Embeddings between partially commutative groups: two counterexamples
aa r X i v : . [ m a t h . G R ] J un Embedddings between partially commutative groups: twocounterexamples ✩ Montserrat Casals-Ruiz a,1,2 , Andrew Duncan b,3 , Ilya Kazachkov a,2,4 a Mathematical Institute, University of Oxford, 24-29 St. Giles’, Oxford, OX1 3LB, UK b School of Mathematics and Statistics, Newcastle University, Newcastle upon Tyne, NE1 7RU, United Kingdom
Abstract
In this note we give two examples of partially commutative subgroups of partially commutativegroups. Our examples are counterexamples to the Extension Graph Conjecture and to the WeaklyChordal Conjecture of Kim and Koberda, [KK]. On the other hand we extend the class of par-tially commutative groups for which it is known that the Extension Graph Conjecture holds, toinclude those with commutation graph containing no induced C or P . In the process, some newembeddings of surface groups into partially commutative groups emerge. Keywords:2000 MSC:
1. Introduction
Partially commutative groups are a class of groups widely studied on account of their simple defini-tion, their intrinsically rich structure and their natural appearance in several branches of computerscience and mathematics. Crucial examples, which shape the theory of presentations of groups,arise from study of their subgroups: notably Bestvina and Brady’s example of a group which is ho-mologically finite (of type
F P ) but not geometrically finite (in fact not of type F ); and Mihailova’sexample of a group with unsolvable subgroup membership problem. More recently results of Wiseand others have lead to Agol’s proof of the virtual Haken conjecture: that every hyperbolic Haken3-manifold is virtually fibred. An essential step in the argument uses the result that the funda-mental groups of so-called “special” cube complexes embed into partially commutative groups. (Itturns out that the manifolds in question have finite covers which are special.) In the light of such ✩ The authors are grateful to the referee for careful reading of, and several improvements to, the text.
Email addresses: [email protected] (Montserrat Casals-Ruiz ), [email protected] (AndrewDuncan), [email protected] (Ilya Kazachkov) Supported by a Marie Curie International Incoming Fellowship within the 7th European Community FrameworkProgramme. Supported by the Spanish Government, grant MTM2011-28229-C02-02, partly with FEDER funds. Partially supported by EPSRC grant EP/F014945. Supported by the Royal Commission’s 1851 Research Fellowship.
Preprint submitted to Elsevier November 16, 2018 esults it is natural to ask which groups arise as subgroups of partially commutative groups, and,in particular, when one partially commutative group embeds in another.Let Γ be a finite (undirected) simple graph, with vertex set X and edge set E , and let F ( X ) bethe free group on X . For elements g, h of a group denote the commutator g − h − gh of g and h by[ g, h ]. Let R = { [ x, y ] ∈ F ( X ) | x, y ∈ X and { x, y } ∈ E } . Then the partially commutative (pc) group with commutation graph
Γ is the group G (Γ) withpresentation h X | R i . The complement of a graph Γ is the graph Γ with the same vertex set asΓ and an edge joining vertices u and v if and only if there is no such edge in Γ. The partiallycommutative group G (Γ) is said to have non-commutation graph Γ.Note that the graph Γ uniquely determines the partially commutative group up to isomorphism;that is two pc groups G (Γ) and G (Λ) are isomorphic if and only if Γ is isomorphic to Λ, see [D87.3].Since isomorphism of pc groups can be characterised in terms of their defining graphs is natural toask for an analogous characterisation of embedding between pc groups. Question 1 (Problem 1.4 in [CSS]) . Is there a natural graph theoretic condition which determineswhen one partially commutative group embeds in another?If Λ and Γ are graphs such that Λ is isomorphic to a full subgraph of Γ, then Λ is said to be an induced subgraph of Γ and we write Λ ≤ Γ. If Λ is not an induced subgraph of Γ then Γ is said to beΛ -free . It is easy to see that if Λ is an induced subgraph of Γ then G (Λ) ≤ G (Γ). However, unless Γis a complete graph, G (Γ) will contain subgroups which do not correspond to induced subgraphs inthis way. Moreover, in general pc groups contain subgroups which are not pc groups. In fact Dromsproved [D87.2] that every subgroup of a partially commutative group G (Γ) is partially commutativeif and only if Γ is both C and P free, where C n is the cycle graph on n vertices and P n is thepath graph with n edges.Significant progress towards answering Question 1 has been made by Kim and Koberda [KK],exploiting the notion of the extension of a graph. The extension graph Γ e of a graph Γ is the graphwith vertex set V e = { g − xg ∈ G (Γ) | x ∈ V (Γ) , g ∈ G (Γ) } and an edge joining u to v if andonly if [ u, v ] = 1 in G (Γ). In [KK] it is shown that if Λ is a subgraph of Γ e then G (Λ) embeds in G (Γ). However, the converse is shown to hold only if Γ is triangle-free ( C free): in which case if G (Λ) embeds in G (Γ), then Λ is a subgraph of Γ e ([KK, Theorem 10]). This suggests the followingconjecture. Conjecture 1.
The Extension Graph Conjecture, [KK, Conjecture 4]] Let Γ and Γ be finitegraphs. Then G (Γ ) < G (Γ ) if and only if Γ < Γ e .In Section 5 we extend the class of graphs for which Conjecture 1 is known to hold, to include inaddition to the triangle-free graphs, a family lying at the opposite end of the spectrum; being “builtout of triangles”. More precisely, we generalise one direction of Droms’ theorem to prove that if Γis both C and P free, then the Extension Graph Conjecture holds for G (Γ). However, in Section3 below we give a counterexample to the Extension Graph Conjecture. We note that Droms alsoshowed that a pc group is coherent if and only if it is chordal [D87.1], i.e. C n -free, for all n ≥ −
1; see [CSS] and the references there). In [DSS] it was shown that many orientablesurface groups embed into pc groups and subsequently Crisp and Wiest [CW] showed that thefundamental group of every compact surface embeds into some pc group, except in the cases of thecompact, closed, non-orientable, surfaces of Euler characteristic − , G ( C ). InSection 4 we establish similar results for the complement P of P , in place of C .Although being an induced subgraph is not a necessary condition for embeddability, for someparticular induced subgraphs it is a useful criterion. For instance, a pc group G (Γ) contains G ( P )(resp. C ), if and only if P (resp. C ) is an induced subgraph of Γ, see [Kambites, KK]. In [GLR],the authors ask if the presence of an induced n -cycle, n ≥
5, in the defining graph Γ is a necessarycondition for the group G (Γ) to contain a hyperbolic surface group. However, counterexamples weregiven in [K08] and [CSS]. For example, the fundamental group of the closed, compact, orientablesurface of genus 2 embeds in the pc group G ( P (6)), although P (6) has no induced n -cycle, for n ≥
5. Note however that in the case when the graph Γ is triangle-free, G ( C n ) ≤ G (Γ) for some n ≥ C m ≤ Γ for some 5 ≤ m ≤ n , see [KK, Corollary 48].In spite of these examples, the question raised by Gordon, Long and Reid survives in the followingform. Question 2.
Is there a partially commutative group G (Γ) that contains the fundamental group ofa closed hyperbolic surface but contains no subgroup isomorphic to G ( C n ), for n ≥ n -cycle, n ≥ G ( C n ), n ≥
5. A possible candidate is the class of weakly chordal graphs: a graphΓ is called weakly chordal if Γ does not contain an induced C n or C n for n ≥
5, where C n denotesthe complement of C n . Since the extension graph of a weakly chordal graph is weakly chordal, see[KK, Lemma 30], one can weaken Conjecture 1 as follows. Conjecture 2 (The Weakly Chordal Conjecture, [KK, Conjecture 13]) . If Γ is a weakly chordalgraph, then G (Γ) does not contain G ( C n ) for any n ≥ P (6) is weakly chordal, and contains hyperbolic surface groups, wewould have a positive answer to Question 2. However, our second example, in Section 4, gives acounterexample to the Weakly Chordal Conjecture.
2. Preliminaries By graph we mean simple, undirected graph. Let Γ be a finite graph with vertices X and let G (Γ)be the partially commutative group with commutation graph Γ.If w is a word in the free group on X then we say that w is reduced in G (Γ) if w has minimal lengthamongst all words representing the same element of G (Γ) as w . If w is reduced in G (Γ) then we3efine alph Γ ( w ) to be the set of elements of X such that x or x − occurs in w . It is well-knownthat all words which are reduced in G (Γ) and represent a particular element g of G (Γ) have thesame length, and that if w = w ′ in G (Γ) then alph Γ ( w ) = alph Γ ( w ′ ).For a word w in the free monoid on X ∪ X − we write alph( w ) for the set of elements of X suchthat x or x − occurs in w . We write u . = v to indicate that u and v are equal as words in the freemonoid on X ∪ X − . By van Kampen’s Lemma (see [BSR07]) the word w represents the trivial element in a fixed group G given by the presentation h A | R i if and only if there exists a finite connected, oriented, based,labelled, planar graph D , where each oriented edge is labelled by a letter in A ± , each boundedregion (2- cell ) of R \ D is labelled by a word in R (up to shifting cyclically or taking inverses) and w can be read on the boundary of the unbounded region of R \ D from the base vertex. Thenwe say that D is a van Kampen diagram for the boundary word w over the presentation h A | R i .Note that any van Kampen diagram can also be viewed as a 2-complex, with a 2-cell attached foreach bounded region. A van Kampen diagram for the word w is minimal if no other van Kampendiagram for w has fewer 2-cells.Following monograph [O], if we complete the set of defining relations adding the trivial relations1 · a = a · a ∈ A , then every van Kampen diagram can be transformed so that its boundaryis a simple curve. In other words, as a 2-complex the van Kampen diagram is homeomorphic to adisc tiled by 2-cells which are also homeomorphic to a disc. We shall assume that all van Kampendiagrams are of this form.From now on we shall restrict our considerations to the case when G is a partially commutativegroup. Let D be a minimal van Kampen diagram for the boundary word w . Given an occurrenceof a letter a ∈ A ± in w , there is a 2-cell C , in the 2-complex D , attached to a . Since every 2-cell ina van Kampen diagram is either labelled by a relation of the form a − b − ab or is a padding · a · · a − , there is just one occurrence of a and one occurrence of a − onthe boundary of C .Since D is homeomorphic to a disc, if the occurrence of a − on the boundary of C is not onthe boundary of D , there exists a unique 2-cell C ′ = C attached to this occurrence of a − in D .Repeating this process, we obtain a unique band in D .Because of the structure of the 2-cells and the fact that D is homeomorphic to a disc, a band neverself-intersects in a 2-cell; indeed, since D is homeomorphic to a disc, a 2-cell corresponding to aself-intersection of a band would be labelled by the word aaa − a − , a contradiction.Thus, since the number of 2-cells in D is finite, in a finite number of steps the band will again meetthe boundary in an occurrence of a − in w .We will use the notation L a to indicate that a band begins (and thus ends) in an occurrence of aletter a ∈ A ± . Fix an orientation of the boundary of D so that the label of the boundary, readfrom the base vertex, with this orientation, is the word w . The band L a determines a boundarysubword w ( L a ) of w : namely the subword of w that is read on the boundary between the ends4f L a , in the direction of fixed orientation. In other words, every band L a induces the followingdecomposition of the boundary word w : w = w a ǫ w ( L a ) a − ǫ w , where ǫ ∈ {± } .Furthermore the band L a determines a band word z ( L a ): namely the word obtained by readingalong the boundary of the band from the terminal end of a ǫ to the initial end of a − ǫ .Note that if two bands L a and L b cross then the intersection 2-cell realises the equality a − b − ab = 1and so a = b and [ a, b ] = 1. It follows that if a letter c occurs in w ( L a ) and c = a ± or [ c, a ] = 1,then the ends of the band L c both occur in w ( L a ), i.e. the band L c is contained in the regionbounded by z ( L a ), and w ( L a ). Notice that since z ( L a ) labels the boundary of the band L a , itfollows that every letter b ∈ z ( L a ) we have that b = a ± and [ b, a ] = 1.The following will be useful in the next section. If L u is a band, where u = a ± , for some a ∈ A such that a / ∈ alph( w ( L u )), then we say that L u is an outside band . Lemma 2.1.
Let D be a van Kampen diagram for the boundary word w . If x ∈ alph( w ) then there exists an outside band L u , u = x ± . Let L v be a band, where v ∈ A ± . If y ∈ alph( w ( L v )) , such that [ y, v ] = 1 or y = v ± , thenthere exists an outside band L z , with z = y ± , which has both ends in w ( L v ) .Proof.
1. Amongst all bands beginning and ending in x ± , choose a band L u , u = x ± , suchthat the length of w ( L u ) is minimal. As no two bands with ends x ± can cross it follows that x / ∈ alph( w ( L u )). Hence L u is an outside band.2. As y ∈ alph( w ( L v )), there exists a band L with one end at an occurrence of y ± in w ( L v ).As [ y, v ] = 1 or y = v ± , bands L v and L cannot cross, so L has both ends in w ( L v ). Thatis, every band with one end at an occurrence of y ± in w ( L v ) has both ends in w ( L v ). Nowchoose such a band L ′ with | w ( L ′ ) | minimal over lengths of boundary words of all bandsending at occurrences of y ± in w ( L v ). As in the proof of the first part of the Lemma, L ′ isan outside band.
3. A counterexample to the Extension Graph Conjecture
In this Section we give a counterexample to Conjecture 1. Let Γ and Γ be the graphs of Figure1a and Figure 1b, respectively. Let G (Γ ) and G (Γ ) be the corresponding partially commutativegroups: G (Γ ) = h a, b, c, d, e | [ a, d ] , [ a, e ] , [ b, e ] , [ c, d ] , [ d, e ] i , G (Γ ) = h a , a , b, c, d, e | [ a , a ] , [ a , c ] , [ a , d ] , [ a , e ] , [ a , b ] , [ a , d ] , [ a , e ] , [ b, e ] , [ c, d ] , [ d, e ] i . aa a bcde (a) Γ PSfrag replacements aa a bcde (b) Γ Figure 1Define a map ϕ = a a a ; b b ; c c ; d d ; e e. Lemma 3.1.
The map ϕ defined above extends to a monomorphism ϕ : G (Γ ) → G (Γ ) .Proof. It is immediate that ϕ ([ a, d ]) = ϕ ([ a, e ]) = ϕ ([ b, e ]) = ϕ ([ c, d ]) = ϕ ([ d, e ]) = 1in G (Γ ), hence ϕ extends to a homomorphism.We show that ϕ is injective. For a contradiction assume that ker( ϕ ) contains a non-trivial element g and let w = w ( a, b, c, d, e ) ∈ G (Γ ) be a minimal length word representing g . Consider a minimalvan Kampen diagram D for ϕ ( w ) = w ( a a , b, c, d, e ) = 1.Since the canonical parabolic subgroup generated by { b, c, d, e } is isomorphic to its image by ϕ , itfollows that the word w must contain at least one occurrence of the letter a (and its inverse a − ).Since ϕ ( a ) = a a , ϕ ( w ) contains at least one occurrence of a and of a and every occurrenceof a ± (or a ± ) in ϕ ( w ) occurs in a subword ( a a ) ± . Moreover, if w has subword of the form( a a ) ǫ u ( a a ) − ǫ , with ǫ ∈ {± } and a / ∈ alph( u ), then it follows that a / ∈ alph( u ) (and if a / ∈ alph( u ) then a / ∈ alph( u )).We claim that if L x is an outside band, with x = a ± , then there exists a band L with one end atan occurrence of c ± in w ( L x ) and its other end not in w ( L x ). Similarly, if L y is an outside band,with y = a ± , then there exists a band L with one end at an occurrence of b ± in w ( L y ) and itsother end not in w ( L y ). As there is an isomorphism of Γ interchanging a and a and b and c , itsuffices to prove the first of these claims.To see that the first claim holds, suppose that L x is an outside band, with x = a ± . Without lossof generality (replacing w with w − if necessary), we may assume that x = a . Then the band L x induces the decomposition ϕ ( w ) . = w a a w a − a − w and, since a / ∈ alph( w ), it follows that6 , a / ∈ alph( w ). Then alph( w ) must contain either b or c . Indeed, otherwise, since w belongsto the subgroup generated by { b, c, d, e } and so w = ϕ ( w ), the subword aw a − of w would notbe reduced in G (Γ ), a contradiction. Also, as w = ϕ ( w ), the word w is reduced in G (Γ ).Assume to begin with that alph( w ) does not contain c ; so does contain b . In this case, since[ a , b ] = 1, Lemma 2.1.2 implies the existence of an outside band L z , where z = b ± , with bothends in w . As L z is an outside band b / ∈ alph( w ( L z )) and hence, since w ( L z ) is a subword of w , we have alph( w ( L z )) ⊆ { d, e } . Moreover, w is reduced in G (Γ ) so w ( L z ) must contain anoccurrence d ± . From Lemma 2.1.2 again, there is an outside band L v , where v = d ± , with bothends in w ( L z ). However, this means that alph( w ( L v )) ⊆ { e } , and so w is not reduced in G (Γ ), acontradiction.Therefore we conclude that c ∈ alph( w ) and there is a band L u , where u = c ± , with one endin w . If both ends of L u are in w then there exists an outside band L z , z = c ± , with bothends in w (Lemma 2.1.2). As w is reduced in G (Γ ), alph( w ( L z )) must contain either b or e . If b ∈ alph( w ( L z )) then, since c / ∈ alph( w ( L z )), we obtain a contradiction, using the argument of theprevious paragraph. Therefore b / ∈ alph( w ( L z )) and e ∈ alph( w ( L z )). This implies the existence ofan outside band L v , v = e ± , with both ends in w ( L z ). However, in this case alph( w ( L v )) ⊆ { d } ,and again w is not reduced in G (Γ ), a contradiction. Thus one end of L u lies in w or w . Thisproves the claim, for L x , with L = L u .Returning to the proof of the lemma, let L x be an outside band, with x = a ± , and let w = w ( a a ) ǫ w ( a a ) − ǫ w , be the corresponding decomposition of w , where w ( L x ) = w , if ǫ = 1and w ( L x ) = a − w a , otherwise. From the claim above, there exists a band L c , where c = c ± ,with one end in w and the other end in w or w . Taking a cyclic permutation of w (and ϕ ( w )),if necessary, we may assume that L c has its other end in w .Let L ′ be the band which has one end at the occurrence of a ± in ( a a ) ǫ w . Since [ a , c ] = 1the bands L c and L ′ cannot cross and, as a / ∈ alph( w ), the other end of L ′ must lie in w (seeFigure 2). Now we find an outside band L y , where y = a ± , with both ends in w ( L ′ ). Hence w decomposes as w . = w c − δ w a − ǫ w a γ w a − γ w , for some γ, δ ∈ {± } , as shown in Figure 2. Thus, the claim above implies there exists a band L b , where b = b ± , with exactly one end in w ( L y ) = w . As [ b, c ] = 1 and [ b, a ] = 1 theband L b cannot cross L c or L x , so the other end of L b lies in in w , w or w . In particular w ( L b ) is a subword of w ( L c ). Moreover, since w ( L b ) contains an occurrence of a ± we have a , a ∈ alph( w ( L b )).Now suppose that we have occurrences of letters c , . . . , c n , b , . . . , b n , with c i = c ± and b i = b ± ,and corresponding bands L c i , L b i , such that w ( L b i ) is a subword of w ( L c i ), for i = 1 , . . . , n , w ( L c i +1 )is a subword of w ( L b i ), for i = 1 , . . . , n −
1, and a , a ∈ w ( L b n ).As [ a , b ] = 1 there exists an outside band L x n +1 , where x n +1 = a ± , with both ends in w ( L b n ).Thus w ( L b n ) decomposes as w ( L b n ) . = u ( a a ) ζ u ( a a ) − ζ u , where L x n +1 has both ends at occurrences of a ± in ( a a ) ζ u ( a a ) − ζ , for some u ∈ h b, c, d, e i and ζ ∈ {± } . From the claim above, there is a band L c n +1 , where c n +1 = c ± , with one end in u w w w w w L x L y L c L ′ Figure 2and the other end not in u . As [ b, c ] = 1, the other end of L c n +1 lies in u or u . In particular, w ( L c n +1 ) is a subword of w ( L b n ).If L c n +1 has one end in u then let L ′′ be the band which has one end at the occurrence of a ± in( a a ) ζ u , as shown in Figure 3. Then w ( L ′′ ) is a subword of u ( a a ) ζ and there exists an outsideband L y n +1 , where y n +1 = a ± , with both ends in w ( L ′′ ).From the claim again, we then have a band L b n +1 , where b n +1 = b ± , with one end in w ( L y n +1 )and the other end outside w ( L y n +1 ). As [ b, c ] = 1, it follows that the other end of L b n +1 lies in w ( L c n +1 ), so w ( L b n +1 ) is a subword of w ( L c n +1 ).If L c n +1 has one end in u then let L ′′ be the band which has one end at the occurrence of a ± in u ( a a ) − ζ ; see Figure 4. A similar argument shows that we can find a band L y n +1 , where y n +1 = a ± , with both ends in w ( L ′′ ), and then a band L b n +1 , where b n +1 = b ± , with exactly one end in w ( L y n +1 ) and the other end in w ( L c n +1 ), so in this case as well, w ( L b n +1 ) is a subword of w ( L c n +1 ).As only one end of L b n +1 lies in w ( L y n +1 ) it follows, as above, that a , a ∈ alph( w ( L b n +1 )). Hencewe can extend the list above, of occurrences c i and b i , to c n +1 and b n +1 .Therefore, if such a diagram exists, there is an infinite sequence of subwords w ( L c i ) of ϕ ( w ), notwo of which are equal, such that w ( L c i +1 ) is a subword of w ( L c i ). Since ϕ ( w ) is of finite length,this is a contradiction. Lemma 3.2.
The graph Γ is not an induced subgraph of the extension graph Γ e .Proof. Suppose, for a contradiction, that Γ is an induced subgraph of Γ e . Then there arevertices v a , v b , v c , v d , v e , v f of Γ e , such that the map sending a to v a , b to v b and so on, in-8Sfrag replacements L b n L b n +1 L c n + L x n + L y n +1 L c L ′′ · · · Figure 3PSfrag replacements L b n L b n + L c n + L x n + L y n +1 L c L ′′ · · · Figure 49uces an embedding of graphs. Therefore, by definition of Γ e there exist canonical generators x a , x b , x c , x d , x e , ∈ { a , a , b, c, d, e } and elements w a , w b , w c , w d , w e ∈ G (Γ ), such that v a = x w a a , v b = x w b b , v c = x w c c , v d = x w d d , v e = x w e e , where the words w − y x y w y , are reduced in G (Γ ), for y ∈ { a, b, c, d, e, f } and the graph spannedby the set { v a , v b , v c , v d , v e } in Γ e is isomorphic to the graph Γ , i.e. the following commutationrelations, and only these, hold between the v y :[ v a , v d ] , [ v a , v e ] , [ v b , v e ] , [ v c , v d ] , [ v d , v e ] . We perform a case-by-case analysis. We repeatedly use the fact that if Y is a subset of the canonicalgenerating set of a pc group G (Γ) and g ∈ h Y i then alph Γ ( g ) ⊆ Y . That is, if w is a reduced wordin G (Γ) and represents g then every letter occurring in w belongs to Y . In particular, if x is agenerator and v is a reduced word in G (Γ) then x ∈ alph Γ ( x v ).Note that the sets of vertices { v a , . . . , v e } and { v wa , . . . , v we } in Γ e , where w ∈ G (Γ ), span isomorphicgraphs, in this case both isomorphic to Γ . Hence, conjugating the vertices of the set { v a , . . . , v e } by w − d we may assume that w d = 1. Case I: x d = c . In this case, using the Centraliser Theorem, [S89, DK93], since [ a, d ] = 1, it followsthat x w a a ∈ C G (Γ ) ( x d ) = h a , c, d i , a free Abelian group, so w a = 1 and x a ∈ { a , d } .Similarly, since [ e, d ] = [ c, d ] = 1, it follows that x c , x e ∈ { a , d } and w c , w e = 1. It is clear that x a , x e and x c need to be pairwise distinct, so this is a contradiction. Case II: x d = b . This case follows from Case I, using the symmetry of the graph Γ (interchanging b and c ). Case III: x d = d . From the Centraliser Theorem again, C G (Γ ) ( d ) = h a , a , c, d, e i , which has centre h a , d i . Since [ d, e ] = 1, we have x w e e ∈ h a , a , c, d, e i , and so, as w − e x e w e is reduced in G (Γ ), theletters a and d do not belong to alph( w e ) and w e ∈ h a , c, e i . This means that w e ∈ C G (Γ ) ( x d ),so we may conjugate again, this time by w − e , and assume that w e = 1. As x d and x e must bedistinct, we must have x e ∈ { a , a , c, e } . If x e = c , then the statement follows from Case I usingthe symmetry of Γ (interchanging d and e ).Hence we may assume that x e ∈ { a , a , e } . Case III.1: x e = a . In this case x e ∈ Z ( C G (Γ ) ( d )). As above, since [ c, d ] = 1, it follows that x w c c ∈ C G (Γ ) ( d ) and so [ v e , v c ] = [ x e , x w c c ] = 1, a contradiction. Case III.2: x e ∈ { a , e } . As above, since [ c, d ] = [ a, d ] = 1, it follows that x c , x a ∈ { a , a , c, e } .If x c = a , then [ v c , v e ] = [ x w c c , x w e e ] = 1, a contradiction. It follows that x c ∈ { a , c, e } . Since[ c, a ] = 1, it follows that x a ∈ { a , c, e } . On the other hand, since [ a, e ] = 1, it follows that[ v a , v e ] = [ x w a a , x e ] = 1, so x w a a ∈ C G (Γ ) ( { a , e } ) = C G (Γ ) ( e ) = h a , a , e, d, b i . Thus x a ∈ { a , e } and w a = 1. However, this means that C G (Γ ) ( v a ) = C G (Γ ) ( { a , e } ) = C G (Γ ) ( v e ), contrary to thehypotheses on the v y ’s. 10 ase IV: x d = e . Follows from Case III using symmetry of the graph Γ (interchanging e and d ). Case V: x d = a . This follows from Case III using the symmetry of Γ (interchanging a and d ). Case VI: x d = a . Follows from Case IV using symmetry of the graph Γ .
4. A counterexample to the Weakly Chordal Conjecture
In this section we give a counterexample to Conjecture 2.
Remark 4.1.
We note that the graph Γ is weakly chordal if and only if the complement graph Γis weakly chordal.Let Γ be the graph C (the 5-cycle)and Γ the graph P (the path of length 7), as shown in Figure5. Let G ( C ) and G ( P ) be partially commutative groups with the underlying non-commutationgraphs Γ and Γ , respectively. That is G ( C ) = h a, b, c, d, e | [ a, c ] , [ a, d ] , [ b, d ] , [ b, e ] , [ c, e ] i (1)and G ( P ) = h a, b, c , c , d , d , e , e | [ a, c ] , [ a, c ] , [ a, d ] , [ a, d ] , [ a, e ] , [ b, c ] , [ b, d ] , [ b, d ] , [ b, e ] , [ b, e ] , [ c , c ] , [ c , d ] , [ c , e ] , [ c , e ] , [ c , d ] , [ c , e ] , [ c , e ] , [ d , d ] , [ d , e ] , [ d , e ] , [ e , e ] i . Remark 4.2.
Observe that any for n ≥ P n , and thus also its complement P n ,are weakly chordal. Indeed, P n is C m free, for all m, n ≥
1, and for n ≥
3, the graph P n has 2vertices of degree n − n − n −
2. Therefore, for n ≥ P n contains noinduced C m ; and so its complement P n contains no C m , m ≥
1. Hence P n is weakly chordal, forall integers n ≥
0. Furthermore, the complement of C is again C . Moreover, G ( P n ) ≤ G ( P m ),as P n ≤ P m , for all n ≤ m , and G ( C n ) ≤ G ( C ), for all n ≥
5, see [KK, Theorem 11].Define a map ϕ = a a ; b b ; c c c ; d d d ; e e e . Proposition 4.3.
The map ϕ defined above induces an embedding of G ( C ) into G ( P ) . ac c b c d ed d e e (a) Γ = C PSfrag replacements ac c bcde d d e e (b) Γ = P Figure 5The following corollary follows immediately from Remark 4.2 and Proposition 4.3.
Corollary 4.4.
For all n ≥ and m ≥ , the group G ( C n ) embeds into G ( C ) = G ( C ) , the group G ( P ) embeds into G ( P m ) and the group G ( C n ) embeds into G ( P m ) . Proposition 4.3 may also be used to show that many surface groups embed into G ( P m ), for m ≥ Corollary 4.5.
The fundamental group of a compact surface of even Euler characteristic at most − embeds into G ( C ) ≤ G ( P m ) , for all m ≥ .Proof. The compact orientable and non-orientable surfaces of Euler characteristic − G ( C ). As observed in [R07] the orientable surface of Eulercharacteristic − G ( C ). The construction of these finite coversappears in [M, Example 2.6] and a similar construction gives a finite cover of the compact non-orientable surface of Euler characteristic − − S , namely genus( S ) = (2 − χ ( S )) / g , such that g ≥
2, embed in G ( P m ), m ≥ Proof of Proposition 4.3. As ϕ maps each of the relators of the presentation (1) of G ( C ) to theidentity of G ( P ), it is immediate that ϕ is a homomorphism. We show that ϕ is injective.Let w be a reduced, non-trivial, word in G ( C ), suppose that ϕ ( w ) = 1, and assume that w is ofminimal length, among all such words. Let D be a minimal van Kampen diagram for ϕ ( w ). As inthe proof of Lemma 3.1, ϕ ( w ) must contain a letter from { c ± i , d ± i , e ± i } . Suppose first that ϕ ( w )contains an occurrence of e ± i . Then ϕ ( w ) contains e ± and, as in the proof of Lemma 3.1, withoutloss of generality we may assume there exists a band L v such that v = e and w ( L v ) = e w e − ,where w contains no occurrences of e ± i , i = 1 ,
2. As w is reduced in G (Γ ) it follows that w ( L v )must contain a letter from { a ± , d ± i } . 12n the case where w ( L v ) contains no letter d ± i , the word w contains an occurrence of a ± and, as[ a, e ] = 1, no band L a can cross L v , so a band with one end at an occurrence of a ± in w musthave both ends in w . Hence there must be a band L x such that x = a ± , w ( L x ) is a subword of w and no letter a ± occurs in w ( L x ). As w is reduced, alph( w ( L x )) must contain a letter from { b, e i } ; but w ( L x ) is a subword of w ( L v ), so this implies b ± must occur in w ( L x ). Now we finda band L y , where y = b ± , w ( L y ) is a subword of w ( L x ) and contains no occurrence of b ± . Weare forced to conclude that w ( L y ) ∈ h c , c i . As w is reduced in G (Γ ) the word w ( L y ) contains anoccurrence of c ± i and, as all such letters occur in subwords ( c c ) ± , it follows that w ( L y ) containsan occurrence of c ± . As [ b, c ] = 1, there is a band L z , where z = c ± , with both ends in w ( L y ). Inthis case w ( L y ) contains the ends c ǫ and c − ǫ of L z , and, as w ( L y ) ∈ h c , c i , w cannot be reduced,a contradictionHence we assume that w ( L v ) contains an occurrence d ± i . As [ d , e ] = 1, we may choose a band L x , such that such that x = d ± , and w ( L x ) is a subword of w ( L v ) which either contains no letter d ± i , if x = d − , or factors as w ( L x ) = d w d − , where no letter d ± i occurs in w , if x = d . Next,we may choose a band L y , such that y = c ± , w ( L y ) is a subword of w ( L x ) and contains no letter c ± i , except possibly the first and last (which may be c and c − ). Finally there must be a band L z , with z = b ± , such that w ( L z ) is a subword of w ( L y ) and there is no occurrence of b ± in w ( L z ). This forces w ( L z ) to be an element of h a i , which cannot be trivial as the ends of L z mustbe separated by at least one letter a ± , if w is reduced. However, there is then a band with ends a ǫ and a − ǫ , both lying in w ( L z ); and so w is not reduced, a contradiction.If e i / ∈ alph( w ), i = 1 ,
2, then begin above with d i instead of e i and the same argument gives theresult. The case where both e i , d i / ∈ alph( w ) follows similarly.
5. Subgroups of C and P free graphs Let us call graphs which are C and P free thin-chordal graphs . The following theorem is ageneralisation of part of the result of Droms [D87.2, Theorem]. Theorem 5.1.
The extension graph conjecture holds for finite, thin-chordal graphs, in the followingstrong form. Let Γ be a finite thin-chordal graph. Then H is a subgroup of G (Γ) if and only if H ∼ = G (Γ ′ ) , for some induced subgraph Γ ′ of Γ e .Proof. Let Γ be a thin-chordal graph. If H ∼ = G (Γ ′ ), for an induced subgraph Γ ′ of Γ e , then it followsfrom [KK, Theorem 2] that H is a subgroup of G (Γ). To prove the converse, we use induction on thenumber of vertices of Γ. If Γ is not connected then G (Γ) is a free product of partially commutativegroups of thin-chordal graphs, each of which has fewer vertices than Γ. Hence the result holds foreach factor, by induction. The result for G (Γ) follows from the Kurosh subgroup theorem.To see this, assume that the connected components of Γ are Γ , . . . , Γ n and let G i = G (Γ i ). From[KK, Lemma 26 (2)], it follows that Γ e is the disjoint union of countably many copies of Γ ei , i = 1 , . . . , n . That is Γ e = n a i =1 a j ∈ N Γ ei,j , ei,j = Γ ei , for all j ∈ N and i = 1 , . . . , n . Let D i be a set of representatives of double cosets Hx G i of H and G i in G (Γ). Then, from the Kurosh subgroup theorem, H ∼ = F ∗ ∗ ni =1 (cid:0) ∗ d ∈ D i H ∩ d G i d − (cid:1) , where F is a free group of finite or countably infinite rank.By induction, we have H ∩ d G i d − ∼ = d − Hd ∩ G i ∼ = G (Λ i,d ) , where Λ i,d ≤ Γ ei , for all d ∈ D i , i = 1 , . . . , n . Let X be a free generating set for F . Then, for each i , there exists a bijection α i : X ` D i → N . Hence we have an embedding of graphs, X a a d ∈ D i Λ i,d ≤ a j ∈ N Γ ei,j , where x ∈ X maps to a vertex of Γ i,α i ( x ) and Λ i,d is embedded in Γ ei,α ( d ) = Γ ei , for all d ∈ D i .Hence F ∗ (cid:0) ∗ d ∈ D i H ∩ d G i d − (cid:1) ∼ = G ( X ) ∗ ( ∗ d ∈ D i G (Λ i,d )) ∼ = G X a a d ∈ D i Λ i,d ! , where X a a d ∈ D i Λ i,d ≤ a j ∈ N Γ ei,j . Therefore H ∼ = G ( X ) ∗ ( ∗ ni =1 ( ∗ d ∈ D i G (Λ i,d ))) ∼ = G X a n a i =1 a d ∈ D i Λ i,d !! , where X a n a i =1 a d ∈ D i Λ i,d ! ≤ n a i =1 X a a d ∈ D i Λ i,d ! ≤ n a i =1 a j ∈ N Γ ei,j = Γ e . Suppose then that Γ is connected and that the result holds for all graphs of fewer vertices, connectedor not. Let H be a subgroup of G (Γ). Then, as shown in [D87.2, Lemma], Γ has a vertex z whichis connected to every other vertex of Γ. Let V (Γ) = Y ∪ { z } and let Λ be the induced subgraph ofΓ with vertices Y . Then Λ is a thin-chordal graph and, from [KK, Lemma 26 (1)], Γ e = { z } ∗ Λ e ,that is, the graph formed from Λ e by adjoining a new vertex z connected to every other vertex.It is shown in [D87.2] that either H ≤ G (Λ) or H ∼ = h z d i × H ′ , where H ′ ≤ G (Λ), and d is apositive integer. By induction on the number of vertices, H in the first case, or H ′ in the secondcase, is a partially commutative group isomorphic to G (Λ ′ ), for some induced subgraph Λ ′ of Λ e .As Λ e is a full subgraph of Γ e this completes the proof in the case H ≤ G (Λ). In the second case H ∼ = h z d i × Γ(Λ ′ ) ∼ = G ( { z } ∗ Λ ′ ) and { z } ∗ Λ ′ ≤ { z } ∗ Λ e = Γ e , so the result holds in this case aswell. 14 eferences [BSR07] N. Brady, H. Short, T. Riley The Geometry of the Word Problem for Finitely GeneratedGroups (Advanced Courses in Mathematics - CRM Barcelona)
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