In this paper we study Lyndon's equation xpyqzr = 1, with x, y, z group elements and p, q, r positive integers, in HNN extensions of free and fully residually free groups, and draw some conclusions about its behavior in Λ-free group

Ciobanu, L.

Fine, B.

Rosenberger, G.

RERO DOC Digital Library

J. Group Theory 14 (2011), 333–339DOI 10.1515/JGT.2010.051Journal of Group Theory( de Gruyter 2011On Lyndon’s equation in some L-free groups andHNN extensionsL. Ciobanu, B. Fine and G. Rosenberger(Communicated by A. V. Borovik)Abstract. In this paper we study Lyndon’s equation xpyqzr ¼ 1, with x, y, z group elementsand p, q, r positive integers, in HNN extensions of free and fully residually free groups, anddraw some conclusions about its behavior in L-free groups.1 IntroductionThe classical result of Lyndon and Schu¨tzenberger ([9]) states that any elements x, yand z of a free group F that satisfy the relation xpyq ¼ zr for p; q; rd 2 necessarilycommute. In the paper of Brady, Ciobanu, Martino and O Rourke ([1]) this resulthas been generalized to L-free groups. In particular, the following result has beenobtained. Let G be a group that acts freely on a L-tree, where L is an ordered abeliangroup, and let x, y, z be elements in G. If xpyq ¼ zr with integers p; q; rd 4, then x, yand z must commute. It has been unclear whether the same conclusion holds for p, q,r not all larger than 4, and in particular the proof in [1] cannot be extended to thesesmaller integer cases. Here we shed light on the behavior of this equation in someHNN extensions and show that for p, q, r not all larger than 4 the conclusion of [1]does not always hold (see Corollary 1). This work complements the results in [5],where Lyndon’s equation is studied in various amalgams of groups.2 ResultsTheorem 1. Let F be a ﬁnitely generated non-cyclic free group, and let u and v be non-trivial elements in F which are not proper powers. Let G ¼ hF ; t j tut1 ¼ vi and rd 2be a given integer. Then for particular choices of u and v there exist non-commutingelements a; b; c A G such that a2b2cr ¼ 1.Proof. The one-relator group H ¼ ha; b; c j a2b2cr ¼ 1i can also be written in termsof the presentation hb; c; d j b1d1b ¼ crdi. This can be seen by letting d ¼ ab andwriting the relation a2b2cr ¼ 1 as db1db1b2cr ¼ 1, which can then be rewritten asb1d1b ¼ crd.Thus in the HNN extension hb; c; d j b1d1b ¼ crdi of the free group generatedby fc; dg, with stable letter b and associated subgroups hdi and hcrdi, the equalitya2b2cr ¼ 1, where a ¼ db1, will be satisﬁed, but none of a, b, c will commute.We can clearly take the HNN extension of any ﬁnitely generated non-cyclic freegroup F with associated cyclic subgroups of the form hxi and hyrxi, where x andy are generators of F , and an equality of the form a2b2cr ¼ 1 will be satisﬁed withoutany of a, b, c commuting. rFor brevity we refer the reader to [2] for a complete account of L-trees and thegroups that act freely on them, called L-free groups.Corollary 1. There exist L-free groups in whicha2b2cr ¼ 1holds for non-commuting a; b; c A G, and rd 2.Proof. In [10, Theorem 3.1] it is shown that groups with a presentation of the formhx; y; x1; . . . ; xn j xyx1y e ¼ wi, where w is any word in fx1; . . . ; xng, act freely onðZ ZÞ-trees, unless e ¼ 1 and w ¼ 1. The groups arising in the proof of Theorem1, i.e. the groups hb; c; d j b1d1b ¼ crdi where rd 2, have this form and willtherefore act on a L-tree. They also satisfy the equation a2b2cr ¼ 1, where a ¼ db1,with non-commuting a, b, c. rBefore we state the next results we need to make the following observations. Let Fbe a free group with basis X . We remind the reader (see for example [8, Chapter I.4])that a Whitehead automorphism of F is an automorphism t of one of the followingtwo kinds:(1) t permutes the elements of XG1 ¼ X UX 1;(2) for some ﬁxed a A XG1, t carries each of the elements x A XG1 into one of x,a1x, xa or a1xa.We now consider the special situation with X ¼ fx1; x2; x3g and F free on X . In Fwe consider the two words(1) w1ðx1; x2; x3Þ ¼ xp1xq2xr3 with pd 2, qd 3, rd 3;(2) w2ðx1; x2; x3Þ ¼ g1x3ua1x13 g2x3ua2x13 . . . gnx3uanx13 with nd1, 10u¼ uðx1; x2Þand u is not conjugate to a power of x1 or x2, ai non-zero integers and g1; . . . ; gnfreely reduced words in hx1; x2i, with gi0 1 for i ¼ 1; . . . ; n.Remark 1. Assume that p; q; rd 3. We ﬁrst note that w10w2 and w1 is minimal(with respect to length) in its automorphic orbit. It can be easily seen that if t is aWhitehead automorphism of F of type (2), if we apply t to w1ðx1; x2; x3Þ, then thelength strictly increases. The minimality of w1 also shows that it cannot be a primitive334 L. Ciobanu, B. Fine and G. Rosenbergerelement by [8, Proposition 4.17]. In fact, w1 is a word of minimal rank, also called aregular word, that is, there is no Nielsen transformation from fx1; x2; x3g to a systemd, f , g with xp1xq2xr3 A fd; f g (see [6]).Remark 2. We now consider w2ðx1; x2; x3Þ. If w2 is minimal, then there is no White-head automorphism t such that the length strictly decreases when applying t. Hence,if w2 is minimal, there is no automorphism taking w1 to w2 by [8, Proposition 4.17],as the only automorphism taking w1 to w2 would be a permutation, and the form ofthe two words does not allow for a permutation to send w1 to w2. If w2 is not mini-mal, then each Whitehead automorphism which decreases the length of w2 will takew2 to a word of the same form. To see this, notice that u contains both x1 and x2. If,for instance, g1 ¼ g 01x2, then an automorphism which replaces x2 by x2x3 ¼ x 02 givesx 02x13 at all other places where x2 occurred, especially inside u.Remark 3. If p ¼ 2 and q; rd 3, then w1 is still minimal, and when we apply a White-head automorphism t to w1ðx1; x2; x3Þ the length strictly increases, except when t isof the form x1 ! x1x12 , x2 ! x2, x3 ! x3, in which case the length stays the same. Ifw2 is minimal, the only automorphisms that could take w1 to w2 are of the form tcomposed with permutations, and one can see that such automorphisms cannot takew1 to a word of the form w2. As in Remark 2, if w2 is not minimal, then each White-head automorphism which decreases the length of w2 will take w2 to a word of thesame form.From the minimality of w1 and the facts about w2 in the above paragraphs we getthe following.Lemma 1. Let F be free with basis fx1; x2; x3g and w ¼ w1ðx1; x2; x3Þ ¼ xp1xq2xr3 withpd 2, qd 3, rd 3. Then there is no automorphism a of F with aðxiÞ ¼ yi, i ¼ 1; 2; 3,such that a1ðwÞ is, written in y1, y2, y3, of the formg1y3ua1y13 g2 y3ua2y13 . . . gn y3uany13with nd 1, 10 u ¼ uðy1; y2Þ and u not conjugate to a power of y1 or y2, all ainon-zero integers, and g1; . . . ; gn freely reduced words in hy1; y2i, with gi0 1 fori ¼ 1; . . . ; n.Lemma 1 states that words of the form w1 and w2 cannot be in the same automor-phic orbit.Theorem 2. Let F be a ﬁnitely generated free group, u and v non-trivial elements inF that are not proper powers, and G ¼ hF ; t j tut1 ¼ vi. Let a; b; c A G satisfyapbqcr ¼ 1 with pd 2, qd 3, rd 3.(i) If u is not conjugate to v1, then a, b, c must commute.(ii) If u is conjugate to v1, then a, b, c either commute or generate the Klein bottlegroup hx; y j xyx1y ¼ 1i.On Lyndon’s equation in some L-free groups and HNN extensions 335Proof. Let H ¼ ha; b; ci. We will consider three cases:(1) u and vG1 are not conjugate;(2) u and v are conjugate;(3) u and v1 are conjugate.In case (1) assume that H is not abelian. Then H cannot be free of rank 2 or 3because the word apbqcr is regular by Remark 1. Thus, by [4, Theorem 2.2], Hmust be a one-relator group, that is, in our case, H ¼ ha; b; c j apbqcr ¼ 1i. Further-more, by the proof of [4, Theorem 2.2], there is a Nielsen transformation fromfa; b; cg to a system fx1; x2; x3g for which we may assume, without loss of generality,that x1; x2 A F , hx1; x2i non-cyclic, x3 ¼ t and H has a presentation of the formH ¼ hx1; x2; x3 jwðx1; x2; x3Þ ¼ 1i;with wðx1; x2; x3Þ ¼ g1x3ha1x13 g2x3ha2 . . . gnx3hanx13 , nd 1, h ¼ ua A hx1; x2i,a0 0, ai0 0 and gi A hx1; x2i non-trivial and freely reduced for i ¼ 1; . . . ; n. Butthis contradicts Lemma 1 if h is not conjugate to a power of x1 or x2. Therefore His abelian if h is not conjugate to a power of x1 or x2.Now suppose that h is conjugate to a power of x1 or x2. Without loss of gener-ality we may assume that h ¼ xg1. Since h ¼ ua and u is not a proper power in Fwe get that x1 ¼ ud in F . With respect to the equation apbqcr ¼ 1 and becauseH ¼ ha; b; ci we may replace x1 by u. Hence, let x1 ¼ u. We may also assume thatx2 is not a proper power in F . Now let both x1 and x2 be not a proper power in F .Using this and the cancellation arguments in [4, Theorems 1 and 2], we see that v isin hx1; x2i because H is not free of rank 2 or 3. Hence, H has a presentation of theform K ¼ hx1; x2; t j tx1t1 ¼ vi with v ¼ vðx1; x2Þ freely reduced in x1 and x2, andt ¼ x3. But in the free group on a, b, c there is no automorphism j withjðaÞ ¼ x1; jðbÞ ¼ x2 and jðcÞ ¼ t such that j1ðapbqcrÞ with 2c p, 3c q, 3c r is,written in x1, x2 and t, of the form tx1t1v1. This gives a contradiction. Hence, H isabelian in this case as well.In case (2), we can assume without loss of generality that u ¼ v. Then G is fullyresidually free, which implies that H is fully residually free. Thus H has the same uni-versal theory as that of free groups. Therefore apbqcr ¼ 1 with p; q; rd 2 implies thata, b and c commute.In addition we remark that, for the case u ¼ v, by the classiﬁcation given in[7, Theorem 5], any non-abelian, non-free rank 3 subgroup K of G is a freerank one extension of centralizers of a free group of rank 2, that is, in our case:K ¼ hx1; x2; x3 j x3hx13 ¼ hi with h A hx1; x2i; additionally, either h is regular andnot a proper power in G or h is not regular, in which case K is isomorphic tohx; y; j xy ¼ yxi ? hz ji (the free product of a free abelian group of rank 2 and theintegers).In case (3), we can assume without loss of generality that u ¼ v1, that is,G ¼ hF ; t j tut1 ¼ u1i. Since t2ut2 ¼ u, one can easily extend the arguments in[7, Theorem 5] (which only rely on the Nielsen cancellation method, and no residual336 L. Ciobanu, B. Fine and G. Rosenbergerproperties, in a group with relation u ¼ v, in order to obtain a classiﬁcation of rank3 subgroups), regarding non-abelian, non-free rank 3 subgroups of fully residuallyfree groups to the case u ¼ v1 and obtain that any non-abelian, non-free rank 3subgroup K of G has a presentation K ¼ hx1; x2; x3 j x3hx13 ¼ hei, with e ¼G1 andh A hx1; x2i; additionally, either h is regular and not a proper power in G or h is notregular, in which case K is isomorphic to hx; y j xyx1 ¼ y ei ? hz ji, where e ¼ 1 ore ¼ 1.We assume now that H ¼ ha; b; ci is not free, not abelian and of rank 3.Then H is isomorphic to a subgroup K of G as described above, that is,HGK ¼ hx1; x2; x3 j x3hx13 ¼ hei. If e ¼ 1, then H is fully residually free and soapbqcr ¼ 1 implies that a, b and c commute.Now let e ¼ 1. In case h is regular, and not a proper power, then, if H is non-freeand not abelian, by the above arguments H has to be a one-relator group, that is, inour case, H ¼ ha; b; c j apbqcr ¼ 1i since apbqcr is regular. By the same arguments asin case (1), there must be a Nielsen transformation from fa; b; cg to a systemfx1; x2; x3g for which, without loss of generality, H has a presentation of the formhx1; x2; x3 j x3hx13 h ¼ 1i. But x3hx13 h is a word of the form w2, and so by Lemma 1this cannot happen.If h is not regular, then HGK ¼ hx; y j xyx1 ¼ y1i ? hz ji. But since a, b, csatisfy apbqcr ¼ 1 and apbqcr is regular, [6, Theorem 5.2] implies that H must infact be a rank 2 subgroup, which is not the case.Finally, let H ¼ ha; b; ci be a non-abelian, non-free rank 2 subgroup. Then by [3,Theorem 1] H must be the Klein bottle groupV ¼ hx; y j xyx1 ¼ y1iG hu; v j u2 ¼ v2i:In V the elements u and v do not commute. Thus by taking, for instance, a ¼ u,b ¼ u1 and c ¼ v1, since u12u8 ¼ v4, one gets that a12b8c4 ¼ 1. rThe groups in Theorem 2 for which the translation lengths of u and v are equalare in fact L-free groups by Bass’ work (see [10, Theorem 2.4.1]). We may extendTheorem 2 to the following result, after reminding the reader that a group G iscalled n-free for a positive integer n if every subgroup of G generated by n elementsis free.Theorem 3. Let L be a non-cyclic, 2-free, fully residually free group, u and v non-trivial elements in L that are not proper powers and u is not conjugate to v1. LetG ¼ hL; t j tut1 ¼ vi. If a; b; c A G satisfy apbqcr ¼ 1 for pd 2, qd 3, rd 3 then a,b, c must commute.Proof. We ﬁrst remark that L is also 3-free by [7].If u is conjugate to v, then we may assume that u ¼ v. Then G is fully residuallyfree and hence Theorem 3 holds.From now on we assume that u is not conjugate to v. In the proof of Theorem 2we used the classiﬁcation of the rank 3 subgroups of G for the case that L is a non-On Lyndon’s equation in some L-free groups and HNN extensions 337abelian free group (see [4]). In [4], in addition to the standard Nielsen cancellationmethod in HNN groups, one only needs three properties of L and G respectively:(1) the subgroups hui and hvi are malnormal in L;(2) L is 3-free;(3) each two-generator subgroup of G is free.Now let L, as in the statement of Theorem 3, be a non-cyclic, 2-free, fully residu-ally free group. We have to show that the properties (1), (2) and (3) also hold in thismore general situation.(1) holds because L is 2-free. Let x A L be such that xuax1 ¼ ub for some integersa; b0 0. Since L is 2-free, the subgroup hx; ui of L is cyclic. Hence x A hui.(2) holds by the above remark that L is 3-free.We now show that (3) also holds. In [3], in the special case that L is a non-abelianfree group we have used, besides the Nielsen cancellation method in HNN groupsand property (1), only the fact that L is 2-free. But this we assume anyway for L.Hence (3) also holds for the more general situation. We may now apply analogousarguments to the ones in the proof of Theorem 2. rCorollary 2. Let S ¼ ha1; b1; . . . ; an; bn j ½a1; b1 . . . ½an; bn ¼ 1i, nd 2, be an orient-able surface group of genusd 2 or S ¼ ha1; . . . ; an j a21 . . . a2n ¼ 1i a non-orientablesurface group of genusd 4. Let u and v be non-trivial elements in S that are not properpowers and u is not conjugate to v1, and let G ¼ hS; t j tut1 ¼ vi. Then if fora; b; c A G and pd 2, qd 3, rd 3 the equality apbqcr ¼ 1 holds, the elements a, b, cmust commute.Proof. In both cases S is a non-cyclic, 2-free, fully residually free group. Hence Cor-ollary 2 holds by Theorem 3. rAcknowledgments. The ﬁrst-named author was supported by the Marie CurieReintegration Grant 230889 and the SNF (Switzerland) through project number200020-121506. The third-named author gratefully acknowledges the hospitality ofthe mathematics department at the University of Fribourg and the support providedby the SNF (Switzerland) through project number 200020-121506.References[1] N. Brady, L. Ciobanu, A. Martino and S. O Rourke. The equation xpyq ¼ zr and groupsthat act freely on L-trees. Trans. Amer. Math. Soc. 361 (2009), 223–236.[2] I. Chiswell. Introduction to L-trees (World Scientiﬁc, 2001).[3] B. Fine, F. Ro¨hl and G. Rosenberger. Two-generator subgroups of certain HNN groups.Contemp. Math. 109 (1990), 19–23.[4] B. Fine, F. Ro¨hl and G. Rosenberger. On HNN-groups whose three-generator subgroupsare free. In Proc. AMS Special Session (Tuscaloosa, AL, 1992), (World Scientiﬁc, 1993),pp. 13–36.338 L. Ciobanu, B. Fine and G. Rosenberger[5] B. Fine, A. Rosenberger and G. Rosenberger. Quadratic properties in group amalgams.Submitted.[6] B. Fine, G. Rosenberger and M. Stille. Nielsen transformations and applications: a sur-vey. In Groups—Korea ’94 (de Gruyter, 1995), pp. 69–105.[7] B. Fine, A. M. Gaglione, A. Myasnikov, G. Rosenberger and D. Spellman. A classiﬁca-tion of fully residually free groups of rank three or less. J. Algebra 200 (1998), 571–605.[8] R. C. Lyndon and P. E. Schupp. Combinatorial group theory (Springer-Verlag, 1977).[9] R. C. Lyndon and M. P. Schu¨tzenberger. The equation aM ¼ bNcP in a free group. Mich-igan Math. J. 9 (1962), 289–298.[10] A. Martino and S. O Rourke. Some free actions on non-Archimedean trees. J. GroupTheory, 7 (2004), 275–286.Received 20 April, 2010; revised 22 June, 2010Laura Ciobanu, University of Fribourg, Department of Mathematics, Chemin du Muse´e 23,1700 Fribourg, SwitzerlandE-mail: laura.ciobanu@unifr.chBenjamin Fine, Fairﬁeld University, Fairﬁeld, CT 06430, U.S.A.E-mail: ﬁne@mail.fairﬁeld.eduGerhard Rosenberger, Heinrich-Barth Str. 1, 20146 Hamburg, GermanyE-mail: gerhard.rosenberger@math.uni-hamburg.deOn Lyndon’s equation in some L-free groups and HNN extensions 339

On Lyndon's equation in some Λ-free groups and HNN extensions

Abstract

Similar works

Full text

Available Versions

RERO DOC Digital Library