description of the set of cyclically reduced ∨ -cocycles as a regular subset of F n . Lemma 6.4.
The set of cyclically reduced ∩ -cocycles of a finitely generated subgroup H of F n is a rational subset of F n , and we can effectively compute a finite automaton recognizingit.Proof. Let Γ = St ( H ) and m = V Γ . Let X = { } × ( V Γ ) m be the set of ( m + 1)-tuples ofvertices in Γ starting at the basepoint; then, for every p = ( p , p , . . . , p m ) ∈ X , K p = T mi =1 L p i − p i ( Γ )is the set of words which are readable between any of the successive vertices in p . Note that S p ∈ X K p is exactly the set of nontrivial words in A ± whose m -th power is readable fromthe basepoint in Γ . Hence, according to Lemma 6.3, the set of cyclically reduced ∩ -cocyclesof H is precisely: Coc ∩ ( H ) ∩ C A = ( C A (cid:114) S p ∈ X K p ) ∪ { } . (6.2) :18 J. Delgado and P. V. Silva
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Since the sets C A and K p are rational (for every p ∈ X ), and X is finite, the claimed resultfollows immediately from (6.2) and the closure properties of rational languages.7. Direct complements
Finally, we address the study of ⊕ -complements of finitely generated subgroups of F n . First,we address the question of their existence (i.e., the existence of ∩ -complements that arealso ∨ -complements). Below we prove that the second requirement does not suppose anyreal restriction (i.e., a finitely generated subgroup admits a ⊕ -complement if and only if itadmits a ∩ -complement, see Proposition 6.2). Theorem 7.1.
Let H be a finitely generated subgroup of F n where H = { } 6 = F n ; then H (cid:54) F n admits a nontrivial ⊕ -complement if and only if it has infinite index in F n . Hence it isdecidable whether a given finitely generated subgroup of F n admits a nontrivial ⊕ -complement.Proof. [ ⇒ ] Since ⊕ -complements are, in particular, ∩ -complements, this is immediate fromProposition 6.2.[ ⇐ ] From the hypotheses, we may assume that St ( H ) is nontrivial, finite, and incomplete(say a -deficient, for some generator a of F n ). Note also that since, for any w ∈ F n , K is a ⊕ -complement of H if and only if wKw − is a ⊕ -complement of wHw − , we may replace H by any conjugate at our convenience.In particular, H has a conjugate which arises from adjoining the arc p a to an a -deficient vertex p of an inverse graph ∆ having no vertices of degree 1. We replace H bythis conjugate.Now, (since H is nontrivial) there must exist a walk p p in ∆ reading some nonemptyreduced word u . Also, since St ( H ) is incomplete, we have n >
1, and hence we may fixsome letter b ∈ A \ { a } . We claim that K = h ( A \ { a } ) ∪ { a − uababa − ua }i is a ⊕ -complement of H .Note that a − ua is reduced because it reads a reduced walk in (the reduced automaton) St ( H ), and, since b = a , then a − uababa − ua is also reduced. It follows that St ( K ) = Fl (( A \ { a } ) ∪ { a − uababa − ua } ) . Suppose that w ∈ F n represents a nontrivial element in H ∩ K . Since w labels a -walkin St ( K ), it has as prefix one of the words in ( A \ { a } ) ∪ { a − uababa − ua } or its inverse.However none of these words can be read off the basepoint of St ( H ). This is obvious for( A \ { a } ) ± . Suppose now that we have a walk a − ua q baba − ua · · · in St ( H ). Since St ( H ) is reduced, we necessarily have that q = . But then we would beunable to read b from the basepoint, since the only arc leaving it has label a − . Similarly, weshow that w cannot have a − u − ab − a − b − a − u − a as a prefix. Therefore H ∩ K = { } .On the other hand, we can obtain St ( H ∨ K ) by identifying the basepoints of St ( H )and St ( K ), followed by a complete folding. Note that every c ∈ A \ { a } labels a loop at thebasepoint of St ( H ∨ K ) because this already happens in St ( K ). Hence it suffices to showthat a also labels a loop at the basepoint of St ( H ∨ K ). ol. 12:1 ON THE LATTICE OF SUBGROUPS OF A FREE GROUP: COMPLEMENTS AND RANK 1:19 Since a − uababa − ua lies in K and is reduced, it is accepted by St ( K ), and thereforeby St ( H ∨ K ). Thus we have a walk a − ua q b q a q b q a − ua in St ( H ∨ K ). Since a − ua is accepted by St ( H ) and St ( H ∨ K ) is inverse, we obtain that q = q = . Similarly, it follows from b being accepted by St ( K ) that q = q = . Thus a labels a loop at the basepoint of St ( H ∨ K ).Hence H ∨ K = F n , and so K is a ⊕ -complement of H as claimed.The following result is an immediate consequence of the above proof. Corollary 7.2.
Every ⊕ -complementable finitely generated subgroup of F n admits a ⊕ -com-plement of any rank greater than or equal to n . Suggestively enough, a kind of dual of the previous situation applies within the subgroupsadmitting ⊕ -complements; namely, the requirement of trivial intersection does not affectthe rank bounds of the possible complements (i.e., they coincide with those obtained for ∨ -complements in Lemma 5.2). Lemma 7.3.
Let H be a ⊕ -complementable subgroup of F n , then max { , n − rk( H ) } ≤ crk ⊕ ( H ) ≤ n . (7.1) Moreover, for n > this is the only general restriction between the rank and the ⊕ -corank ofa subset of F n . Namely, for every pair ( r, c ) ∈ [1 , ∞ ] × [1 , n ] satisfying n − r ≤ c ≤ n , thereexists a subgroup of F n with rank r and ⊕ -corank c .Proof. Since crk ∨ H ≤ crk ⊕ H , the bounds in (7.1) are immediate from those in (5.1) andCorollary 7.2. For the second claim it is enough to consider again the families of subgroups H ( r, c ) and K ( r, c ) introduced in the proof of Lemma 5.2 and check that H ( r, c ) ∩ K ( r, c ) = { } .Indeed, suppose that u ∈ H ( r, c ) ∩ K ( r, c ). If we write u as a reduced word on thegenerators of K ( r, c ), say u = u . . . u m , then the generators a , . . . , a c must be absentbecause they don’t label edges in St ( H ( r, c )). Suppose that u i ∈ { a ( a a ) r , ( a − a − ) r a − } for some i . Since there is no cancellation between consecutive u j , it follows that either( a a ) r or ( a − a − ) r is a factor of u . But neither of these words labels a path in St ( H ( r, c ));hence u = ( a a ) m for some m ∈ Z . The only such word accepted by St ( H ( r, c )) is u = 1;thus H ( r, c ) ∩ K ( r, c ) = { } as required.So, whenever the direct corank exists, the restriction of having trivial intersection doesnot affect the range of values it can take. In Question 7.9 we ask whether the same thinghappens for the coranks themselves.7.1. Direct cocycles.
Throughout this section, H will denote a finitely generated subgroupof F n . Recall that C n denotes the set of cyclically reduced words in F n , and the set ofdirect cocycles of H is Coc ⊕ ( H ) = { u ∈ F n : H ⊕ h u i = F n } . Since Coc ⊕ ( H ) = Coc ∨ ( H ) ∩ Coc ∩ ( H ), the next result follows immediately from Lemmas 5.7 and 6.4. Corollary 7.4.
The set of cyclically reduced ⊕ -cocycles of a finitely generated subgroup H of F n (namely, Coc ⊕ ( H ) ∩ C n ) is a rational subset of F n and we can effectively compute afinite automaton recognizing it. :20 J. Delgado and P. V. Silva
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However, the following example shows that rationality is no longer ensured for the fullset of ⊕ -cocycles. Example 7.5.
Let H = h a, b i (cid:54) F { a,b } = F ; then Coc ⊕ ( H ) is not a rational subset of F . Proof.
By Benois Theorem [2], a subset S of F { a,b } is rational if and only if it constitutes arational language of { a, b } ∗ . We show that a ∗ b ( a − ) ∗ \ Coc ⊕ ( H ) = { a n ba − n : n ≥ } . (7.2)Let n ≥
0. Clearly, ( a n ba − n ) = a n b a − n ∈ H ∩ h a n ba − n i . It follows that a n ba − n ∈ a ∗ b ( a − ) ∗ \ Coc ⊕ ( H ).Conversely, assume that m, n ≥ a m ba − n / ∈ Coc ⊕ ( H ). From theStallings automaton of H : ba it is clear that H ∨ h a m ba − n i = F (since both extremes of the words a m ba − n would becollapsed by the loop reading a , leaving a loop reading b also attached to basepoint). Hence( a m ba − n ) k ∈ H for some k ≥
1. But ( a m ba − n ) k = a m ( ba m − n ) k − ba − n . Since m = n , thereduced word a m ( ba m − n ) k − ba − n is not accepted by St ( H ), a contradiction. Therefore a ∗ b ( a − ) ∗ \ Coc ⊕ ( H ) ⊆ { a n ba − n : n ≥ } and (7.2) holds.Now { a n ba − n : n ≥ } is a classical example of a non-rational language (see e.g. [8,Exercise 4.1.1.c]). Since rational languages are closed under boolean operations, it followsfrom (7.2) that Coc ⊕ ( H ) is not a rational subset of F .We recall now the definition of a context-free language. This can be done through theconcept of a context-free grammar, a particular type of rewriting system. Let A be a finitealphabet. A context-free A -grammar is a triple G = ( V, P, S ) such that V is a finite setdisjoint from A , S ∈ V and P is a finite subset of V × ( V ∪ A ) ∗ . The language generatedby G is L ( G ) = { w ∈ A ∗ : S ∗ ⇒ w } , that is the set of all words on A obtained from S by successively replacing a letter on theleft-hand side of some element of P by its right-hand side. A language L ⊆ A ∗ is context-free if L = L ( G ) for some context-free A -grammar G . A subset X of F n is said to be context-freeif the set of reduced forms of X constitutes a context-free A ± -language. Proposition 7.6.
Let H be a finitely generated subgroup of F n ; then Coc ⊕ ( H ) is a context-free subset of F n , and we can effectively compute a context-free grammar generating it.Proof. Let Γ = St ( H ). It is clear that the claim holds whenever Γ is a bouquet; so, wemay assume that Γ has at least two vertices. Then, for every r ∈ V Γ , we denote by H r theconjugate u − Hu , where u ∈ L r ( Γ ). It is easy to see that St ( H r ) can be obtained from St ( H ) by making r the new basepoint and successively erasing vertices of degree 1 distinctfrom r . Let K r = Coc ⊕ ( H r ) ∩ C n . We show that:Coc ⊕ ( H ) = (cid:16)[ r ∈ V Γ n vwv − : v ∈ L r ( Γ ) , w ∈ K r o(cid:17) ∩ R n . (7.3)Let vwv − belong to the right hand side as prescribed. Then w ∈ Coc( v − Hv ) ∩ C n and so v − Hv ⊕ h w i = F n , yielding H ⊕ h vwv − i = F n . Thus vwv − ∈ Coc ⊕ ( H ).Conversely, let u ∈ Coc ⊕ ( H ). As we saw in the proof of Lemma 5.7, there exists some( p , q ) ∈ P and paths q u u p in St ( H ) such that u = u u . Write u = vwv − with w ∈ C n . Since vwv − = u u , v must be a prefix of u or u − (or both). So we get v ∈ L r ( Γ ) for some r ∈ V Γ . It remains to prove that w ∈ K r = Coc( v − Hv ) ∩ C n ; but H ⊕ h vwv − i = F n yields v − Hv ⊕ h w i = F n and so (7.3) holds.Since the class of context-free languages is closed under finite union and intersectionwith rational languages (see, for example [5, Section 10.5]), it suffices to show that thelanguage { vwv − : v ∈ L r ( Γ ) , w ∈ K r } is context-free for every r ∈ V Γ .Let G = ( V, P, S ) be the context-free ( A ± ∪ { $ } )-grammar defined by V = { S } and P = { S } × ( { $ } ∪ { aSa − : a ∈ A ± } ); then L ( G ) = { v $ v − : v ∈ ( A ± ) ∗ } is context-free.Since L r ( Γ )$ L r ( Γ ) − is rational, it follows that L ( G ) ∩ L r ( Γ )$ L r ( Γ ) − is context-free aswell.Define a homomorphism ϕ from ( A ± ∪ { $ } ) ∗ into the monoid of rational A ± -languages(under product) defined by aϕ = a ( a ∈ A ± ) and $ ϕ = K r . This is an example of a rationalsubstitution . Since n vwv − : v ∈ L r ( Γ ) , w ∈ K r o = ( L ( G ) ∩ L r ( Γ )$ L r ( Γ ) − ) ϕ and the class of context-free languages is closed under rational (in fact, context-free)substitution (see [20, Theorem 4.1.1]), we obtain the desired result.7.2. The cocyclic case and a question.
A subgroup is said to be ( ∨ or ⊕ )- cocyclic if itadmits a cyclic complement of the corresponding type. Obviously, a ⊕ -cocyclic subgroupis necessarily ∨ -cocyclic, but the converse fails for finite index proper subgroups in view ofTheorem 7.1. Theorem 7.7.
Let H be a finitely generated subgroup of infinite index of F n ; then H is ⊕ -cocyclic if and only if it is ∨ -cocyclic.Proof. We prove the nontrivial converse implication. Let Γ = St ( H ) and let V = V Γ .Assume that H ∨ h u i = F n . In view of Remark 4.3, we may assume that every vertex of Γ has degree strictly greater than 1. We may also assume that Γ = B n − , a trivial case.Suppose that u is not cyclically reduced. Write u = vwv − as a reduced product with w cyclically reduced. If v / ∈ L V ( Γ ) then Γ remains a subautomaton of Γ + u = B n afterfolding, a contradiction. Thus there exists a walk v r in Γ . By replacing H by itsconjugate H v it follows that H v ∨ h w i = F n . Therefore, in view of Remark 4.3, we mayassume that u is cyclically reduced.We may assume that H ∩ h u i 6 = { } . Hence u k ∈ H for some k ≥ Γ of the form= q u q u . . . u q k = Case 1:
Suppose that L q i − V ( Γ ) = L q i V ( Γ ) for some i . Replacing u by u − if necessary, wemay assume that L q i − V ( Γ )
6⊆ L q i V ( Γ ). Replacing the basepoint by q i − through conjugacyif needed, we may assume that L q V ( Γ )
6⊆ L q V ( Γ ). We claim that then there exists somereduced v ∈ L q V ( Γ ) (cid:114) L q V ( Γ ).Indeed, if v ∈ L q V ( Γ ) (cid:114) L q V ( Γ ) is not itself reduced, we may write v = xaa − y where xa is reduced and a ∈ A ± . Then, if xa / ∈ L q V ( Γ ), we can take v to be xa . Otherwise,we can replace v by xy ∈ L q V ( Γ ) \ L q V ( Γ ) (note that xa, xy ∈ L q V ( Γ ) would imply v = xaa − y ∈ L q V ( Γ ) since Γ is inverse). Iterating this procedure, we end up finding somereduced v ∈ L q V ( Γ ) \ L q V ( Γ ). Therefore we may assume that v is itself reduced. :22 J. Delgado and P. V. Silva
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Thus there exists a walk γ : v q in Γ reading the (reduced) word v . We claim thatthere exists some walk q w such that vw is reduced. Indeed, since every vertex of Γ hasoutdegree >
1, we can extend γ as much as needed keeping its label reduced. But since Γ isfinite, we must reach a point when one of the new vertices has already appeared before inthe walk. Let p be the first such repeated vertex; then we have paths x p y p where xy is reduced and has v as a prefix. Now xyx − ∈ L q V ( Γ ) \ L q V ( Γ ) and is reducedby the minimality of p . Writing xyx − = vw , we get the desired w .We would like to have vwu cyclically reduced, but that may fail. To overcome thisdifficulty, we will prove the claim below. Claim 7.8. If a, b ∈ A ± ∩ L V ( Γ ) are distinct, then L ( Γ ) ∩ C n ∩ a ( A ± ) ∗ b − = ∅ .Once again, we may extend the arc a q to a walk x p with reduced label, andassume that p is the first repeated vertex within the added arcs; then we use the samebacktracking technique as before to get some y ∈ L ( Γ ) ∩ R n ∩ a ( A ± ) ∗ . If y ends in b − ,we are done; otherwise, we may extend the arc b q to a walk x p with reducedlabel, and assume that p is the first repeated vertex within the added arcs. Again, theprevious backtracking technique provides some y ∈ L ( Γ ) ∩ R n ∩ b ( A ± ) ∗ b − . But then yy ∈ L ( Γ ) ∩ C n ∩ a ( A ± ) ∗ b − . Therefore Claim 7.8 holds.Since every vertex of Γ has degree strictly greater than 1, it follows from Claim 7.8 thatthere exist cyclically reduced words z , z ∈ L ( Γ ) such that z vwz u ∈ L q ( Γ ) ∩C n . However,we cannot ensure that z vwz u / ∈ L q V ( Γ ). Therefore we consider z m !1 vwz u ∈ L q ( Γ ) ∩ C n ,where m = V . Suppose that z m !1 vwz u ∈ L q V ( Γ ). In view of Lemma 6.3, we get that vwz u ∈ L q V ( Γ ), contradicting v / ∈ L q V ( Γ ). Thus we may replace v (respectively w ) by z m !1 v (respectively wz ) to assume that vwu is cyclically reduced.Clearly, H ∨ h vwu i = H ∨ h u i = F n . Since v / ∈ L q V ( Γ ) and vwu is cyclically reduced, wehave that ( vwu ) / ∈ L V ( Γ ). Hence H ∩ h vwu i = { } and so H is ⊕ -cocyclic in case 1. Case 2:
Suppose now that Case 1 does not hold, that is L q i V ( Γ ) = L q j V ( Γ ) for all i, j ∈ [0 , k ]. We claim that this is incompatible with our assumptions. Since B n = Γ + u = Γ / = q ,then in view of Lemma 3.8, R n ⊆ L ( Γ / = q ).Let Γ be the automaton obtained from Γ by adding the arcs q and q (wemay admit arcs labelled by the empty word in automata with all the obvious adaptations).It is straightforward to see that L ( Γ / = q ) = L ( Γ ). We show that L V ( Γ ) ⊆ L V ( Γ ) byinduction on the number t of arcs labelled by 1 appearing in a walk v in Γ . Then v ∈ L V ( Γ ) holds trivially for t = 0. Assume that t > t − p q be the last such arc occurring in the walk; then, we can split our walk into x p q y r where p , q ∈ { , q } and are different. Then y ∈ L qV ( Γ ) = L pV ( Γ ) and so there exists awalk p y r in Γ . Gluing this walk to x p , we obtain a walk labelled by v whichhas t − v in Γ . Thus, L V ( Γ ) ⊆ L V ( Γ ).From the previous discussion we have that: R n ⊆ L ( Γ / = q ) = L ( Γ ) ⊆ L V ( Γ ) ⊆ L V (Γ) ⊆ L V (Γ) ol. 12:1 ON THE LATTICE OF SUBGROUPS OF A FREE GROUP: COMPLEMENTS AND RANK 1:23 (where the last inclusion follows from Γ being inverse) contradicting the fact that Γ isincomplete (since H has infinite index). Therefore Case 2 is excluded and the theorem isproved.So, we have just seen that, whenever the ⊕ -corank is defined, crk ∨ H = 1 if and only ifcrk ⊕ H = 1. This fact, together with the coincidence in the possible ranges for both coranks(see Lemmas 5.2 and 7.3), makes it natural to ask whether the same holds for any possiblevalue of the ⊕ -corank. Question 7.9.
Do the ⊕ -corank and the ∨ -corank of a finitely generated subgroup H F n coincide whenever they are both defined (i.e., when | F n : H | = ∞ )? Acknowledgements
Both authors were partially supported by CMUP (UID/MAT/00144/2019), which is fundedby FCT (Portugal) with national (MCTES) and European structural funds through theprograms FEDER, under the partnership agreement PT2020.
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