Stallings Graphs, Algebraic Extensions and Primitive Elements in F2
aa r X i v : . [ m a t h . G R ] O c t Stallings Graphs, Algebraic Extensionsand Primitive Elements in F Ori Parzanchevski † and Doron Puder ‡ October 30, 2018
Abstract
This paper studies the free group of rank two from the point of view of Stallingscore graphs. The first half of the paper examines primitive elements in this group,giving new and self-contained proofs for various known results about them. Inparticular, this includes the classification of bases of this group. The second half ofthe paper is devoted to constructing a counterexample to a conjecture by Miasnikov,Ventura and Weil, which seeks to characterize algebraic extensions in free groups interms of Stallings graphs.
Let F be a finitely generated free group. A subgroup J of F is said to be an algebraicextension of another subgroup H , if H ≤ J and there does not exist an intermediatesubgroup H ≤ M (cid:12) J such that M is a proper free factor of J . We denote this by H ≤ alg J . This notion, which was formulated independently by several authors (and H ≤ alg J already appears in [Tak51]), is central to the understanding of the lattice of subgroupsof F . For example, it can be shown that every extension H ≤ J of free groups admitsa unique intermediate subgroup H ≤ alg M ≤ ff J (where ≤ ff denotes a free factor).Moreover, if H ≤ F is a finitely generated subgroup, it has only finitely many algebraicextensions in F . Thus, every group containing H is a free extension of one of the algebraicextensions of H , which is a well known theorem of Takahasi [Tak51]. For proofs of thementioned facts, as well as a general survey of algebraic extensions, we refer the readerto [MVW07].Given a basis X of F and H ≤ F , we denote by Γ X ( H ) the Stallings core graph of H Γ X ( H ) with respect to X . This is a pointed, directed, X -labeled graph, such that the wordsformed by closed paths around the basepoint are precisely the elements of H , and whichis minimal with respect to this property. One way to construct this graph is by taking theSchreier right coset graph of H in F w.r.t. X and then deleting all “hanging trees”, i.e., all † Supported by an Advanced ERC Grant. ‡ Supported by the Adams Fellowship Program of theIsrael Academy of Sciences and Humanities. H = (cid:10) ab − a, a − b (cid:11) for X = { a, b } and F = F ( X ) . Werefer to [Sta83, KM02, MVW07, Pud11] for further background on Stallings graphs.Figure 1: The core graph Γ X ( H ) where X = { a, b } and H = (cid:10) ab − a, a − b (cid:11) ≤ F ( X ) . ⊗ a / / •• a / / b O O • a _ _ ❅❅❅❅❅❅❅❅❅❅❅ b O O Given the basis X , and two subgroups H, J ≤ F , there is a graph morphism (whichpreserves the basepoint, directions and labeling) from Γ X ( H ) to Γ X ( J ) if and only if H ≤ J . Such a morphism is unique, when it exists. Given H, J ≤ F , we say that HX -covers J if H ≤ J and the morphism from Γ X ( H ) to Γ X ( J ) is onto. We denote this X -covers by H ≤ ։ X J . (In [MVW07] this is indicated by saying that J is a “ X -principal overgroup” ≤ ։ X of H , and by the notation J ∈ O X ( H ) .)It is not hard to see (e.g. [MVW07, prop. 3.7], or [PP12, claim 3.2]) that if H ≤ alg J ,then H ≤ ։ X J for every basis X of F . The following conjecture, raised in [MVW07], askswhether the converse also holds. Conjecture ([MVW07, §5(1)]) . If H ≤ J ≤ F and H ≤ ։ X J for every basis X of F then J is an algebraic extension of H . The main result of this paper is a counterexample to this conjecture:
Proposition (Prop. 4.1) . Let F = F ( a, b ) be the free group on two generators, H = (cid:10) a b (cid:11) , and J = (cid:10) a b , ab (cid:11) . Then H ≤ ։ X J for every basis X of F , but J is not analgebraic extension of H . The relation “ H ≤ ։ X J ” is basis-dependent, while the relation “ H ≤ ։ X J for every basis X ” is intrinsic, as is “ H ≤ alg J ”. Proposition 4.1 means that the latter two relations aredifferent, and this raises the intriguing question of understanding the algebraic significanceof “covering with respect to all bases”.The proof of Proposition 4.1 follows from a thorough analysis of Stallings graphs, usingclassical results (e.g. [Nie17, Coh72, CMZ81, OZ81]) on primitive elements and bases of F . It turns out that these results can also be proven by appealing solely to Stallingsgraphs, and we use the opportunity to provide self-contained proofs for them in Section3. Section 2 recalls some basic facts about Stallings graphs and foldings, and presents twoauxiliary lemmas which will be used later on. Finally, the proof of the counterexample(Proposition 4.1) is given in Section 4, and some concluding remarks in Section 5. We assume that the reader is familiar with the theory of Stallings foldings, but recallthe basic facts. If Γ is a pointed, directed, X -labeled graph, we denote by π X (Γ) the π X (Γ) F = F ( X ) consisting of the words which appear as closed loops around thebasepoint of Γ . The operators π X and Γ X constitute a bijection between subgroups of F ( X ) and X -labeled core graphs, which matches f.g. subgroups to finite graphs.If Γ is a finite (pointed, directed) X -labeled graph, and π X (Γ) = H , then Γ X ( H ) isobtained from Γ by repeatedly performing one of the following operations, in any order,until neither of them is possible:(1) Folding - merging two edges with the same label, and the same origin or terminus(and thus merging also the other ends).(2)
Trimming - deleting a leaf which is not the basepoint, and the edge which leads toit.The following lemma shows that under certain conditions only foldings are necessary inthis process:
Lemma 2.1.
Let Γ be a finite, pointed, directed, X -labeled graph such that at everyvertex, except possibly the basepoint, there are at least two types of edges (the type of anedge consists of its label and direction). Then the core graph Γ X ( H ) of H = π X (Γ) isobtained from Γ by foldings alone (i.e. without trimming).Proof. Evidently, Γ cannot have leaves, except for possibly the basepoint. Folding stepsdo not decrease the number of types of edges at a vertex, so that the property in thestatement still holds after every folding step, and no new leaves are created throughoutthe process.This simple lemma will prove out to be extremely useful. It already plays a role in Lemma2.3, which characterizes X -covering in simple extensions. Definition 2.2.
Let Γ be a pointed and directed X -labeled graph and let w ∈ F . We saythat w appears in Γ if there exist paths p , p in Γ such that p starts at the basepoint, p appears in terminates at the basepoint, and w = p p (i.e. p p is the presentation of w as a reducedword in X S X − ).For example, for H in Figure 1, a and a ba − appear in Γ X ( H ) , but a b does not.Notice that if w appears in Γ , s.t. π X (Γ) = H , and Γ satisfies the conditions of Lemma2.1, then w appears in Γ X ( H ) as well. This will play a significant part in Section 4. Lemma 2.3.
Let H ≤ F , w ∈ F and J = h H, w i . Then H ≤ ։ X J iff w appears in Γ X ( H ) .Proof. Assume first that w appears in Γ X ( H ) , and let p , p be as in Definition 2.2.Denote by Γ the graph obtained from Γ X ( H ) by identification of p ’s endpoint and p ’sstart-point. We have π X (Γ) = J , and the (pointed, directed, labeled) map from Γ X ( H ) to Γ is onto. Since Γ satisfies the conditions of Lemma 2.1, Γ X ( J ) is obtained from it byfoldings alone. We have now that Γ X ( H ) maps onto Γ , which maps onto Γ X ( J ) , and bytransitivity it follows that H ≤ ։ X J . 3 Γ X ( H ) ❴❡❦r②☎☞✓✙✤✪ ✰ ✷ ✿ ❉ ▲ ❙ ❨ ❴ ❡ ❦ r ③ ☎ ☞ ✓ ✙✤✪✰✷✿❉▲❙❨ • v • v (cid:30) (cid:30) (cid:30) (cid:30) p m m m m p (cid:15) (cid:15) (cid:15) (cid:15) p w ′ Figure 2:
Γ = Γ X ( H ) S p w ′ Assume now that w does not appear in Γ X ( H ) .Let p be the maximal path beginning at the base-point of Γ X ( H ) which is a prefix of w , and denoteby v its endpoint. Let p be the maximal pathending at the basepoint of Γ X ( H ) which is a suf-fix of w , and v its beginning. If w = p w ′ p ,take Γ = Γ X ( H ) S p w ′ where p w ′ is a path la-beled by w ′ , whose beginning is attached to v ,and whose endpoint to v (see Figure 2). Now π X (Γ) = J and Γ has no foldable edges norleaves, i.e. Γ = Γ X ( J ) . Thus Γ X ( H ) is a sub-graph of Γ X ( J ) , and in particular does not maponto it. (In fact, since the map from Γ X ( H ) to Γ X ( J ) is injective, H is a free factor of J .) Remark . With some further work, the basic idea of Lemma 2.3 can lead to an algo-rithm to detect primitive words and free factors in F . See [Pud11, Thm 1]. F In this section we give new proofs for the classical theorems on primitive words and basesof F [Nie17, Coh72, CMZ81, OZ81]. Throughout the section X denotes the basis { a, b } of F = F ( a, b ) .We start with the following lemma, which reduces the classification of bases of F to thatof cyclically reduced (henceforth: CR) bases. CR Lemma 3.1.
Let Y = { u, v } be any basis of F .(1) Write u = xux − † and v = yvy − with u, v CR. Then either x is a prefix of y or y is a prefix of x .(2) Assume that x is a prefix of y , and write u = xux − and v = xwvw − x − . Then w is a prefix of some power of u or of u − (which implies that w − uw is a cyclicrotation of u ).(3) The basis ( xw ) − Y xw = (cid:8) w − uw, v (cid:9) is CR.Therefore, any basis of F is of the form (cid:8) xux − , xwvw − x − (cid:9) where w is a prefix of somepower of u ± , (cid:8) w − uw, v (cid:9) is a CR basis, and x is any word s.t. xux − and xwvw − x − are reduced.Proof. The graph
Γ = • v " " ⊗ x / / y o o • u | | satisfies π X (Γ) = h u, v i = F . It also sat-isfies the conditions of Lemma 2.1, and must therefore fold into Γ X ( F ) = ⊗ a b { { . † By “write” we mean that xux − is a reduced expression of u - no cancellation is needed. This conventionwill repeat throughout the paper, and we will not mention it again. ⊗ , and after the first identi-fication of edges the only possible folding place is at the identified termini. Continuing inthis manner shows that for Γ to fold into Γ X ( F ) , the shorter word among x, y must becompletely merged with a prefix of the longer one, giving (1) . By the same arguments,the graph Γ ′ = • v " " ⊗ u { { w o o must fold into Γ X ( F ) , and for this to happen w mustwind itself completely around u , or around u − (i.e. w must be a prefix of some power of u or of u − ). It follows that w − uw is a cyclic rotation of u , and (cid:8) w − uw, v (cid:9) is a CRbasis.Moving on to CR bases, we have the following: Proposition 3.2.
Let { u, v } be a CR basis of F , such that | u | + | v | ≥ † , and | u | ≤ | v | .Then either u is a prefix or a suffix of v , or u − is.Proof. Since | v | ≥ and v is not a proper power, v contains both a and b , and thus h v i ≤ ։ X F . Since F = h u, v i , Lemma 2.3 implies that u appears in Γ X ( h v i ) , which isjust a cycle labeled by v as a word in X ∪ X − .Let p ′ be the maximal prefix of u which is a path emanating from the basepoint of Γ X ( h v i ) . Since | u | ≤ | v | , this means that p ′ is a prefix of v or of v − . By inverting v ifnecessary we assume that p ′ is a prefix of v . Let s ′ be the maximal suffix of u which isa path ending at the basepoint of Γ X ( h v i ) . Since u is CR, s ′ cannot be a suffix of v − ,and must be a suffix of v .Let m be the middle part of u where p ′ and s ′ overlap (it may be empty: | m | = | p ′ | + | s ′ | − | u | ≥ ). Write p ′ = pm , s ′ = ms , which means that u = pms (see Figure 3). Thus,if p is empty then u = pms = s ′ is a suffix of v , and if s is empty then u is a prefix of v .We proceed to show that they cannot be both nonempty.Figure 3: Illustration of the de-composition of v . vp ′ s ′ p t m sp m sp q r q r . . . q r q s Let
Γ = ⊗ u v { { . Since π X (Γ) = h u, v i = F and Γ satisfies the conditions ofLemma 2.1, it must fold into Γ X ( F ) = ⊗ a b { { , and we will show that this cannothappen if p, s = 1 .Assume therefore that p, s = 1 . Since | v | ≥ | u | we can write v = ptms , and t = 1 sinceotherwise u = v (this shows, in particular, that | v | > | u | ). Since pm is a prefix of v , m is a prefix of tm , which means that m is a prefix of some (positive) power of t (see againFigure 3). We consider two cases: † Here | w | is the length of w as a reduced word in X ∪ X − . ase ( i ) : m is not a power of t . In this case t = qr with q, r = 1 , and m = ( qr ) n q with n ≥ (see Figure 3; n = 0 corresponds to the possibility that p ′ and s ′ do not overlap in v ). Since u = p ( qr ) n qs and v = p ( qr ) n +1 qs , Γ folds into Γ ′ = • q (cid:15) (cid:15) ⊗ p ❦❦❦❦ • s i i ❙❙❙❙ r h h .We aim to show that no folding can occur in Γ ′ , but let us first introduce the followingnotations: for w ∈ F ( X ) , we denote by w the first letter of w as a reduced word in X ∪ X − , and for two words w, w ′ we write w ⊥ w ′ to indicate that w = ( w ′ ) . Namely, ⊥ w ⊥ w ′ implies that no folding occurs in • w o o w ′ / / .Returning to Γ ′ , we have p ⊥ s − since u (or equivalently v ) is CR, so no folding occursat ⊗ . Since v = p ( qr ) n +1 qs is reduced, p − , r − ⊥ q and q − ⊥ r, s . We also have r ⊥ s , for otherwise p ′ s = p ( qr ) n qr would be a common prefix of u = p ′ s and v = p ( qr ) n +1 qs , contradicting the maximality of p ′ . Finally, r − ⊥ p − follows in the sameway from the maximality of s ′ , and we conclude that Γ ′ cannot be folded any further, i.e. Γ ′ = Γ X ( h u, v i ) , which contradicts Γ X ( h u, v i ) = Γ X ( F ) . Case ( ii ) : m equals a power of t , m = t n ( n ≥ ), so that u = pt n s and v = pt n +1 s .This time Γ folds into Γ ′ = ⊗ p " " • t b b s c c . We have p ⊥ s − as before; p − ⊥ t and t − ⊥ s follows from v = pt n +1 s being reduced; s ⊥ t holds, since otherwise pt n s = pt n t would be a common prefix of u and v , contradicting the maximality of p ′ = pt n ; likewise, p − ⊥ t − by the maximality of s ′ . Now, if n > then t ⊥ t − since v = pt n +1 s isreduced, and if n = 0 then p − ⊥ s since u = ps is reduced. In either case, Γ ′ cannot foldinto Γ X ( F ) : For n = 0 , assuming that Γ ′ folds at all, Γ X ( h u, v i ) = ⊗ p " " • s c c r / / • t ′ b b ,where t = rt ′ r − with t ′ CR. Thus, Γ X ( h u, v i ) = Γ X ( F ) . For n > , Γ ′ folds into ⊗ p ′ (cid:29) (cid:29) • s ′ ^ ^ r / / • t b b where r is the common suffix of p and s − , so that p = p ′ r , s = s ′ r .If p ′ , s ′ = 1 we are done, and p ′ = s ′ = 1 is impossible since p ⊥ s − . If p ′ = 1 = s ′ thenthe graph folds into • s ′′ " " ⊗ r / / y o o • t | | where s ′′ is CR and s ′ = ys ′′ y − , and likewisefor p ′ = 1 = s ′ . Definition.
A word w ∈ F ( a, b ) is monotone if for every letter ( a or b ) all the exponents monotone of this letter in w have the same sign. Proposition 3.3.
A CR primitive word in F is monotone.Proof. Let u be a CR primitive. By Lemma 3.1, possibly applying some cyclic rotationto u , we can complete it to a CR basis { u, v } . We show that both u and v are monotone,by induction on | u | + | v | . The base case | u | + | v | = 2 is trivial. Assume that | u | ≤ | v | .Using Proposition 3.2, and perhaps replacing u , v , or both of them by their inverses(which does not affect monotonicity), we can write v = ut . Now { u, t } is a basis with | u | + | t | < | u | + | v | , and we claim that t is CR as well. Otherwise, t = rt ′ r − with t ′ CR and r = 1 , and we have Γ X ( h u, t i ) = ⊗ r / / u • t ′ | | , as u ⊥ u − since u is CR, ( t ′ ) − ⊥ t ′ since t ′ is, and all other relevant pairs since v = urt ′ r − is. This, of course,contradicts h u, t i = F .Therefore, by the induction hypothesis u and t are monotone. Assume first that | u | ≤ | t | .Since v = ut is CR, u − cannot be neither a prefix nor a suffix of t . Thus, by Proposition6.2 u must be a prefix or a suffix of t , and in either case v is monotone. The sameargument applies to the case | u | > | t | .We stress the following observation made in the proof: Corollary 3.4.
Let { u, v } be a CR basis of F with u a prefix of v , and write v = ut .Then { u, t } is again a CR basis. This leads to a constructive description of all CR bases of F : Proposition 3.5.
Any CR basis of F is obtained as follows: given a pair of positiveco-prime integers ( p, q ) , there is a unique sequence of pairs ( p, q ) = ( p , q ) , ( p , q ) , . . . , ( p ℓ , q ℓ ) = (1 , (3.1) which is the result of applying the Euclidean g.c.d. algorithm (i.e. if p i < q i then p i +1 = p i and q i +1 = q i − p i , and vice-versa). Let X ℓ = { u ℓ , v ℓ } be one of the four bases (cid:8) a ± , b ± (cid:9) ,and define X i = { u i , v i } iteratively for i = ℓ − . . . by ( u i , v i ) = ( ( u i +1 , v i +1 u i +1 ) p i < q i ( u i +1 v i +1 , v i +1 ) q i < p i . (3.2) Finally, take X , conjugate its elements by any common prefix or suffix (thus cyclicallyrotating both of them), and possibly replace one of them by its inverse.Proof. This construction certainly gives a CR basis of F (CR follows from monotonicity),and it remains to show that every CR basis is thus obtained. This is done by reversing theprocess, as follows. Let { x , y } be a CR basis. Discarding the trivial bases (cid:8) a ± , b ± (cid:9) ,we can assume (by inversion if necessary) that one of x or y is a prefix/suffix of theother (by Proposition 3.2). Lemma. x , y can be rotated by a a common prefix/suffix, so that a sequence of CRbases { x , y } , . . . , { x ℓ , y ℓ } with the following properties is obtained:(1) For every ≤ i ≤ ℓ − , the shorter of x i , y i is a suffix of the longer one.(2) Each basis is obtained from the previous one by ( x i +1 , y i +1 ) = ( ( x i , t ) | x i | < | y i | and y i = tx i ( t, y i ) | y i | < | x i | and x i = ty i . (3.3) (3) { x ℓ , y ℓ } is one of the four bases (cid:8) a ± , b ± (cid:9) .Proof of the Lemma. If we do not perform the rotation of x , y by a common pre-fix/suffix, the same holds, but with the exception that at each stage the shorter among x i , y i may be a prefix of the longer one, and not a suffix: this follows from by Proposition3.2, and Corollary 3.4, which ensures that all of these bases are CR. Assume that the7rocess first fails at step m , i.e. the shorter among x m , y m is not a suffix of the longerone, and assume for simplicity that x m is shorter, so that y m = x m t .Since x , y are products of positive powers of x m and y m = x m t , it follows that x m is acommon prefix of both of them. Therefore, { x ′ , y ′ } = (cid:8) x − m x x m , x − m y x m (cid:9) is a cyclicrotation of { x , y } . Let { x ′ i , y ′ i } i =0 ...m be the bases obtained by (3.3) from { x ′ , y ′ } . Sincethe expression of x ′ i , y ′ i as words in x ′ m , y ′ m is the same as that of x i , y i as words in x m , y m ,we still have that at every step until m the shorter of x ′ i , y ′ i is a suffix of the longer one.In fact, x ′ i = x − m x i x m and likewise for y ′ i , for all i ≤ m . Now, assertion (1) holds for step m as well, as x ′ m = x m and y ′ m = x − m ( x m t ) x m = tx m = tx ′ m .We continue in this manner: at the next step at which assertion (1) fails we replace { x ′ , y ′ } by { x ′′ , y ′′ } which resolves that step, and note that x ′′ , y ′′ is still a cyclic rotationof the original x , y by a common prefix/suffix, and that no new failures of (1) wereintroduced by this change for the previous steps. Repeating this for every failure of (1) guarantees that it hold throughout the process.We continue the proof of the proposition, assuming that x , y were inverted and rotatedaccording to the Lemma, and { x , y } , . . . , { x ℓ , y ℓ } are the bases obtained by (3.3). Thesequence of integer pairs ( | x | , | y | ) , . . . , ( | x ℓ | , | y ℓ | ) = (1 , is then the sequence obtainedby the Euclidean algorithm for ( | x | , | y | ) (as in (3.1)), and in particular this shows that | x | , | y | are co-prime. Thus, if one takes ( p, q ) = ( | x | , | y | ) and ( u ℓ , v ℓ ) = ( x ℓ , y ℓ ) ,and follows (3.2) as explained in the statement of the proposition, the process in (3.3) isreversed, and one obtains ( u , v ) = ( x , y ) . Corollary 3.6.
For a CR basis { u, v } , regard u and v as cyclic words, and assume (byinverting u if necessary) that one of them is a subword of the other. Then one of a, b always appears (in both u and v ) with exponent ε for some fixed ε ∈ { , − } , and theother letter always appears with exponent m or m + 1 for some m ∈ Z .Proof. Let X i = { u i , v i } ( ≤ i ≤ ℓ ) be the bases constructed in Proposition 3.5 to give X = { u, v } . Assume, for simplicity, that X ℓ = { u ℓ , v ℓ } = { a, b } , and that in the firststep the first option in (3.2) holds, so that X ℓ − = { a, ba } . Let r be the number of timesthe first option in (3.2) holds before it fails, i.e. X ℓ − = (cid:8) a, ba (cid:9) , . . . , X ℓ − r = { a, ba r } (possibly r = 1 ). If r = ℓ then the statement holds. Otherwise, X ℓ − r − = { aba r , ba r } ,and now every cyclic word which is a product of the elements of X ℓ − r − (with positiveexponents only) clearly satisfies the statement of the corollary with ε = 1 and m = r .Since u and v are such words, we are done. Let X = { a, b } and F = F ( X ) . In this section we prove the following: Proposition 4.1.
Let H = (cid:10) a b (cid:11) , and J = (cid:10) a b , ab (cid:11) . Then H ≤ ։ Y J for every basis Y of F , but J is not an algebraic extension of H . roof. First, as H is a free factor of J (since J = H ∗ h ab i ), it is clear that J is not analgebraic extension of H , and it is left to show that H covers J with respect to everybasis Y = { u, v } . For any automorphism ϕ of F , H ≤ ։ X J iff ϕ ( H ) covers ϕ ( J ) w.r.t.the basis ϕ ( X ) = { ϕ ( a ) , ϕ ( b ) } . As ϕ ( X ) achieves all bases of F , what we seek to showis equivalent to the assertion that (cid:10) u v (cid:11) ≤ ։ X (cid:10) u v , uv (cid:11) for every basis { u, v } .By Lemma 2.3, showing that (cid:10) u v (cid:11) X -covers (cid:10) u v , uv (cid:11) is equivalent to verifying that uv appears in Γ X (cid:0)(cid:10) u v (cid:11)(cid:1) . For the case where u and v are CR this is shown in Lemma4.3, and the case where only one of them is CR is handled in Lemma 4.4. For the generalcase, let Y = { u, v } be the base at hand, and write u = xux − and v = yvy − with u, v CR. By Lemma 3.1, x is a prefix of y , or vice-versa. Thus, if w is the shorter among x, y then w − Y w is a basis with one CR element, which was already handled. Inferring fromthis the result for the original Y is done in Lemma 4.2. For this we need an additionaltechnical assumption on Γ X (cid:0)(cid:10) w − u v w (cid:11)(cid:1) , which is seen to hold in Lemmas 4.3 and4.4. Lemma 4.2.
Let { u, v } be a basis of F such that u and v share a common prefix w anda common suffix w − , and write u = wuw − and v = wvw − . If(1) uv appears in Γ X (cid:0)(cid:10) u v (cid:11)(cid:1) , and(2) either u or (cid:0) v − (cid:1) emanates from the basepoint of Γ X (cid:0)(cid:10) u v (cid:11)(cid:1) ,then uv appears in Γ X (cid:0)(cid:10) u v (cid:11)(cid:1) .Proof. Observe the graph Γ , which is obtained by attaching a path labeled by w to thebasepoint of Γ X (cid:0)(cid:10) u v (cid:11)(cid:1) and moving the basepoint to the origin of the w -path: ⊗ Γ X (cid:0)(cid:10) u v (cid:11)(cid:1) ❴❞❥♣④✡✖✧✲ ❀ ❍ ◗ ❲ ❭ ❜ ❣ ♠ ✈ ✄✑✜✭✹❈◆❚❩❴ = ⇒ • Γ ❴❞❥♣④✡✖✧✲ ❀ ❍ ◗ ❲ ❭ ❜ ❣ ♠ ✈ ✄✑✜✭✹❈◆❚❩❴ ⊗ < < w The graph Γ folds into Γ X (cid:0) π X (Γ) (cid:1) = Γ X (cid:0)(cid:10) u v (cid:11)(cid:1) , since if satisfies the conditions ofLemma 2.1: the only vertex that needs checking is the gluing place, and there the con-ditions hold by assumption (2) and the fact that w − ⊥ u, v − (as u, v are reduced).Finally, since uv appears in Γ X (cid:0)(cid:10) u v (cid:11)(cid:1) , uv = wuvw − appears in Γ , and thus also inits folding Γ X (cid:0)(cid:10) u v (cid:11)(cid:1) . Lemma 4.3. If { u, v } is a CR basis of F then(1) uv appears in Γ X (cid:0)(cid:10) u v (cid:11)(cid:1) , and(2) u or (cid:0) v − (cid:1) emanates from the basepoint of Γ X (cid:0)(cid:10) u v (cid:11)(cid:1) . roof. If | u | + | v | = 2 then the claims hold, so assume that | u | + | v | ≥ . Since Γ X (cid:0)(cid:10) u v (cid:11)(cid:1) = Γ X (cid:0)(cid:10) v − u − (cid:11)(cid:1) and Γ X (cid:0)(cid:10) u v , uv (cid:11)(cid:1) = Γ X (cid:0)(cid:10) v − u − , v − u − (cid:11)(cid:1) , one canlook at (cid:8) v − , u − (cid:9) instead of { u, v } (this also does not affect assertion (2) ), and thus itis enough to handle the cases where | u | < | v | .Observe the graph Γ = • u ( ( ◗◗◗◗ ⊗ u ❧❧❧❧ ◦ v v v ♠♠♠♠ • v h h ❘❘❘❘ , which obviously satisfies π X (Γ) = (cid:10) u v (cid:11) . At theblack vertices there can be no folding, as u − ⊥ u follows from u being CR, and likewisefor v . In what follows we will continue to mark by • vertices at which we already knowthat no folding can occur, and by ◦ vertices at which we do not know this.Assume first that u − is not a prefix of v , and m is the maximal common prefix of u − and v . Writing u = u ′ m − and v = mv ′ , Γ folds into • u ′ ( ( ◗◗◗◗ ⊗ u ❧❧❧❧ • v ′ v v ♠♠♠♠ • m o o • v h h ❘❘❘❘ . After trimming oneobtains Γ ′ = • u ′ ( ( ◗◗◗◗ ⊗ u ❧❧❧❧ • v ′ v v ♠♠♠♠ • v h h ❘❘❘❘ , which satisfies the conditions of Lemma 2.1: u − ⊥ u ′ since u is CR and u ′ is a prefix of u , and likewise for v and v ′− ; u ′− ⊥ v ′ by the maximality of m . Therefore, Γ ′ folds into Γ X (cid:0)(cid:10) u v (cid:11)(cid:1) . Since uv appears in Γ ′ , and both u and (cid:0) v − (cid:1) leave its basepoint, the same holds after any foldings, and in particular in Γ X (cid:0)(cid:10) u v (cid:11)(cid:1) .Assume now that u − is a prefix of v and write v = u − t . Now Γ folds and trims into ◦ t ( ( ◗◗◗◗ ⊗ u ❧❧❧❧ •• t h h ❘❘❘❘ u ♠♠♠♠ , as u ⊥ t and t − ⊥ u − follow from v = u − t being CR. Let m be themaximal common prefix of u − and tu − , and write u = um − , tu − = mq (note that (cid:12)(cid:12) tu − (cid:12)(cid:12) > (cid:12)(cid:12) u − (cid:12)(cid:12) ≥ | m | ). The last graph then folds and trims into Γ ′ = • q (cid:15) (cid:15) ⊗ u ❦❦❦❦ • t i i ❙❙❙❙ (with u possibly empty, in which case Γ ′ = ⊗ q • t a a ). This graph satisfies the conditions of Lemma2.1: q − ⊥ t since tu − = mq is CR, being a cyclic rotation of v , and if u is not emptythen u − ⊥ q by maximality of m . Since uv = t appears in Γ ′ and (cid:0) t − (cid:1) = (cid:0) v − (cid:1) leavesits basepoint, the same holds for Γ X (cid:0)(cid:10) u v (cid:11)(cid:1) , which is obtained from it by foldings. Lemma 4.4. If { u, v } is a basis of F with u or v CR then(1) uv appears in Γ X (cid:0)(cid:10) u v (cid:11)(cid:1) , and(2) u or (cid:0) v − (cid:1) emanates from the basepoint of Γ X (cid:0)(cid:10) u v (cid:11)(cid:1) .Proof. If both u and v are CR then we are done by Lemma 4.3. Again, by replacing u and v with v − and u − respectively we can assume that u is CR and v is not (here u is not necessarily shorter). Writing v = wvw − with v CR (and w = 1 ), w is a prefix ofsome power of u or u − by Lemma 3.1. The graph formed by a single u v -loop folds andtrims into Γ = • u / / ◦ w ( ( ◗◗◗◗ ⊗ u ❧❧❧❧ w ( ( ❘❘❘❘ • v v v ♠♠♠♠ • • v o o , where at all the black vertices there can be no foldingsince u and v are CR and v is reduced. 10f w is a prefix of a (positive) power of u then at ◦ there is no folding as well since u isCR, so that Γ X (cid:0)(cid:10) u v (cid:11)(cid:1) is obtained from Γ by foldings. Therefore to establish that uv appears in Γ X (cid:0)(cid:10) u v (cid:11)(cid:1) it is enough to show that it appears in Γ . This is not obvious infirst sight, but it is true: w is a prefix of a positive power of u , so that uw is a prefix of u w , hence uv = uwvw − indeed appears in Γ . Finally, both u and (cid:0) v − (cid:1) = w leavethe basepoint of Γ and thus also of Γ X (cid:0)(cid:10) u v (cid:11)(cid:1) .We assume now that w is a prefix of some power of u − , and observe six cases. Case ( i ) : | u | ≤ | w | , so that w = u − w with w possibly empty. In this case Γ foldsand trims into Γ ′ = • v / / • v ( ( ◗◗◗◗ ⊗ w ❧❧❧❧ •• u h h ❘❘❘❘ • w ♠♠♠♠ u o o , which satisfies Lemma 2.1 (even if w = 1 ). Now uv = u − wvw − u appears in Γ ′ : vw − u is a suffix of Γ ′ oriented clockwise, and u − w is a prefix of Γ ′ oriented counterclockwise since w is a prefix of u − w . In addition, (cid:0) v − (cid:1) = w = (cid:0) u − (cid:1) leaves ⊗ . Case ( ii ) : | u | < | w | < | u | , so that we can write u = qr and w = r − q − r − with q, r = 1 . Now Γ folds and trims into Γ ′ = ◦ v / / • v ( ( ◗◗◗◗ ⊗ q ❧❧❧❧ • r v v ♠♠♠♠ • r h h ❘❘❘❘ • q o o , and no folding can occur atthe black vertices. If at ◦ there is no folding as well, then uv = r − vrqr appears in Γ ′ and (cid:0) v − (cid:1) = w = (cid:0) r − (cid:1) leaves ⊗ , yielding the same for Γ X (cid:0)(cid:10) u v (cid:11)(cid:1) .Assume now that there is folding at ◦ , so that q − and v have a common prefix. Ifthis prefix is shorter than | v | than the vrqr part in Γ ′ survives in Γ X (cid:0)(cid:10) u v (cid:11)(cid:1) , and thus uv = r − vrqr still appears in it. Otherwise, v is a prefix of q − , so that r − v is a prefixof Γ ′ oriented CCW, and rqr is a suffix of Γ ′ oriented CW, so that uv appears already inthe lower half of Γ ′ . Finally, this half survives in Γ X (cid:0)(cid:10) u v (cid:11)(cid:1) since q − cannot overlapwith r , since u = qr is monotone by Proposition 3.3. In both cases (cid:0) v − (cid:1) = (cid:0) r − (cid:1) stillleaves ⊗ . Case ( iii ) : w = u − . The reasoning here is as in the previous case with r = 1 .In cases ( iv ) − ( vi ) w is a proper prefix of u − , and we write u = qr and w = r − (with r, q = 1 ). Here Γ folds and trims into Γ ′ = • q / / ◦ v ) ) ❙❙❙❙ • r ❦❦❦❦ • v v v ♠♠♠♠ ⊗ q h h ❘❘❘❘ (cid:7) r o o , with folding possibleonly at ◦ , and the folding cannot reach past (cid:7) due to the monotonicity of u = qr . Thisalready shows that (cid:0) v − (cid:1) = (cid:0) r − (cid:1) must leave the basepoint of the final Stallings graph Γ X (cid:0)(cid:10) u v (cid:11)(cid:1) . It is left to show that the folding and trimming at ◦ does not prevent uv = qvr from appearing in Γ X (cid:0)(cid:10) u v (cid:11)(cid:1) . If the lower half of Γ ′ survives the folding andtrimming then uv certainly appears in it. We assume therefore that there is folding at ◦ ,and that it encompasses either all of v to the right of ◦ or all of rq to the left of it (i.e. itreaches the lower half of Γ ′ ). Case ( iv ) : | v | ≤ | q | . By our assumption, v is a prefix of q − , so q = qv − ( q maybe empty). Γ ′ then folds and trims into Γ ′′ = • v u u ❦❦❦❦ r / / • q ) ) ❙❙❙❙ • ◦ v v v ♠♠♠♠ ⊗ q h h ❘❘❘❘ (cid:7) r o o . Now uv = qvr = qr and since the folding from ◦ downward must stop at (cid:7) (or earlier), the • ⊗ q o o (cid:7) r o o partsurvives in Γ X (cid:0)(cid:10) u v (cid:11)(cid:1) (since | v | < (cid:12)(cid:12) q − r − v (cid:12)(cid:12) ) and we are done.11 ase ( v ) : | q | < | v | ≤ | rq | . Now we can assume that v is a prefix of q − r − , so that r = st and v = q − t − (possibly with s = 1 ). In this case uv = qvr = t − r alreadyappears in the ⊗ (cid:7) r o o part of Γ ′ , which always survives due to monotonicity. Case ( vi ) : | rq | < | v | . Now we can assume that q − r − is a prefix of v . Therefore, uv = qvr is a suffix of vr , and thus appears in the ⊗ (cid:7) r o o • v o o part in Γ ′ . If v is nota prefix of q − r − q − then the folding from ◦ downward stops before reaching this part,and we are done. We thus add the assumption that v is a prefix of q − r − q − . Since | rq | < | v | we can write q = st so that v = q − r − t − = t − s − r − t − . Now Γ ′ foldsand trims into Γ ′′ = • t u u ❦❦❦❦ • s o o ◦ • r i i ❙❙❙❙ ⊗ s h h ❘❘❘❘ (cid:7) r o o t ♠♠♠♠ , and uv = r − t − r appears in ⊗ (cid:7) r o o t / / • , whichsurvives any further folding since | s | < (cid:12)(cid:12) t − s − r − (cid:12)(cid:12) . While the original conjecture [MVW07, §5(1)] fails, it is plausible that some modificationof it holds. One possible option is the following:
Conjecture 5.1.
Let H ≤ J be subgroups of the free group F . Then H ≤ alg J iff H ≤ ։ X J for every free extension F ′ of F , and every basis X of F ′ . Since the relation H ≤ alg J does not depend on the ambient group, one direction holds asbefore. But in contrast with the original conjecture, the example in Section 4 is no longera counterexample: let F = F ( a, b ) , H = (cid:10) a b (cid:11) and J = (cid:10) ab, a b (cid:11) . For F ′ = F ( a, b, c ) and X = (cid:8) a, cb − , cbc − (cid:9) , H does not X -cover J : denote x = a , y = cb − and z = cbc − .Then written in this basis, H = (cid:10) x y − z y (cid:11) and J = (cid:10) x y − z y, xy − zy (cid:11) . By Lemma2.3, H ≤ ։ X J iff xy − zy appears in Γ { x,y,z } (cid:0)(cid:10) x y − z y (cid:11)(cid:1) , which is not the case.Another plausible option is that the original conjecture from [MVW07, §5(1)] holds forfree groups of rank three or more, as it is clear that the counterexample exploits manyidiosyncrasies of F . If this is true, then Conjecture 5.1 follows as well. References [CMZ81] M. Cohen, W. Metzler, and A. Zimmermann,
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