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Key words. regular rooted tree, automorphisms of trees, finite automata, ... the fact that finite automata have either polynomial or exponential growth lead us to.
Geometriae Dedicata 108: 193–204, 2004. Ó 2004 Kluwer Academic Publishers. Printed in the Netherlands.

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Finite Automata of Polynomial Growth do Not Generate A Free Group SAID SIDKI Departamento de Matema´tica, Universidade de Brası´lia, 70910-900 Brası´lia DF, Brazil. e-mail: [email protected] (Received: 21 September 2003; accepted in final form: 2 March 2004) Abstract. A free subgroup of rank 2 of the automorphism group of a regular rooted tree of finite degree cannot be generated by finite-state automorphisms having polynomial growth. This result is in fact proven for rooted trees of infinite degree under some natural additional conditions. Mathematics Subject Classifications (2000). Primary 20F05, 20F10; secondary 20H05, 68Q68. Key words. regular rooted tree, automorphisms of trees, finite automata, polynomial growth, free groups.

1. Introduction The known noncyclic free groups generated by finite automata – for example, those constructed in [1, 2] – have generators of exponential growth. This observation and the fact that finite automata have either polynomial or exponential growth lead us to conjecture in [3] that free groups of rank 2 cannot be generated by finite automata of polynomial growth. Our purpose here is to prove this conjecture more generally for automata with finite number of states and possibly infinite alphabet, under natural additional hypotheses. We treat the question in the language of automorphisms of rooted regular trees. This is justified by the fact that a so-called synchronous automaton with input and output alphabet Y can be interpreted as a level preserving endomorphism of a rooted regular tree T generated by Y ; the cardinality of Y is the degree of the tree T . The interpretation is also true in the other direction, where the states of a level preserving endomorphism a of T is a set QðaÞ of level preserving endomorphisms and where a is the initial state. Automorphisms a of T , which have a finite number of states constitute a group denoted by F . To a finite-state automorphism a, we associate the directed graph DðaÞ corresponding to the automaton which represents this automorphism. The graph has the finite set QðaÞ for vertices. If the identity automorphism 1 is a state of a, we remove it from DðaÞ, together with the edges incident with it, to produce the subgraph DðaÞ0 . An a 2 F is said to be strongly finite-state provided DðaÞ0 has a finite number of directed edges. The set of strongly finite-state automorphisms form a group which we denote by F 0 . The growth function of a counts the number of

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directed paths in the graph DðaÞ0 starting at the initial state a and having length n. Finally, we define the level closure of a group H of automorphisms of the tree to be the group generated by the permutations induced by H on the levels of the tree. The general result is: THEOREM 1. Let T be a regular rooted tree of arbitrary degree and H a strongly finite-state group of automorphisms of polynomial growth of T. If the level closure of H does not contain free groups of rank 2, then neither does H. It is clear that if the generating set Y of T is finite, or more generally, if the permutation groups induced by H on the levels of the tree are locally solvable-byfinite, then the level closure of H does not contain free groups and therefore the theorem applies to these situations. The proof of the theorem depends ultimately on the existence of an n-syllable length function defined on the group PolðnÞ of strongly finite-state automorphisms having polynomial growth of degree at most n.

2. Preliminaries 2.1.

TREES AND THEIR AUTOMORPHISMS

Let Y be a nonempty set and #Y be its cardinality. The regular rooted tree T ¼ T ðY Þ is identified with the free monoid M ¼ [iP0 Y i generated by Y , with the order relation vOu if and only if u is a prefix of v. For any u 2 M, the subset uM corresponds to the subtree Tu headed by u, which is isomorphic to the tree T via uw ! w. Let A denote the group of automorphisms of T and stabA ðkÞ be the point-wise stabilizer of the set Y k in A. The permutation group SymðY Þ embeds in A by r : yu ! y r u for all r 2 SymðY Þ; y 2 Y and u 2 M. An automorphism a 2 stabA ð1Þ induces an automorphism ay on each of the subtrees Ty , y 2 Y . Using the isomorphism between T and Ty we identify ay with an element from A and thus identify a 2 stabA ð1Þ with f ðaÞ 2 F ðY ; AÞ, the set of functions from Y into A; thus, ay ¼ f ðaÞy . From the above, an a 2 A has a first level development as a ¼ f ðaÞrðaÞ where f ðaÞ 2 F ðY ; AÞ and rðaÞ 2 SymðY Þ. Generally, for level kP1, the automorphism has the corresponding development a ¼ f ðkÞ ðaÞrðk1Þ ðaÞ where f ðkÞ ðaÞ 2 F ðY k ; AÞ; rðk1Þ ðaÞ 2 hF ðY i ; SymðY ÞÞj0 O i O k  1i; for i ¼ 0, F ðY 0 ; SymðY ÞÞ ¼ SymðY Þ. We write r/ ðaÞ for rðaÞ and then from ay ¼ f ðay Þry ðaÞ we define ayz ¼ f ðay Þz for all y; z 2 M. In this manner, we produce for all u 2 M, au 2 A and ru ðaÞ 2 SymðY Þ. If a 2 F ðY ; AÞ, then SupðaÞ denotes the support set fi 2 Y jai 6¼ 1g and pi ðaÞ denotes the value of f ð1Þ ðaÞ at i 2 Y . For u 2 M of length kP1 and a 2 A, we let u  a denote the function in F ðY k ; AÞ whose value at u is a and is the identity 1 for the

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other indices of length k. Define for every level kP1 the set F 0 ðY k ; AÞ ¼ fa 2 F ðY k ; AÞjSupðaÞ is finiteg. Moreover, define A0 ¼ fa 2 Ajf ðkÞ ðaÞ 2 F 0 ðY k ; AÞ for all kP1g; F

0

¼ F \ A0 :

The following result has a straightforward proof. LEMMA 1. The sets A0 ; F

0

A0 ¼ F 0 ðY ; A0 ÞSymðY Þ;

are subgroups of A and they decompose as F 0 ¼ F 0 ðY ; F 0 ÞSymðY Þ:

Let H be a subgroup of A. It is clear that H is strongly finite-state if and only if H OF 0 . For each level k of the tree, let rk ðH Þ denote the permutation group induced by H on Y k . As rk ðH Þ is a subgroup of A, the group rðH Þ ¼ hrk ðH ÞjkP0i is welldefined and is the level closure of H . We say that H is an R-group if and only if H OF 0 and rðH Þ does not contain free groups of rank 2. We will make frequent use of PROPOSITION 1. Let J be a subdirect product of J1  J2 . If J is a free group of rank 2 then either J1 or J2 is a free group of rank 2. Also, if J1 or J2 contains a free subgroup of rank 2, then so does J. Proof. Let a ¼ ða1 ; a2 Þ, b ¼ ðb1 ; b2 Þ be the generators of J . Then, J1 ¼ ha1 ; b1 i and J2 ¼ ha2 ; b2 i. (i) Suppose both groups J1 , J2 are not free. Then there exist nontrivial reduced words u ¼ uða; bÞ, v ¼ vða; bÞ such that uða; bÞ ¼ ð1; uða2 ; b2 ÞÞ, vða; bÞ ¼ ðvða1 ; b1 Þ; 1Þ and uða2 ; b2 Þ 6¼ 1 6¼ vða1 ; b1 Þ. Now as uða; bÞ, vða; bÞ are nontrivial commuting elements of the free group J , there exist w ¼ wða; bÞ 2 J and r; s 2 N such that u ¼ w r , v ¼ w s and therefore, wða1 ; b1 Þr ¼ 1, wða2 ; b2 Þs ¼ 1; thus, wða; bÞrs ¼ 1 and wða; bÞ ¼ 1 follows, which is a contradiction. (ii) If c ¼ ðc1 ; c2 Þ, d ¼ ðd1 ; d2 Þ 2 J1  J2 are such that hc1 ; d1 i is a free group of rank 2, then clearly, hc; di is a free group of rank 2. ( The following inheritance properties hold for R-groups. LEMMA 2. Let H be an R-group. Then subgroups of H and the projections pi ðstabH ð1ÞÞ are R-groups for all i 2 Y. Proof. If K is a subgroup of H , then KOF 0 and rðKÞOrðH Þ; therefore, K is an R -group. Now let K ¼ stabH ð1Þ. Clearly, pi ðKÞOF 0 for all i 2 Y . Suppose rðpj ðKÞÞ contains a free group L of rank 2. There exists a finitely generated subgroup M of pj ðKÞ such that LOrðMÞ. Likewise, there exists a finitely generated subgroup U of K such that MOpj ðU Þ. As SupðU Þ is finite, U is a subdirect product of a finite product of nontrivial pi ðU Þs. Therefore, rðU Þ is a subdirect product of a finite product of nontrivial rðpi ðU ÞÞs and consequently we can apply the previous proposition to reach a contradiction. (

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2.2.

SAID SIDKI

AUTOMATA AND THEIR GROWTH

The automaton associated to a has the set Y for input and output alphabet, the set QðaÞ ¼ fau ju 2 Mg for set of states and a/ ¼ a for initial state. For any y 2 Y , the state transition function is au ! auz where z is the image of y under ru ðaÞ and the output function is ðau ; yÞ ! z. The directed graph DðaÞ corresponding to this automaton has for vertices the set of states QðaÞ and for labeled directed edges au ! y=z auz . We will suppress the qualification that the directed edges are labeled. A directed path is a sequence of directed edges: au ! y=z auz followed by auz !z=t auzt , etc. The length of a directed path in this graph is the number of directed edges on the path, taking into account the multiplicity of occurrences of these edges. A circuit in the graph is a finite directed path with equal initial and end vertices. An automorphism a is finite-state provided QðaÞ is a finite set; we use this term even though Y may be infinite. Since QðabÞ  QðaÞQðbÞ and Qða1 Þ ¼ fa1 u jau 2 QðaÞg, the set F of finite-state automorphisms is a subgroup of A. If a 2 F 0 then the graph DðaÞ has a finite number of vertices and also finite number of edges, except possibly for those incident with the identity state 1. Thus, the subgraph D 0 ðaÞ obtained from DðaÞ by removing 1 and the edges incident with it, has a finite number of vertices and edges. The adjacency matrix ½a of a is defined by ½acd ¼ #fy 2 Y jd ¼ cy g for all c; d 2 QðaÞnf1g; in ordering the states QðaÞnf1g, we let a be the first element. The geometric significance of the growth of finite automata derives from the following well-known facts which can be found in [4]. PROPOSITION 2. Let D be a finite directed graph and v be a vertex of D from which all other vertices can be reached by directed paths of D. Then the growth of directed paths in D starting at v is either polynomial or exponential. The growth is exponential if and only if there are two distinct minimal directed circuits with a common vertex. The growth is polynomial of degree n provided it is not exponential, the number of distinct minimal circuits on any directed path is at most n þ 1 and there is at least one path with this number of circuits. Given a 2 F 0 we define its growth function hðaÞ: N ! ZP0 as that of D0 ðaÞ. The growth function hðaÞ is determined by hða; kÞ ¼ #Supðf ðkÞ ðaÞÞ which is also the first entry of ½ak e where e is the transpose of the vector ð1; 1; . . . ; 1Þ. Denote the level closure of F 0 by Polð1Þ. If a has polynomial growth of degree n, we write degðaÞ ¼ n. Note that degðaÞ ¼ 1 if and only if it is an element of Polð1Þ. The set of finite-state automorphisms of polynomial growth at most nP0 is PolðnÞ ¼ fa 2 F 0 jhða; kÞOck n ; for some c > 0 and for all kg; and Pol ¼ [nP1 PolðnÞ. If a 2 Pol, then degðaÞ ¼ nP0, provided a 2 PolðnÞn Polðn  1Þ. To illustrate the above notions, let T2 be the regular rooted tree generated by Y ¼ f0; 1g and consider the automorphisms: the extension r of the permutation

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ð0; 1Þ, r1 ¼ ð1; rÞ, s ¼ ð1; sÞr, c ¼ ðc; sÞ, d ¼ ðd; dÞr. The adjacency matrices for the directed subgraphs D0 ðaÞ associated to these automorphisms are:     1 1 0 1 ; ½d ¼ ð2Þ: ; ½s ¼ ð1Þ; ½c ¼ ½r ¼ ð0Þ; ½r1  ¼ 0 1 0 0 All these automorphisms except d have polynomial growth: degðrÞ ¼ degðr1 Þ ¼ 1;

degðsÞ ¼ 0;

degðcÞ ¼ 1:

PROPOSITION 3. For all nP  1, the set PolðnÞ is a group and PolðnÞ ¼ F 0 ðY ; PolðnÞÞSymðY Þ: Proof. Let nP0 and a, b be automorphisms such that hða; kÞOc1 k n , hðb; kÞOc2 k n ðk1Þ

for some c1 ; c2 > 0 and all k. Since f ðkÞ ða1 Þ ¼ ðf ðkÞ ðaÞ1 Þr ðaÞ and f ðkÞ ðabÞ ¼ ðk1Þ 1 f ðkÞ ðaÞf ðkÞ ðbÞr ða Þ , it follows that hða1 ; kÞ ¼ hða; kÞOc1 k n and hðab; kÞO hð f ; kÞ þ hðg; kÞOðc1 þ c2 Þk n ; thus PolðnÞ is a group. The second assertion follows simply from: if a 2 PolðnÞ and y 2 Y , then y  a 2 PolðnÞ. ( Remark 1. The group PolðnÞ is an infinitely iterated wreath product and therefore does not satisfy any nontrivial group law. Let a 2 PolðnÞ, nP0. Then a is said to be a pure n-nerve provided it lies on a circuit in its own graph. It is an n-nerve provided its states which are pure n-nerves lie on the same circuit. If a is an n-nerve, then there is a unique index i 2 Y such that ai is an nnerve, whereas aj has degree less than n for all j 2 Y , j 6¼ i. Given a 2 PolðnÞ, nP0, there is a first level k of the tree such that each entry of f ðkÞ ðaÞ is either a pure n-nerve or has degree less than n. This fact allows us to introduce a syllable length function on PolðnÞ, which plays a crucial role in our analysis. Suppose that for this a and k, the pure n-nerve entries of f ðkÞ ðaÞ are d1 ; d2 ; . . . ; ds and they occur respectively at indices u1 ; u2 ; . . . ; us . Then, a ¼ b1 b2 . . . bs h where bj ¼ uj  dj ð1 O i O sÞ are n-nerves and h 2 Pol ðn  1Þ. This s is a canonical number associated to a, in the following sense: if a ¼ cq1 c0 q2 . . . c00 qt c000 where q1 ; q2 ; . . . ; qt are n-nerves and c; c0 ; . . . ; c00 ; c000 2 Polðn  1Þ, then sOt. This is so, because the qj ’s may be assumed to be in stabF ðkÞ and each qj contributes a unique n-nerve to the entries of f ðkÞ ðaÞ. We call s the n-syllable length of a and we write it as P jajn ¼ s. Note that i2Y jai jn Ojajn . Since jajn ¼ jf ðaÞjn and jy  ajn ¼ jajn , it follows P that jajn O i2Y jai jn and thus we obtain the formula jajn ¼

X

jai jn :

i2Y

At times, when the context is clear, we will drop the index from the summation notation. Also, if we are concerned with elements of PolðnÞ for a fixed n, we will write jaj instead of jajn ; in this case, jaj ¼ 0 means that a 2 Polðn  1Þ or Polð1Þ.

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To illustrate these notions, consider the automorphism a ¼ ðc; lÞr of the binary tree where c ¼ ð1; sÞ; l ¼ ðl; sÞ; s ¼ ð1; sÞr. Then, c is a 0-nerve, s is a pure 0-nerve and l is a pure 1-nerve. Going down to level k ¼ 2, we produce f ð2Þ ðaÞ ¼ ð1; s; l; sÞ and therefore, a ¼ bh where b ¼ ð1; 1; l; 1Þ is a 1-nerve, h ¼ ð1; s; 1; sÞr 2 Polð0Þ. We have, jaj1 ¼ jbj1 ¼ 1 and jhj0 ¼ 2.

3. Proof of the Main Result The proof consists of several steps which reduce the type of action of the free group on successive levels of the tree. This leads in the end to essentially the case where T is the binary tree. (1) Let H be an R-subgroup of PolðnÞ and suppose by contradiction that H contains a free subgroup of rank 2. Let a ¼ f ðaÞrðaÞ, b ¼ gðbÞrðbÞ 2 H generate a free group G of rank 2. Then, G is an R-group. Let G be minimal, in the sense that maxfdegðaÞ; degðbÞg ¼ n is minimal. If degðaÞ ¼ m; degðbÞ ¼ n then minimize m. Furthermore, for these choices of m; n, minimize jajm , jbjn . If m ¼ n then let jajn Ojbjn . LEMMA 3. The group PolðG; 1Þ ¼ G \ Polð1Þ is trivial. Proof. Since Polð1Þ does not contain free groups of rank 2, it follows that PolðG; 1Þ is cyclic, say generated by c. Suppose c 6¼ 1 and let k be an element in Gnhci. Then L ¼ hc; ki is a free R-group of rank 2. There is a minimal finite level k of the tree such that c ¼ rðk1Þ ðcÞ. Write k ¼ f ðkÞ ðkÞrðk1Þ ðkÞ. Then, as hrðk1Þ ðcÞ; rðk1Þ ðkÞi is not a free group, there exists a nontrivial reduced word w in two variables such that wðrðk1Þ ðcÞ; rðk1Þ ðkÞÞ ¼ 1. Note that v ¼ wðc; kÞ is a nontrivial element of stabG ðkÞ with finite support. Choose a nontrivial element l of stabL ðkÞ with minimum s ¼ #SupðlÞ. If SupðlÞ is not invariant under the action of c then #Supð½l; lc Þ < s and therefore, ½l; lc  ¼ 1, which is absurd. Therefore, c leaves SupðlÞ invariant and induces on it a permutation of finite order t. As ½l; ct  ¼ 1, we have a contradiction. ( (2) Suppose Y is partitioned as Y1 [ Y2 where Y1 , Y2 are subsets of Y , invariant under the action of hrðaÞ; rðbÞi. Then, f ðaÞ ¼ ðaY1 ; aY2 Þ; rðaÞ ¼ rY1 ðaÞrY2 ðaÞ, where aY1 and rY1 ðaÞ are the restrictions of f ðaÞ and rðaÞ to Y1 ; aY2 and rY2 ðaÞ are defined similarly. Also write f ðbÞ ¼ ðbY1 ; bY2 Þ; rðbÞ ¼ rY1 ðbÞrY2 ðbÞ. Define the following elements of PolðnÞ: a0 ¼ ðaY1 ; 1ÞrY1 ðaÞ;

a00 ¼ ð1; aY2 ÞrY2 ðaÞ;

b0 ¼ ðbY1 ; 1ÞrY1 ðbÞ;

b00 ¼ ð1; bY2 ÞrY2 ðbÞ:

Then, a ¼ a0 a00 ; b ¼ b0 b00 . Let GY1 ¼ ha0 ; b0 i; GY2 ¼ ha00 ; b00 i be called the Yi -factors of G, i ¼ 1; 2, respectively. Then, GY1 ; GY2 are subgroups of PolðnÞ. Moreover, ½GY1 ; GY2  ¼ 1; GY1 \ GY2 ¼ 1 and G is a subdirect sum of GY1  GY2 . It follows from

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Proposition 1 that one of GY1 , GY2 is a free group. Moreover, by using Lemma 2, we conclude that GY1 and GY2 are R-groups. Now, by the minimality of the syllable lengths of a, b, one and only one of GY1 , GY2 is free (say GY1 is free); thus, ja0 jm ¼ jajm ; jb0 jn ¼ jbjn , ja00 jm ¼ 0; jb00 jn ¼ 0. (3) We consider below a certain extreme situation where G fixes an an infinite path in the tree. LEMMA 4. Suppose G fixes an infinite path C in the tree T. Then there exists u 2 C such that hau ; bu i is not a free group. Proof. Suppose by contradiction that hau ; bu i is a free group of rank 2 for all u 2 C. Then, jau jm ¼ jajm ; jbu jn ¼ jbjn for all u 2 C and jaw jm ¼ 0; jbw jn ¼ 0 for all w 2j C. Now, since a is finite-state, there exist u; v 2 C such that v < u and av ¼ au ; clearly, avi ¼ aui for all i 2 Y . Also, there exists a unique i 2 Y such that ui 2 C and jaui jm ¼ jajm . As avi ¼ aui , we have vi 2 C. The same considerations apply to b; thus, there exist u0 ; v0 2 C such that v0 < u0 and bv0 ¼ bu0 . It is easy to show that there exists a pair u00 ; v00 2 C such that v00 < u00 and av00 ¼ au00 ; bv00 ¼ bu00 . We may suppose a ¼ au ; b ¼ bu and, in order to simplify the notation, let u ¼ jj . . . j ¼ jh for some j 2 Y ; h 2 N. There exists a nontrivial reduced word w such that wða00 ; b00 Þ ¼ 1 and therefore, wða; bÞ ¼ ðwðaj ; bj Þ; 1Þ 6¼ 1. Thus, ha; wða; bÞi is a free group of rank 2, wða; bÞu ¼ wða; bÞ and jwða; bÞv jn ¼ 0 for all v 2j C. We may replace wða; bÞ by ½w; a ¼ ð½wðaj ; bj Þ; aj ; 1Þ in the previous group, where again ½w; au ¼ ½w; a, j½w; av jn ¼ 0 for all v 2j C. After repeating this reduction a finite number of times, we produce elements c, d such that hc; di is a free group of rank 2, with cu ¼ c and c fixes all indices of the tree outside the subtree headed by u. Hence, c ¼ 1 and we have a contradiction. ( (4) We continue the analysis in (2). It follows from the previous lemma that there exists an orbit Y1 of hrðaÞ; rðbÞi such that GY1 is free. Suppose #Y1 ¼ k > 1. LEMMA 5. If c 2 stabGY1 ð1Þ, c 6¼ 1 then SupðcÞ ¼ Y1 and as c 2 F 0 , k is finite. Proof. Suppose SupðcÞ is a proper subset of Y1 and choose c to minimize #SupðcÞ. Since hrðaÞ; rðbÞi is transitive on Y1 and k > 1, there exists d 2 GY1 such that Supðcd Þ ¼ SupðcÞrðdÞ 6¼ SupðcÞ. Therefore, Supð½c; cd Þ is a proper subset of SupðcÞ and consequently, ½c; cd  ¼ 1. The argument continues as in the proof of Lemma 3, reaching c ¼ 1, which is a contradiction. ( LEMMA 6. The permutations rY1 ðaÞ, rY1 ðbÞ are k-cycles. Proof. Let j 2 Y1 and Oa ðjÞ, Ob ðjÞ be the orbits of j under rY1 ðaÞ, rY1 ðbÞ, respectively. Furthermore, let #Oa ðjÞ ¼ a, #Ob ðjÞ ¼ b; then a or b is greater than 1. Suppose that a < k and let a_ ¼ ða0 Þa , b_ ¼ ðb0 Þb . We claim that ha_ j ; b_ j i is free of rank

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2. For otherwise, there exists c 2 GY1 such that c 6¼ 1 and cj ¼ 1; but then, we conclude that there exists r 2 N such that cr fixes j, ðcr Þj ¼ 1 and since oðcÞ is infinite, cr 6¼ 1, which contradicts the previous lemma. Since a_ j is a product of the automorphisms ai such that i 2 Oa ðjÞ, it follows from the minimality of G that X X fjai jm ji 2 Y g ¼ jajm Ojða00 Þj jm ¼ fjai jm ji 2 Oa ðjÞg and so, jai jm ¼ 0 for all i 2j Oa0 ðjÞ. But, as we have assumed that Y1 partitions into more than one orbit under rY1 ðaÞ, the same argument applies to the other orbits and thus jai jm ¼ 0 for all i 2 Y1 . We conclude that jajm ¼ 0, contradicting degðaÞ ¼ m. ( (5) To summarize, we have: there exists a unique orbit Y1 of hrðaÞ; rðbÞi such that the Y1 -factor GY1 is free of rank 2 and if #Y1 ¼ k > 1 then rY1 ðaÞ; rY1 ðbÞ are cycles of equal length k. LEMMA 7. If #Y1 ¼ k > 1 then k ¼ 2. Proof. Suppose by contradiction that k > 2. To simplify the notation, we rewrite G ¼ GY1 , Y1 ¼ f1; 2; . . . ; kg and suppress the n from the syllable length symbol. As hak ; bk i is a free subgroup of stabG ð1Þ, then the projections on its coordinates are also free groups. Rewrite rðaÞ ¼ p;

rðbÞ ¼ q; ðaÞ ¼ ða1 ; . . . ; a1 k Þ;

f ðbÞ ¼ ðb1 ; . . . ; bk Þ:

We will use the following formulas, 1 1 a1 ¼ ða1 1p ; . . . ; akp Þp ; 1 ðaj Þi ¼ ai aip1 . . . aipðj1Þ ; ðaj Þi ¼ a1 ip aip 2    aip j ;

ðbaj Þi ¼ bi aiq1 aiq1 p1    aiq1 pðj1Þ ; 1 1 ðbaj Þi ¼ bi a1 iq1 p aiq1 p 2    aiq1 pj

for 1O i; j O k  1. (i) The p and q are k-cycles where k ¼ #Y1 . Therefore, given i 2 Y1 , there exists a unique jðiÞ 2 Y1 such that 1O jðiÞ O k=2 and qpjðiÞ or qpjðiÞ fixes i; write j ¼ jðiÞ. We have pi hak ; bajðiÞ i ¼ ha1 a1p1    a1pðk1Þ ; bi aiq1 aiq1 p1    aiq1 pðj1Þ i is a free group of rank 2 and ja1 a1p1    a1pðk1Þ j ¼ ja1 j þ ja2 j þ    þ jak j ¼ jaj; therefore the minimality of the pair jaj; jbj implies jbjOjbi aiq1 aiq1 p1    aiq1 pðj1Þ j; jb1 j þ jb2 j þ    þ jbk jOjbi j þ jaiq1 j þ jaiq1 p1 j þ    þ jaiq1 pðj1Þ j    ðÞ:

FINITE AUTOMATA OF POLYNOMIAL GROWTH

201

Let s; t 2 f1; 2; . . . ; kg be such that jas j ¼ min1OiOk jai j, jbt j ¼ min1OiOk jbi j. Furthermore, let 1 O j O k  1 be such that baj fixes t. If jbt j < jas j then, inequality ðÞ implies jbj < ja1 j þ ja2 j þ    þ jak j ¼ jaj; a contradiction. On considering the pair ak , we obtain jas j ¼ jbt j; jaj ¼ jbj: Now, if jbt j 6¼ 0 then ðÞ implies that j ¼ k  1. But as was already observed we can choose jOk=2, by interchanging a with a1 , if necessary. We reach k=2 ¼ k  1, k ¼ 2 which is a contradiction and so, jbt j ¼ 0. (ii) Set U ¼ fi 2 Y1 jjai j 6¼ 0g; U 0 ¼ fi 2 Y1 jjai j ¼ 0g; V ¼ fi 2 Y1 jjbi j 6¼ 0g; V 0 ¼ fij 2 Y1 jbi j ¼ 0g and u ¼ #U ; v ¼ #V , u0 ¼ #U 0 , v0 ¼ #V 0 . Then we conclude from jbj ¼ jajOjatq1 j þ jatq1 p1 j þ    þ jatq1 pð j1Þ j that jPk  u0 . Since j O k=2 we arrive at u0 P k=2 if k is even and u0 Pðk þ 1Þ=2 if k is odd. (iii) Let c be a permutation of f1; 2; . . . ; kg. Then, ac ¼ ða1c ; a2c ; . . . ; akc Þpc ;

bc ¼ ðb1c ; b2c ; . . . ; bkc Þqc ;

Gc is a free R-group and jac j ¼ jaj, jbc j ¼ jbj. Moreover, let t be such that jbt j ¼ 0 and j such that 1 O j O k  1, qp j fixes t and choose c to fix t. Then,   ðÞ ðbaj Þc ¼ bt atq1 c atq1 p1 c . . . atq1 pðj1Þ c . . . t

We will choose c such that ftq1 c; tq1 p1 c; . . . ; tq1 pðj1Þ cg \ U is minimal. Suppose k is odd. Then, U 0 \ V 0 is nonempty; choose t 2 U 0 \ V 0 . Then uOðk  1Þ=2 and we can choose c to fix t and send U into U 0  ftg. Then all the factors of the right-hand side of ðÞ have n-syllable length zero; so, degðððbaj Þc Þt Þ < n which is a contradiction. Suppose k is even, k > 2. If u0 or v0 is greater than k=2 then the argument of the previous paragraph applies, where we may need to interchange the roles of a and b. Therefore, u0 ¼ v0 ¼ k=2. If t 2 U \ V 0 then we can choose a c which maps U  ftg into U 0 and then reduction follows. Thus, we may assume U \ V 0 is empty; that is, U ¼ V . Choose i0 2 U and choose c to fix i0 and map U  fi0 g into U 0  ftg. In this case, u ¼ 1 and k ¼ 2 and again we have a contradiction. ( (6) We are at the following stage in the proof: there exists a unique orbit Y1 of hrðaÞ; rðbÞi such that the Y1 -factor GY1 is a free R-group of rank 2 and furthermore, if #Y1 ¼ k > 1 then k ¼ 2. Suppose #Y1 ¼ k ¼ 2. Then, rY1 ðaÞ ¼ rY1 ðbÞ ¼ r, a cycle of length 2. We rewrite GY1 ¼ G; Y1 ¼ f0; 1g, r ¼ ð0; 1Þ, a ¼ ða0 ; a1 Þr, b ¼ ðb0 ; b1 Þr. (i) We will prove

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LEMMA 8. ja0 jn ¼ ja1 jn ¼ jb0 jn ¼ jb1 jn . Proof. We already have jajn ¼ jbjn . The group stabG ð1Þ ¼ ha2 ; ab; bai and its images under the projections p0 , p1 on the first and second coordinates produce free R-groups of rank 3. Any two different elements in the following list a2 ¼ ða0 a1 ; a1 a0 Þ;

b2 ¼ ðb0 b1 ; b1 b0 Þ;

ab ¼ ða0 b1 ; a1 b0 Þ;

ba ¼ ðb0 a1 ; b1 a0 Þ;

1 ab1 ¼ ða0 b1 0 ; a1 b1 Þ;

1 a1 b ¼ ða1 1 b1 ; a0 b0 Þ

generate a free R-group of rank 2. We consider the projections of these groups on their first coordinates. From p0 : ha2 ; bai ! ha0 a1 ; b0 a1 i, we obtain jbjn ¼ jb0 jn þ jb1 jn Ojb0 a1 jn Ojb0 jn þ ja1 jn and thus, jb1 jn Oja1 jn . From p0 : ha2 ; abi ! ha0 a1 ; a0 b1 i, we obtain jbjn ¼ jb0 jn þ jb1 jn Oja0 b1 jn Oja0 jn þ jb1 jn and thus, jb0 jn Oja0 jn : From p0 : ha2 ; ab1 i ! ha0 a1 ; a0 b1 0 i, we obtain jbjn ¼ jb0 jn þ jb1 jn Oja0 b1 0 jn Oja0 jn þ jb0 jn

and thus, jb1 jn Oja0 jn :

It follows that jb1 jn Oja1 jn , jb0 jn Oja0 jn . By symmetry, we reach jb1 jn ¼ ja1 jn , jb0 jn ¼ ja0 jn and so, ja1 jn Ojb0 jn . Therefore, ja1 jn Oja0 jn . On exchanging a with a1 and b with b1 we arrive at ja0 jn ¼ ja1 jn ¼ jb0 jn ¼ jb1 jn :

(

1 (ii) Let LðcÞ ¼ ha0 a1 ; ci where c 2 C ¼ fb0 b1 ; a0 b1 ; b0 a1 ; a0 b1 0 ; a1 b1 g. Then, LðcÞ is a free R-group of rank 2 and ja0 a1 jn ¼ jcjn ¼ jajn ¼ jbjn . The set Y is a union of orbits under the action of L ¼ LðcÞ. There is a unique orbit Z 0 ¼ ZðcÞ of L such that LZ 0 is free; also, #Z 0 O 2 and rZ 0 ða0 a1 Þ ¼ rZ 0 ðcÞ. If Z 0 ¼ fsg then jða0 a1 Þs j ¼ jajn and jða0 a1 Þi j ¼ 0 for all i 6¼ s. If #Z 0 ¼ 2 and Z 0 ¼ fs; tg then jða0 a1 Þs j ¼ jða0 a1 Þt j, jajn ¼ jða0 a1 Þs j þ jða0 a1 Þt j and jða0 a1 Þi j ¼ 0 for all i 6¼ s; t. Hence, ZðcÞ ¼ Z 0 is the same set for all c 2 C.

LEMMA 9. #Z0 ¼ 1 and a; b : os $ 1sa0 . Proof. Let Z 0 ¼ fs; tg. Then, t ¼ sa 0 a 1 ¼ sc ; a0

s ¼t

a1 1

¼t

b1 1

s ¼ ta0 a1 ¼ tc ; b0

¼t ;

a0

t ¼s

for all c 2 C; a1 1

1

¼ s b1 ¼ s b0 :

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Therefore, ðosÞa ¼ 1sa0 ;

2

ðosÞa ¼ 0sa0 a1 ¼ 0t;

3

ðosÞa ¼ 1ta0 ;

4

ðosÞa ¼ 0ta0 a1 ¼ 0s;

a2 : os $ 0t; 1sa0 $ 0ta0 ; ðosÞb ¼ 1sb0 ¼ 1ta0 ;

2

ðosÞb ¼ 0ta0 b1 ¼ 0s;

ðotÞb ¼ 1tb0 ¼ 1sa0 ;

2

ðotÞb ¼ 0sa0 b1 ¼ 0t; b2 : os $ os; ot $ 0t: Therefore, a0 a1 : s $ t. Yet, as b0 b1 : s $ s; t $ t, we conclude that s ¼ t. The assertion a; b: os $ 1sa0 follows. ( (iii) Generalizing the previous item, stabG ð1Þ fixes a finite path in the tree B ¼ f/; 0; 0s; . . . ; 0s . . . tug such that for any x 2 B we have a; b : x $ xa ¼ x0 , stabG ð1Þ projects at x onto hax ax0 ; ax bx0 ; bx ax0 i and jax j ¼ jax0 j ¼ jbx j ¼ jbx0 j ¼ 12 jaj. 1 Furthermore, any two elements from fax ax0 ; bx bx0 ; ax bx0 ; bx ax0 ; ax b1 x ; ax0 bx0 g generate a free group. Inductively, we produce an infinite path C which is fixed by ha0 a1 ; b0 b1 i and such that hax ax0 ; bx bx0 i is a free group of rank 2 for all x 2 C. Now, applying Lemma 4 gives us a contradiction which finishes the proof of the theorem.

4. Concluding Remarks Let Tn be the regular rooted tree generated by Y ¼ f1; 2; . . . ; ng where nP2 and let r be the permutation ð1; 2; . . . ; nÞ. The automorphisms s ¼ ð1; . . . ; 1; sÞr; l ¼ ð1; . . . ; 1; l1 Þr of Tn are called adding machines. The group Kn ¼ hs; li is a subgroup of Polð0Þ and therefore does not contain free subgroups of rank 2. However, Kn has involved structure, for it is torsion-free and just-nonsolvable (see, [5]); the group K2 was first studied in [6]. Beside such notable groups, Pol contains also the just-infinite torsion groups defined by R. Grigorchuk on T2 and by N. Gupta and the author on Tp for any odd prime p; see, [7], [8]. Thus, in spite of our main result, a description of the isomorphism types of subgroups of Pol, analogous to that which J. Tits gave for linear groups, will be a complicated task, especially since the set of subgroups of Pol is closed under certain generalized wreath products (see, [9, 10]).

Acknowledgement The author gratefully acknowledges support from the Swiss National Science Foundation and the Brazilian Conselho Nacional de Pesquisa.

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3. Sidki, S.: Automorphisms of one-rooted trees: growth, circuit structure and acyclicity, J. Math. Sci. 100 (2000), 1925–1943. 4. Ufnarovski, V. A.: On the use of graphs for calculating the basis, growth and Hilbert series of associative algebras, Math. USSR-Sb. 68 (1991), 417–428. 5. Sidki, S. and Silva, E. F.: A family of just-nonsolvable torsion-free groups defined on n-ary trees, In: Atas da XVI Escola de A´lgebra, Brası´ lia, Matema´tica Contemporaˆnea 21 (2001), 255–274. 6. Brunner, A., Sidki, S. and Vieira, A. C.: A just-nonsolvable torsion-free group defined on the binary tree, J. Algebra 211 (1999), 99–114. 7. Grigorchuk, R. I.: On the Burnside problem on periodic groups, Funct. Anal. Appl. 14 (1980), 41–43. 8. Gupta, N. and Sidki, S.: On the Burnside problem on periodic groups, Math. Z. 182 (1983), 385–388. 9. Brunner, A. and Sidki, S.: Wreath operations in the group of automorphisms of the binary tree, J. Algebra 257 (2002), 51–64. 10. Sidki, S.: Tree-wreathing applied to generation of groups by finite automata, in press.