Examples of retracts in a free group that are not the fixed subgroup of any group of automorphisms

EXAMPLES OF RETRACTS IN A FREE GROUP THAT ARE NOT THE FIXED SUBGROUP OF ANY GROUP OF AUTOMORPHISMS LAURA CIOBANU AND WARREN DICKS Abstract. Let Fbe a free group of rank at least three. We show that some retracts of Fpreviously studied by Martino-Ventura are not equal to the fixed subgroup of any group of automorphisms of F. This shows that, in F, there exist subgroups that are equal to the fixed subgroup of some set of endomorphisms but are not equal to the fixed subgroup of any set of automorphisms. Moreover, we determine the Galois monoids of these retracts, where, by the Galois monoid of a subgroup Hof F, we mean the monoid consisting of all endomorphisms of Fthat fix H. 1. Two Galois connections Throughout, let Fbe a free group and let Hbe a subgroup of F. 1.1. Notation. For x, y ∈F, ¯xdenotes x−1, [x, y] denotes xy¯x¯y, and we write x∼yif xand yare conjugate in F; thus, x∼¯yxy. Let End(F) denote the monoid of endomorphisms of F. Let EndH(F) denote the submonoid consisting of all endomorphisms of Fwhich fix every element of H. Let Aut(F) denote the group of invertible elements in End(F), that is, the group of automorphisms of F. Let AutH(F) denote the subgroup consisting of all automorphisms of Fwhich fix every element of H. For any subset Sof End(F), let Fix(S) denote the set of elements of F that are fixed by every element of S. An F-retraction is an idempotent element of End(F), and an F-retract is the image, or set of fixed elements, of an F-retraction.  We think of AutH(F) as a ‘Galois group’ and EndH(F) as a ‘Galois monoid’. 2000 Mathematics Subject Classification. 20E05, 20E36. Key words and phrases. free group, automorphism, retract, fixed subgroup. 1 2 LAURA CIOBANU AND WARREN DICKS Now Aut(−)(F) is a function from the set of subgroups of Fto the set of subsets of Aut(F), and Fix(−) is a function in the reverse direction. This pair of functions is a Galois connection, and the images of the functions are the sets of closed subsets. Thus His Aut(F)-closed (in F) if H= Fix(S) for some subset Sof Aut(F), and the Aut(F)-closure of H(in F) is Fix(AutH(F)), the smallest Aut(F)-closed subgroup of Fcontaining H. Replacing Aut with End everywhere in the previous paragraph gives another Galois connection. Notice that all F-retracts are End(F)-closed. The End(F)-closure of H, Fix(EndH(F)) is a subgroup of the Aut(F)-clo- sure of H, Fix(AutH(F)). In general, the relation between the two closures is not well understood. Enric Ventura [4, Theorem 3.9] showed that if the rank of Fis at most two, then Fix(EndH(F)) = Fix(AutH(F)). We now show that if the rank of Fis greater than two, then there exist examples where Fix(EndH(F)) 6= Fix(AutH(F)). 1.2. Hypotheses. Let Fbe a free group of rank three, and let {a, b, c}be a basis of F. Here, we denote an element αof End(F) by the triple (aα, bα, cα). Notice that we write endomorphisms on the right of their arguments.  1.3. Example. Suppose that Hypotheses 1.2 hold. Let d=ba[c, b]¯a∈Fand let H=ha, di=ha, ba[c, b]i ≤ F. Let ψ= (a, d, 1) ∈End(F) and let φ= (a, b, cb)∈Aut(F). Notice that, for each n∈Z,φn= (a, b, cbn) and φnψ= (a, d, dn). Using normal-form arguments, we find, in Corollary 3.3 below, that EndH(F) = {φn, φnψ|n∈Z}=hφi∪hφiψand, hence, AutH(F) = hφi. It is then straightforward to verify the following. Every non-invertible element of EndH(F) is an F-retraction with image H. The multiplication in EndH(F) is described by the monoid presentation EndH(F) = hφ, ¯ φ, ψ |ψφ =ψ2=ψ, ¯ φφ =φ¯ φ= 1imonoid. Let K=ha, b, cb¯ci. The two closures of Hare Fix(AutH(F)) = Fix({φ}) = Kand Fix(EndH(F)) = Fix({ψ}) = H. Here, K=H∗ hbi 6=H. The closure of {ψ}in End(F) is EndFix({ψ})(F) = hφi ∪ hφiψ. There is another type of behaviour we want to mention. 1.4. Example. Suppose that Hypotheses 1.2 hold. Let d=ba[c2, b]¯a∈Fand let H=ha, di=ha, ba[c2, b]i ≤ F. EXAMPLES OF NON-AUTOMORPHISM-CLOSED RETRACTS IN A FREE GROUP 3 Let ψ= (a, d, 1) ∈End(F) and let φ= (a, b, cb)∈Aut(F). Notice that, for each n∈Z,φnψ= (a, d, dn). In Corollary 3.4 below, we find that EndH(F) = {1, φnψ|n∈Z}={1} ∪ hφiψand, hence, AutH(F) = {1}. It is then straightforward to verify the following. Every non-identity element of EndH(F) is an F-retraction with image H. The multiplication in EndH(F) is described by the monoid presentation EndH(F) = h{φnψ|n∈Z} | {φnψ·φmψ=φnψ|m, n ∈Z}imonoid. The two closures of Hare Fix(AutH(F)) = Fix({1}) = Fand Fix(EndH(F)) = Fix({ψ}) = H. Here, F6=H. The closure of {ψ}in End(F) is EndFix({ψ})(F) = {1} ∪ hφiψ. By adjoining a free-group free factor simultaneously to Hand to F, one obtains examples where Fhas arbitrary rank greater than two. 2. Relation with work of Martino-Ventura 2.1. Definitions. We say that His one-auto fixed (in F) if H= Fix(S) for some one-element subset Sof Aut(F). Thus a one-auto-fixed subgroup is Aut(F)-closed. We define one-endo fixed analogously. Thus an F-retract is one-endo fixed, and a one-endo-fixed subgroup is End(F)-closed.  Ventura [4, Theorem 3.9] showed that, if the rank of Fis at most two, then the four concepts, one-endo fixed, End(F)-closed, one-auto fixed, and Aut(F)-closed, all coincide. Now suppose that the rank of Fis finite and greater than two. It is not known whether or not the Aut(F)-closed subgroups are just the one-auto-fixed subgroups, nor whether or not the End(F)-closed subgroups are just the one-endo-fixed subgroups. In [2, Corollary 3.4], A. Martino and E. Ventura showed that if His Aut(F)-closed then His a free factor of some one-auto-fixed subgroup K of F. By [2, Proposition 5.4], there exists an example where His not Aut(F)-closed but His a free factor of some one-auto-fixed subgroup Kof F. We find it interesting that the latter phenomenon appears unbidden in Example 1.3, above. Later, Martino-Ventura [3] gave a family of ingeniously chosen F-retracts none of which is one-auto fixed, thus showing that not all one-endo-fixed subgroups are one-auto fixed. Explicitly, they show the following. 4 LAURA CIOBANU AND WARREN DICKS Suppose that Hypotheses 1.2 hold, let i,j,kbe integers, and let Hi,j,k =ha, baicjbck¯ bi. Then [3, Proposition 18] shows that Hi,j,k is an F-retract, but that Hi,j,k is one-auto fixed if and only if ijk = 0. The six-tuple (i, j, k, a, b, c) in our notation corresponds to (−s, r, t, b, cbs, a) in their notation. What we do in this article is to describe elementary techniques that give satisfactorily complete information about H1,1,−1and H1,2,−2. It is straightforward to check that the same techniques apply to Hi,j,−j, for ij 6= 0, and to verify that no new types of behaviour arise. For k6=−jand ijk 6= 0, we have been unable to calculate the Aut(F)- closure of Hi,j,k. 3. Some Galois monoids Our argument is concentrated in the next result. 3.1. Lemma. Suppose that Hypotheses 1.2hold. For all d∈ hciand all x∈F, the following implications hold. (i) If ¯ d¯abdx ∼¯a¯xbx in F, then x∈ {1, b ¯ d¯ bad¯a}. (ii) If x∼¯a¯xba[d, b]xin F, then x∈ {b, ba[d, b]¯a}. Proof. (i). Suppose that (1) ¯ d¯abdx ∼¯a¯xbx. Let B={a, b, c, ¯a,¯ b, ¯c}. We view the elements of Fas reduced monoid words in B. Write x=bmyanwhere m,n∈Z,y∈F,ydoes not begin with bor ¯ b, and ydoes not end with aor ¯a. By abelianizing (1) we see that the derived subgroup of F, denoted F′, contains x=bmyan; thus (2) yF′= ¯an¯ bmF′ in F/F′, a free abelian group with basis {aF′, bF′, cF′}. Also, notice that (1) can be rewritten as (3) ¯ d¯abdbmyan∼¯a¯yby, and that the right-hand side of (3) is now cyclically reduced. Let |y|adenote the number of times aoccurs in the expression of yas a reduced monoid word in B, and similarly for the other five elements of B. On applying |−|ato (3) we see that (4) 0 + |y|a+ max{n, 0} ≥ 0 + |y|¯a+ 0 + |y|a, EXAMPLES OF NON-AUTOMORPHISM-CLOSED RETRACTS IN A FREE GROUP 5 and, moreover, equality holds in (4) if and only if no ais cancelled in the cyclic reduction of the left-hand side of (3). Now (4) amounts to max{n, 0} ≥ |y|¯a, and, by (2), it is clear that equality holds. Thus |y|¯a= max{n, 0}, and no ais cancelled in the cyclic reduction of the left-hand side of (3). Similarly, by applying |−|¯ato (3), we find that |y|a= max{−n, 0}. Also, by applying |−|band |−|¯ bto (3), we find that |y|¯ b= max{m, 0}, |y|b= max{−m, 0}, and no bis cancelled in the cyclic reduction of the left-hand side of (3). We now consider five non-pairwise-disjoint cases. Case 1: m=n= 0. Here |y|a=|y|¯a=|y|b=|y|¯ b= 0. Thus, each side of (3) has no occurrence of aor ¯ b, has a unique occurrence of ¯a, and has a unique occurrence of b. In each side, we can equate the cyclic subword between ¯aand band we find that 1 = ¯y. Hence x= 1, as desired. Case 2: m≤ −1. Here |y|b=−mand |y|¯ b= 0. Write y=y′by′′ as a reduced monoid word such that y′has no occurrence of b, and, hence, y′′ has −m−1 occurrences of b. Notice that y′6= 1 since ydoes not begin with b. In each side of (3) there are −m+ 1 occurrences of bfollowed, cyclically, by −moccurrences of ¯ b. In each side, we take the first of the −m+ 1 occurrences of bas the terminating point. For the left-hand side we get y′′an¯ d¯abdbmy′b, that is, y′′an¯ d¯abd¯ b−my′b. For the right-hand side we get y¯a¯yb, that is, y¯a¯y′′¯ b¯y′b. Since there is no cancellation of bor ¯ b, we can equate the cyclic subwords between the last ¯ band the first b, and we find that y′= ¯y′. Hence y′= 1, which is a contradiction, as desired. Case 3: m≥1. Here |y|¯ b=mand |y|b= 0. In each side of (3), there are m+1 occurrences of bfollowed, cyclically, by moccurrences of ¯ b. In each side, we take the last of the m+ 1 occurrences of bas the terminating point, and find that yan¯ d¯abdbm=y¯a¯yb. Rearranging, we find that y=¯ bm−1¯ d¯ bad¯an+1. Since ydoes not begin with bor ¯ b, and does not end with aor ¯a, we see that m= 1, n=−1, y=¯ d¯ bad, and x=b¯ d¯ bad¯a, as desired. Case 4: n≤ −1. This is similar to Case 3. Here |y|a=−nand |y|¯a= 0. In each side of (3) there are −n+1 occurrences of ¯afollowed, cyclically, by −noccurrences of a. In each side, we take the first of the −n+ 1 occurrences of ¯aas the starting 6 LAURA CIOBANU AND WARREN DICKS point, and find that ¯a−n¯ d¯abdbmy= ¯a¯yby. Thus y=¯ bm−1¯ d¯ bad¯an+1, and, as in Case 3, x=b¯ d¯ bad¯a, as desired. Case 5: n≥1. This is similar to Case 2. Here |y|a= 0 and |y|¯a=n. Write y=y′¯ay′′ as a reduced monoid word such that y′′ has no occurrence of ¯a, and, hence, y′ has n−1 occurrences of ¯a. Notice that y′′ 6= 1 since ydoes not end with ¯a. In each side of (3) there are n+ 1 occurrences of ¯afollowed, cyclically, by noccurrences of a. In each side, we take the last of the n+ 1 occurrences of ¯aas the starting point, and find ¯ay′′an¯ d¯abdbmy′= ¯a¯yby = ¯a¯y′′a¯y′by. Equating the cyclic subwords between the last ¯aand the first a, we find that y′′ = ¯y′′. Hence y′′ = 1, which is a contradiction, as desired. This completes the proof of (i). (ii). Suppose that x∼¯a¯xba[d, b]x. Let y:= ¯ bx. Then x=by and by ∼¯a¯yadb ¯ dy. Applying φ:= (a, ¯ d¯abd, c)∈Aut(F) and letting z:= yφ, we find that ¯ d¯abdz ∼¯a¯zbz. By (i), z∈ {1, b ¯ d¯ bad¯a}. Applying φ−1= (a, adb ¯ d, c), we see that y=zφ−1∈ {1, a[d, b]¯a}. Left multiplying by bwe find that x=by ∈ {b, ba[d, b]¯a}, as desired.  We can now calculate some Galois monoids. 3.2. Theorem. Suppose that Hypotheses 1.2hold. Let jbe a positive integer, let d=ba[cj, b]¯a, and let H=ha, di=ha, ba[cj, b]i.For all (x, y, z)∈ EndH(F), the following hold. (i) y∈ {d, b}. (ii) If y=dthen there exists an integer nsuch that z=dn. (iii) If y=band j= 1 then there exists an integer nsuch that z=cbn. (iv) If y=band j≥2then z=c. Proof. (i). Here x=aand yx[zj, y] = ba[cj, b]. Hence yazjy¯zj¯y=ba[cj, b] and zjy¯zj= ¯a¯yba[cj, b]y. Thus y∼¯a¯yba[cj, b]y. By Lemma 3.1(ii), y∈ {b, ba[cj, b]¯a}={b, d}. This proves (i). (ii). Let y=d. Let Cdenote the centralizer of zjin F. Now zjd¯zj= ¯a¯ dba[cj, b]d= ¯a(a[cj, b]¯a¯ b)ba[cj, b]d=d. Hence d∈C. EXAMPLES OF NON-AUTOMORPHISM-CLOSED RETRACTS IN A FREE GROUP 7 To show that z∈ hdi, we may assume that z6= 1. Recall that C=hz′ifor some z′∈F; see [1, Proposition I.2.19]. (A trivial normal-form argument shows that there exists some z′∈Fsuch that z′is not a proper power and zj=z′nfor some positive integer n. An almost-as-trivial normal-form argument shows that C=hz′i. Alternatively, Cis a free (sub)group with a non-trivial center, and hence cyclic.) Thus d=z′mfor some integer m. Since bF′=dF′=z′mF′= (z′F′)min F/F′, a free abelian group with basis {aF′, bF′, cF ′}, we see that m=±1. Thus d=z′±1. Now z∈C=hz′i=hdi. This proves (ii). (iii) and (iv). Let y=b. Then zjb¯zj= ¯a¯ bba[cj, b]b=cjb¯cj. Hence ¯cjzjcommutes with b. A trivial normal-form argument shows that there exists n∈Zsuch that ¯cjzj=bn. If j= 1, then z=cbn. This proves (iii). If j≥2, then the equation cjbn=zjclearly implies that n= 0 and z=c. This proves (iv).  The case of Theorem 3.2 where j= 1 gives the key part of Example 1.3. 3.3. Corollary. Suppose that Hypotheses 1.2hold, let d=ba[c, b]¯a, and let H=ha, di=ha, ba[c, b]i. If (x, y, z)∈EndH(F), then there exists n∈Z such that either (x, y, z) = (a, b, cbn)or (x, y, z) = (a, d, dn). Conversely, all these endomorphisms fix aand d, and, hence, fix H. The case of Theorem 3.2 where j= 2 gives the key part of Example 1.4. 3.4. Corollary. Suppose that Hypotheses 1.2hold, let d=ba[c2, b]¯a, and let H=ha, di=ha, ba[c2, b]i. If (x, y, z)∈EndH(F), then either (x, y, z) = (a, b, c)or there exists n∈Zsuch that (x, y, z) = (a, d, dn). Conversely, all these endomorphisms fix aand d, and, hence, fix H. Acknowledgments The first-named author was supported by Marie Curie Intra-European Fellowship 515027 while a guest at the Centre de Recerca Matem`atica. She is grateful for the hospitality she received there. The research of the second-named author was funded by the DGI (Spain) through Project BFM2003-06613. References 1. Roger C. Lyndon and Paul E. Schupp, Combinatorial Group Theory, Ergeb. Math. Grenzgeb. 89, Springer-Verlag, Berlin, 1977. 2. A. Martino and E. Ventura, On automorphism-fixed subgroups of a free group, J. Algebra 230(2000), 596–607. 3. A. Martino and E. Ventura, Examples of retracts in free groups that are not the fixed subgroup of any automorphism, J. Algebra 269(2003), 735–747. 8 LAURA CIOBANU AND WARREN DICKS 4. Enric Ventura, On fixed subgroups of maximal rank, Comm. in Algebra 25(1997), 3361–3375. Laura Ciobanu, Department of Mathematics, University of Auckland, Private Bag 92019, Auckland, New Zealand E-mail addresses:[email protected], LCioban[email protected] Warren Dicks, Departament de Matem` atiques, Universitat Aut` onoma de Barcelona, E-08193 Bellaterra (Barcelona), Spain E-mail address:[email protected] URL:http://mat.uab.es/∼dicks/