We exhibit free-by-cyclic groups containing non-free locally-free subgroups, including some word hyperbolic examples. We also show that these groups are not subgroup separable. We use Bestvina-Brady Morse theory in our arguments

Leary, I.J.

Niblo, G.A.

Wise, D.T.

English

Southampton (e-Prints Soton)

Some free-by-cyclic groupsIan J. Leary ∗ Graham A. Niblo Daniel T. Wise †January 8, 2008A group is said to be locally free if every finitely generated subgroup of it isfree. One example is the additive group of the rationals. We exhibit a finitelygenerated group G that is free-by-cyclic and contains a non-free, locally freesubgroup. The smallest such example that we have found is of the formG ∼= FnoZ for n = 3. We also construct word-hyperbolic examples for largervalues of n, and show that the groups are not subgroup separable. We usedBestvina and Brady’s ‘Morse theory for cube complexes’ in the constructionof these groups. The authors thank Jim Anderson, who posed a questionconcerning 3-manifolds that led to these examples. This work was started ata conference at Southampton, immediately before Groups St. Andrews, whichwas funded by EPSRC visitor grants GR/L06928 and GR/L31135, and by agrant from the LMS.Throughout this note, Fn denotes a free group of rank n, x¯ denotes x−1,and xy = y¯xy.Proposition 1. The group G given by the presentationG =〈a, b, t : at = b, bt = aba¯〉contains a non-free, locally free subgroup, and is isomorphic to a split exten-sion F3 o Z.Proof The given presentation expresses G as an ascending HNN extension,with base group freely generated by a and b and with stable letter t. Defineφ:G → Z by φ(t) = 1, φ(a) = φ(b) = 0, and let K be the kernel of φ. ThenK is a strictly ascending union of 2-generator free groups〈a, b〉 ⊆ 〈a, b〉t¯ ⊆ 〈a, b〉t¯2 ⊆ · · · ⊆ K.Any such group is locally free and not free, since it is not finitely generatedand the rank of its abelianization is at most two. (In fact, the abelianizationof K is infinite cyclic.)∗Partially supported by EPSRC grant no. GR/L69398†Supported by NSF grant no. DMS-96275061It remains to show that G is free-by-cyclic. Define ψ:G → Z by ψ(t) =ψ(a) = ψ(b) = 1. It will be shown that the kernel of ψ is free of rank 3. Apresentation 2- complex Y for G may be constructed by attaching two 2-cellsto a rose with edges a, b, and t according to the maps given in figure 1. SinceY is obtained from the presentation of G as an HNN extension with free basegroup, it follows that Y is an Eilenberg-Mac Lane space for G. Represent thethree 1-cells of Y as unit intervals, and represent the two 2-cells of Y as aunit square in and a 2× 1 rectangle, as indicated in figure 1. This makes Yinto an affine cell complex in the sense of Bestvina and Brady ([1], Def. 2.1).Figure 1.Now take S1 = R/Z, viewed as a cell complex with one vertex and oneedge of length 1, as an Eilenberg-Mac Lane space for the integers. A cellularmap g:Y → S1 may be defined that induces the homomorphism ψ:G → Zon fundamental groups and is affine on each cell—in figure 1 this map isrepresented by ‘height’. The inverse image of the vertex of S1 is a roseconsisting of one vertex and three 1-cells (the dotted lines on figure 1). Nowlet X be the cover of Y corresponding to the subgroup H = ker(ψ). The mapg lifts to a map f :X → R. X is an affine cell complex, and f is a Morsefunction in the sense of [1], Def. 2.2. By construction, X is an Eilenberg-Mac Lane space for H, and for any integer t, Xt = f−1(t) consists of adisjoint union of copies of a 3- petalled rose.Bestvina and Brady’s Morse theory allows one to compare X and Xt: bylemma 2.5 of [1], a space homotopy equivalent to X may be obtained fromXt by coning off a subspace homeomorphic to a copy of the descending link(resp. ascending link) at v for each vertex v of X such that f(v) > t (resp.f(v) < t). All vertices of X have isomorphic links, since Y has only onevertex, and the ascending and descending links at each vertex are as shownin figure 2.2The descending link The ascending linkFigure 2.Both the ascending and descending link are contractible. Since coning offa contractible subspace does not change the homotopy type of a space, itfollows that X is homotopy equivalent to Xt. But it is already known thatX is an Eilenberg-Mac Lane space for H, and that Xt is a disjoint union of3-petalled roses. It follows that Xt is connected, and that H is free of rankthree.With the benefit of hindsight, a shorter proof that G as above is free-by-cyclic may be given—see Proposition 2 below. Such a proof gives noindication as to how G was discovered however. Moreover, the techniquesof Proposition 1 generalize easily to more complicated presentations such asthose given in Proposition 3.Proposition 2. Let H ′ be freely generated by x, y and z, and define anautomorphism θ of H ′ byθ(x) = y, θ(y) = z, θ(z) = y2x¯.The group G of Proposition 1 is isomorphic to H ′o〈t〉, where the conjugationaction of t on H ′ is given by θ.Proof First, check that the endomorphism θ is an automorphism of H ′ byexhibiting an inverse:θ−1(z) = y, θ−1(y) = x, θ−1(x) = z¯x2.Now, eliminate b from the given presentation for G to obtainG = 〈a, t : t¯t¯attat¯a¯ta¯〉.Substitute a = xt, and eliminate a, obtainingG = 〈x, t : t¯t¯(xt)tt(xt)t¯(t¯x¯)t(t¯x¯)〉 = 〈x, t : t¯2xt3xt¯x¯2〉.Add new generators y = xt and z = xt2, obtainingG = 〈x, y, z, t :xt = y, yt = z, ztxy¯2t¯〉= 〈x, y, z, t :xt = y, yt = z, zt = y2x¯〉.Thus G is seen to be isomorphic to H ′ o 〈t〉 as claimed.Next, we show how to construct a word-hyperbolic group having similarproperties to the group G.3Proposition 3. For s ≥ 3 define a word W (x, y) byW (x, y) = xy4xy5x · · ·xy4+sx.The group Gs with presentationGs = 〈a, b, t : at = b, bt = W (b, a)b(W (a, b))−1〉is free-by-cyclic and contains a non-free, locally free subgroup. For s suffi-ciently large, Gs is word-hyperbolic.Proof As in Proposition 1, Gs is a strictly ascending HNN-extension withbase group freely generated by a and b, so contains a non-free, locally freesubgroup. As in Proposition 1, an Eilenberg-Mac Lane space for Gs with anaffine cell structure can be made by attaching a unit square and an m × 1rectangle to a rose with three petals of length 1. (Here m is one more thanthe length of the word W , as shown in figure 3.) The argument given inProposition 1 shows that Gs is expressible as Fn o Z, where n is the totalarea of the two 2-cells in figure 3, i.e., n = m+ 1 = s+ 4 + (8 + s)(s+ 1)/2.Figure 3.It remains to show that Gs is word-hyperbolic for s sufficiently large. Forthis, it suffices to show that some presentation for Gs satisfies the C′(1/7)small cancellation condition (see [3]). Eliminate b from the presentation forGs. This leaves a 1-relator group, with relator4btab4ab5a · · · ab4+sab¯2a¯4+sb¯ · · · b¯a¯5b¯a¯4b¯t¯= t¯at2a(t¯a4ta)(t¯a5ta) · · · (t¯a4+sta)t¯a¯2(ta¯4+st¯a¯)(ta¯3+st¯a¯) · · · (ta¯5t¯a¯)(ta¯4t¯a¯).The total length of this relator as a word in a and t may be seen to be8 + (14 + s)(s+ 1). (The bracketing of the word is intended to facilitate thischeck.) For any 4 ≤ r ≤ 4+s, the word t¯art occurs exactly once as a subwordof either the relator or its inverse (but not both). Any subword of the relatorof length at least 2s + 15 contains a subword of the form (t¯art)±1 for somesuch r. (The worst case is the subword a4+stat¯a¯2ta¯4+s, of length 2s + 14.)Hence any subword of the relator or its inverse of length 2s + 15 occursin a unique place. It follows that Gs is word-hyperbolic whenever 2s + 15 ≤1/7 (8 + (14 + s)(s+ 1)). This inequality is satisfied for all sufficiently large s.(In fact, s ≥ 9 suffices.)P. Scott asked if free-by-cyclic groups are necessarily subgroup separable.An example due to Burns, Karass and Solitar showed that this is not the case(see [2]). The groups constructed above give another, simpler, argument toshow this.Proposition 4. The groups G and Gs, constructed in Propositions 1 and 3,are not subgroup separable.Proof In each case, let L1 be the subgroup generated by a and b, and letL2 = tL1t¯. Then L1 and L2 are free of rank two and are conjugate in G (resp.in Gs). Moreover, L1 is a proper subgroup of L2. It follows that L1 cannotbe closed, since it cannot be separated from any element of L2 \ L1: in anyfinite quotient, the images of L1 and L2 have the same order, since they areconjugate, and so must be equal since the image of L1 is a subgroup of theimage of L2.5References.[1] M. Bestvina and N. Brady, Morse theory and finiteness properties ofgroups, to appear in Inventiones Math.[2] R. G. Burns, A. Karrass,and D. Solitar, A note on groups with separablefinitely generated subgroups. Bull. Austral. Math. Soc. 36 (1987), no. 1, 153–160.[3] R. Strebel, Small cancellation groups, Appendix to: Sur les groupes hy-perboliques d’apre´s Mikhael Gromov, 227–273, Progr. Math., 83, Birkha¨userBoston, Boston, MA, 1990.Authors’ addresses:Faculty of Mathematical Studies,University of Southampton,Southampton,SO17 1BJ.Department of Mathematics,Cornell University,Ithaca, NY148536

Some free-by-cyclic groups

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