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Every Group Is an Outer Automorphism Group of a Finitely Generated Group

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Every Group Is an Outer Automorphism Group of a Finitely Generated Group View metadata, citation and similar papers at core.ac.uk brought to you by CORE provided by Elsevier - Publisher Connector Journal of Pure and Applied Algebra 200 (2005) 137–147 www.elsevier.com/locate/jpaa Every group is an outer automorphism group of a finitely generated groupଁ Inna Bumagina,b,∗, Daniel T. Wisea aDepartment of Mathematics, McGill University, Montreal, Que., Canada H3A 2K6 bSchool of Mathematics and Statistics, Carleton University, Ottawa, Ont., Canada K1S 5B6 Received 3 October 2002; received in revised form20 December2004 Communicated by J. Rhodes Available online 25 February 2005 Abstract We show that every countable group Q is isomorphic to Out(N) where N is a finitely generated C( 1 ) subgroup of a countable 6 small-cancellation group G. Furthermore, when Q is finitely presented, we can choose G to be finitely presented and residually finite. © 2005 Elsevier B.V. All rights reserved. MSC: 20F28; 20F06; 20F67; 20E26; 20E22 1. Introduction While many groups arise as automorphisms of algebraic objects, it is not true that every group Q arises as Aut(N) for some group N. Indeed, it is well-known that a nontrivial cyclic group of odd order is not isomorphic to Aut(N) for any group N. The situation for Out(N) is markedly different and the question of whether a given group Q can be realized as an outer automorphism group of some group N has attracted some recent attention. In [11] it was shown that any group Q is isomorphic to Out(N) where N is the fundamental group of a graph of groups but the cardinality of the generating set of N is greater than the ଁ Research supported by NSERC grant. ∗ Corresponding author. School of Mathematics and Statistics, Carleton University, Ottawa, Ont., Canada K1S 5B6. Fax: +1 613 520 3536. E-mail addresses: [email protected] (I. Bumagin), [email protected] (D.T. Wise). 0022-4049/$ - see front matter © 2005 Elsevier B.V. All rights reserved. doi:10.1016/j.jpaa.2004.12.033 138 I. Bumagin, D.T. Wise / Journal of Pure and Applied Algebra 200 (2005) 137–147 cardinality of the generating set of Q. The problem becomes more complicated if N is required to belong to a special class of groups. Any Q is isomorphic to Out(N) where N is simple [2], or where N is torsion-free metabelian with trivial center [4]. Any countable Q is isomorphic to Out(N) where N is a locally finite p-group [3]. Finally, when Q is finite, N can be chosen to be the fundamental group of a closed hyperbolic 3-manifold [8]. The first main result in this paper is Theorem 11, where we show that any countable Q is (N) C( 1 ) isomorphic to Out , where N is a finitely generated subgroup of a finitely generated 6 group G. The second main result is Theorem 15 where we show that if Q is finitely presented, N can be chosen to be a finitely generated subgroup of a finitely presented residually finite C( 1 ) 6 group. We note that in analogy with Kojima’s result, when Q is finite, the group N C( 1 ) produced by our construction is the fundamental group of a finite 6 complex. The main idea is to use a variant of Rips’s construction [12] in order to obtain a short → N → G → Q → C( 1 ) exact sequence 1 1, where G is a 6 group and N is finitely generated. As observed in [17], the natural homomorphism Q → Out(N) is easily shown to be injective in this case. The main difficulty is to choose G so that there are no “surprise” outer automorphisms of N, and so Q → Out(N) is surjective. A criterion for establishing this surjectivity is given in Lemma 9, which is the central underlying technical result in this paper. We close this introduction by discussing the extent to which our results can be generalized. Since there are uncountably many finitely generated groups Q, it is not true that every such Q arises as Out(N) where N is a subgroup of a finitely presented group. Nevertheless we pose the following problemthat would generalize our two mainresults: Problem 1. Is every countable group Q isomorphic to Out(N) where N is finitely generated and residually finite? 2. Review of small-cancellation theory A piece in a presentation a1,a2,... | R1,R2,... is a word U such that US1 = US2 ±1 ±1 and both (US1) and (US2) are cyclic permutations of relators. The presentation satisfies C() if for any piece U which is a subword of the cyclic word URj ,wehave|U| < |Rj | where we use |X| to denote the length of the word X. We will also say that the set of words {R1,R2,...} in the free group a1,a2,... |−satisfies C () if the presentation above satisfies C(). We say that G is a C() group if G has a presentation that satisfies the C() condition. We will need the following well-known facts about the metric small cancellation groups whose presentations satisfy a C() condition. The first theoremis the key result in small cancellation theory. Proposition 2 (Lyndon and Schupp [10, Chapter V,Theorem 4.4]). If W is a freely reduced C( 1 ) word which represents the identity element in a 6 group, then W contains a subword V such that VS is a conjugate of some relator or its inverse, and S is the concatenation of at most three pieces. I. Bumagin, D.T. Wise / Journal of Pure and Applied Algebra 200 (2005) 137–147 139 C( 1 ) A proof of the following theoremfor 8 groups, based on an idea of Lipschutz [9] C( 1 ) was given by Lyndon and Schupp [10, Chapter V, Theorem10.1] .For 6 groups it was proved by Greendlinger [5]. C( 1 ) Proposition 3. Every torsion element in a 6 small-cancellation group is a power of a conjugate of an element W, where W n is a relator. C( 1 ) C( 1 ) Greendlinger proved for 8 groups [6], and later for 6 groups [7], that in these groups, any two elements that commute necessarily belong to a cyclic subgroup. Truffault [15] and Seymour [13] independently, strengthened the result of Greendlinger and showed C( 1 ) that in 6 groups, centralizers of non-trivial elements are cyclic. From this latter result we observe that C( 1 ) G. Corollary 4. LetGbea 6 group, and let N be a non-cyclic subgroup of Then the centralizer of N in G is trivial. Proof. Let N =h1,h2,.... Then CG(N) ⊂ CG(h1) ∩ CG(h2) ∩···Let d ∈ CG(h1) ∩ CG(h2) ∩···hence hi ∈ CG(d) for all i.Ifd = 1, then CG(d) is cyclic so that H is cyclic, a contradiction. Thus, CG(N) ={1}. 2.1. Construction of special small-cancellation words Definition 5. An x,y word is special if it is a positive word in x,y with no x2 or y3 subwords. It is easy to provide sets of special words satisfying stringent small cancellation con- ditions. The following lemma provides us with examples of such sets; we will use these examples in our proof. Lemma 6. Consider the infinite word W = xy(xy2)xy(xy2)2xy(xy2)3 ··· (1) for each infinite sequence {n1,n2,...} of natural numbers, one can find an infinite family W ={w ,w ,...} C( 1 ) |w |n ∞ 1 2 of words which satisfies the 20 condition, so that i i for each i. (2) for each positive integer m and a large enough integer p one can find a family Vm = {v ,v ,...,v } C( 1 ) 1 2 m of words which satisfies the 20 condition, so that given a subset I of {1, 2,...,m}, the lengths of the elements of Vm are as follows: |vi|=p for each i ∈ I and |vi|=p + 1 for each i/∈ I. Proof. To prove the first assertion, let 2 ki 2 ki +1 2 2ki wi = xy(xy ) xy(xy ) ···xy(xy ) , (1) k { , 2 n } k { k , 2 n } W . where 1 max 60 3 1 and i max 2 i−1 3 i be the words in ∞ Obviously, w x,y |w |= 9 k2 + k + >n , i is a special word of length i 2 i 5 i 1 i and the length of the maximal 140 I. Bumagin, D.T. Wise / Journal of Pure and Applied Algebra 200 (2005) 137–147 k − k piece pi = y(xy2)2 i 1xy(xy2)2 i xy of wi equals |pi|=12ki + 2. Hence W∞ satisfies the C( 1 ) 20 condition, and we have proven the first assertion. To prove the second assertion, consider the finite subset Wm ={w2,w3,...,wm+1}⊂W∞ chosen according to part (1) of this lemma (we set ni = 1 for all i). We modify the words wi ∈ Wm as follows. Let p>|wm+1| be an integer number. A “correcting” exponent for i =2, 3,...,m+1, is the maximal integer ci which does not exceed (p −|wi|)/(2(ki +1)). Define (i) (i) (i) ci 2 ki ci 2 ki +1 ci 2 2ki 2 vi = (xy) (xy ) (xy) (xy ) ···(xy) (xy ) x 1 (xy) 2 (xy ) 3 , (i) (i) ∈{, } (i) |v |=p. where 1 , 3 0 1 and 2 is a nonnegative integer, so that i Note that the (i) (i) (i) length of the correcting term x 1 (xy) 2 (xy2) 3 does not exceed 2(ki + 1). V ={v ,v ,...,v } C( 1 ) We claimthat m 1 2 m satisfies the 20 condition.
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