On Lie rings of torsion groups

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1 Bull. Mah. Sci. (2016) 6: DOI /s On Lie rings of orsion grous Consuelo Marínez 1 Efim Zelmanov 2,3 Received: 16 Aril 2016 / Acceed: 23 May 2016 / Published online: 10 June 2016 The Auhor(s) This aricle is ublished wih oen access a Sringerlink.com Absrac We rove ha he Lie ring associaed o he lower cenral series of a finiely generaed residually- orsion grou is graded nil. Keywords Torsion grou Nil Lie algebra Mahemaics Subjec Classificaion 20F40 20F45 20F50 1 Inroducion Le G be a grou. A descending sequence of normal subgrous G = G 1 > G 2 > is called a cenral series if [G i, G j ] G i+ j for all i, j 1. The direc sum of abelian grous L(G) = i 1 G i /G i+1 is a graded Lie ring wih Lie bracke [a i G i+1, b j G j+1 ]=[a i, b j ]G i+ j+1 ; a i G i, b j G j. Of aricular ineres are he lower cenral series: G 1 = G, G i+1 =[G i, G], i 1, and, for a fixed rime number, he Zassenhaus series (see [7,8]). Communicaed by S. K. Jain. The firs auhor has been arially suored by MTM C3-1-P and GRUPIN and he second one by he NSF. B Consuelo Marínez cmarinez@uniovi.es 1 Dearameno de Maemáicas, Universidad de Oviedo, C/ Calvo Soelo s/n, Oviedo, Sain 2 Dearmen of Mahemaics, UCSD, La Jolla, CA , USA 3 KAU, Jeddah, Saudi Arabia

2 372 C. Marínez, E. Zelmanov Le be a rime number. We say ha a grou G is residually- if he inersecion of all normal subgrous of indices i, i 1, is rivial. A graded Lie ring L = L 1 + L 2 + is called graded nil if for an arbirary homogeneous elemen a L i he adjoin oeraor ad(a) : x [a] is niloen. The main insrumen in he sudy of he Burnside roblem in he class of residually -grous is he connecion beween orsion in he grou G and graded nilness in he Lie algebra L(G) of he Zassenhaus series (see [6 9]). In his aer we rove his connecion for he lower cenral series and for an arbirary cenral series of G. Theorem 1 Le Ɣ be a finiely generaed residually- orsion grou. Le Ɣ = Ɣ 1 > Ɣ 2 > be he lower cenral series of Ɣ. Then he Lie ring L(Ɣ) = i 1 Ɣ i /Ɣ i+1 is graded nil. Noe ha he known imoran classes of orsion grous: Golod Shafarevich grous (see [2,3]), Grigorchuck grous [4] and Gua Sidki grous [5] are residually. We say ha a (ossibly infinie) grou Ɣ is a -grou if for an arbirary elemen g Ɣ here exiss k 1 such ha g k = 1. Clearly, for a residually -grou being orsion and being a -grou are equivalen. Theorem 2 Le Ɣ be a -grou. Le Ɣ = Ɣ 1 >Ɣ 2 > be a cenral series. Then he Lie ring L(Ɣ) = i 1 Ɣ i /Ɣ i+1 is locally graded nil. In oher words, we claim ha an arbirary finiely generaed graded subalgebra of L(Ɣ) is graded nil. 2 Definiions and resuls Le Ɣ be a residually- grou and le G be is ro- comleion (see [1]). For an elemen y G le y G denoe he closed normal subgrou of G generaed by y. Le [y G, y G ] denoe he closed commuaor subgrou of y G. For elemens g 1, g 2,...,g n G le [g 1, g 2,...,g n ]=[g 1, [g 2, [...,g n ]]...] be heir lef-normed commuaor. We will need he following equaliies which can be found in [1]: (1) For an arbirary ineger k 1wehave[y] k =[y k ] mod [y G, y G ]; (2) [y k ]=[y] (k 1) [y] ( 2) k...[y,...x ] (k k) mod [y G, y G ]. k Le be a rime number. The equaliies (1) and (2) imly ha (3) c = [y ][y,...,x] 1 = [y ( 1) ][y ( 2) ]...[y ( 1),...x ] mod 1 [y G, y G ]. Hence [y ] =c[[y ]] ( 2)/...[[y ],...x] ( 1)/ mod [y G, y G ]. 2 Ieraing his rocess and he use of equaliies (1) and (2) we conclude ha here exiss an infinie sequence of nonnegaive inegers k i 0 such ha

3 On Lie rings of orsion grous 373 (4) [y ] =c[c] k 1[c] k 2...mod [y G, y G ]. Le ρ be a lef-normed grou commuaor, ρ =[g 1,...g m ], where each elemen g i is eiher equal o y or o x k for some k 1. Le d x (ρ) denoe he sum of owers of x involved in ρ and d y (ρ) he number of imes he elemen y occurs in ρ. Lemma 1 Le ρ =[y i1 i2,...,x ik ],d y (ρ) = 1, 0 i 1,...,i k l 1. Then ρ is a converging roduc of commuaors σ of yes: (i) σ =[y j1 j2,...,x js ], where d y (σ ) = 1, no more han one ineger of j 1,..., j s is differen from 0 and d x (σ ) d x (ρ); (ii) σ =[y j1,...,y,...], where d y (σ ) 2 and (d y (σ ) 1) l + d x (σ ) d x (ρ). Proof Suose ha i α, i β 1, 1 α = β k. We will reresen ρ as a roduc of commuaors of yes (i) and (ii) and of commuaors of ye (iii) ρ =[y j1 j2,...,x js ], where 0 j 1,..., j s l 1 and 0 d x (ρ )> d x (ρ) and ye (iv) ρ =[y j1 j2,...,x js ], where 0 j 1,..., j s l 1 and d x (ρ ) = d x (ρ), s > k. Ieraing we will ge rid of commuaors (iii) and (iv). Wihou loss of generaliy we will assume ha α = 1, β = 2 and 1 i 1 i 2. By (3) we have [y i1 ]=[y ( 1) i 1 1 ]...[y ( 1) i 1 1,...,x i 1 1 ][y i 1 1,...,x i 1 1 ] 1 mod [y G, y G ]. Now ρ is a roduc of commuaors of he form [y ( ) i 1 1,...,x i 1 1 i2,...,x ik ],1 1; of he commuaor [y i 1 1,...,x i 1 1 i2,...,x ik ] and of commuaors ha involve a leas wo occurrences of y and he owers x i 1 1 i2,...,x ik (see [7]). The laer commuaors are commuaors of ye (ii). The commuaor [y i 1 1,...,x i 1 1 i2,...,x ik ] saisfies condiion (iv). Hence i remains o consider commuaors of he form [y ( ) i 1 1,...,x i 1 1 } {{ } i2,...,x ik ]. Since ( ),1 1, we will consider he commuaor [y i 1 1,...,x i 1 1 i2,...,x ik ].

4 374 C. Marínez, E. Zelmanov Modulo longer commuaors we can move he ower x i2 o he lef. By (4) we ge [y i2 ]=σ [σ] s 1 [σ] s 2... where σ =[y i 2 +1 ][y i2,...,x i 2 ] 1 mod [y G, y G ]. The commuaors [y i 2 +1 i 1 1,...,x i 1 1,...x ik ] and [y i2,...,x i 2 i 1 1,...,x i 1 1,...x ik ] are of ye (iii) since i 2 i 1 and herefore i2+1 + i1 1 > i 2 + i 1. This finishes he roof of he lemma. Consider again he commuaor ρ =[y i1,...,x ik ]. Suose ha x l = 1. Consider he l-ule ind(ρ) = (k l 1,...,k 0 ), k i Z 0, where k i is he number of imes i occurs among i 1,...,i k. Clearly, k 0 + k 1 + k l 1 = k. Consider he lengh-lex order in Z l 0: (α 1,...,α l )>(β 1,...,β l ) if eiher αi > β i or α i = β i and (α 1,...,α l )>(β 1,...,β l ) lexicograhically. Lemma 2 Le x, y G l = 1, y s = 1. A commuaor ρ =[y i1,...,x ik ] such ha d x (ρ) (s + 1) l can be reresened as a roduc of commuaors σ = [y j1,...,y,...,x jq ], where d y (σ ) 2 and (d y (σ ) 1) l + d x (σ ) d x (ρ). Proof We will show ha ρ is a (converging) roduc of commuaors of he form σ 1 and σ 2, where d y (σ 1 ) 2, (d y (σ 1 ) 1) l + d x (σ 1 ) d x (ρ) for commuaors of he form σ 1, whereas commuaors of he form σ 2 look as σ 2 =[y j1,...,x j ] wih d x (σ 2 )>d x (ρ) or d x (σ 2 ) = d x (ρ) and ind(σ 2 )>ind(ρ). Then, alying his asserion o commuaors of he form σ 2 and ieraing we will ge rid of commuaors σ 2. We claim ha a leas one i, 0 i l 1, occurs in i 1,...,i k no less han imes. Indeed, oherwise d x (ρ) ( 1)( l 1 ), which conradics our assumion ha d x (ρ) (s + 1) l. Suose ha i occurs in i 1,...,i k no less han imes and i is he smalles in {i 1,...,i k } wih his roery. Moving he occurrences of i o he lef, modulo longer commuaors, we assume i 1 = =i = i. By (2) we have [y i,...,x }{{ i ]=[y } i+1 ][y ( 1) i ] 1...[y ( 1) i,...,x i ] 1 τ 1...τ q, 1 where τ j are commuaors ha involve y a leas wice. The commuaor σ =[y i+1 i +1,...,x ik ] has greaer index han ρ. Indeed, d x (σ ) = d x (ρ), buind(σ ) is lexicograhically greaer han ind(ρ).

5 On Lie rings of orsion grous 375 For a commuaor τ j =[τ j i +1,...,x ik ],wehave d x (τ j ) d x(ρ) ( 1) i. Hence, l (d y (τ j ) 1) + d x(τ j ) l + d x (ρ) ( 1) i > d x (ρ). Consider now he commuaor ρ =[y i i +1,...,x ik ]. We claim ha here exiss j {i +1,...,i k } such ha j i. Indeed, oherwise all inegers in {i +1,...,i k } are smaller han i and herefore occur ( 1) imes. Hence, d x (ρ) i + ( 1)( i 1 ) = i+1 + i 1 < 2 l (s + 1) l, which conradics he assumion of he lemma. Moving x j o he righ end in ρ modulo longer commuaors we will assume ha i k = j i. Consider he commuaor ρ =[y i i +1,...,x ik+1 ].Wehaved x (ρ ) = d x (ρ) ( 1) i j s l. By he inducion assumion on s he commuaor ρ is a roduc of commuaors w in y and x, each commuaor involves μ = μ(w) 2 elemens y and (μ 1) l +d x (w) d x (ρ ). We will assume ha w =[w 1,...,w μ ], w j =[y,...],1 j μ. Remark Any commuaor ha has degree μ + 1iny and degree d x (w) in x fis he requiremens of he lemma since μ l + d x (w) d x (ρ ) + l d x (ρ). The commuaor [w 1,...,w μ j ] is equal o a roduc [[w 1 j ],w 2,...,w μ ][w 1, [w 2 j ],...]...[w 1,...,[w μ j ]] modulo longer commuaors (see he Remark above). Consider [w 1,...,[w ν j ],...,w μ ]. In [w ν j ] move x j o he lef osiion nex o y modulo longer commuaors (see he Remark above). By (4), [y j ]=c[c] k 1[c] k 2...τ 1...τ q, where c =[y j+1 ][y j,...,x j ] 1 ; τ 1,...τ q [y G, y G ]; d x (τ 1 ),..., d x (τ q ) j. If he commuaor [y j ] is relaced by one of τ 1,...,τ q hen see he Remark. If [y j ] is relaced by c hen d x ([w 1,...,w ν 1, c,w ν+1,...,w μ ]) d x ([w 1,...,w μ j ]) + ( 1) j d x (w) + j+ 1. Hence, (μ 1) l + d x ([w 1,...,w μ j ]) (μ 1) l + d x (μ) + j+ 1 d x (ρ )+ j+1 = d x (ρ) ( 1) i j + j+1 = d x (ρ)+( 1)( j i ) d x (ρ).

6 376 C. Marínez, E. Zelmanov Since d y ([w 1,...,w μ j ]) 2, his commuaor saisfies he requiremens of he lemma. If he commuaors [y j ] is relaced by [c,...,x] k, hen (μ 1) l + d x ([w 1,...,w ν 1, [c,...,x] k 1,w ν+1,...,w μ ])>d x (ρ). This finishes he roof of he lemma. Lemma 3 Le x G i,x l = 1, y G j,y s = 1. Suose ha j 2i l. Then (yg j+1 )ad(xg i+1 ) (s+1)l = 0 in he Lie algebra L = k=1 G k /G k+1. Proof By Lemma 2 he grou commuaor ρ =[y,...,x] can be reresened as a (s+1) l roduc of commuaors w =[w 1,...,w μ ],μ 2, where each w k is a commuaor of he ye w k =[y j1,...,x jr ], (μ 1) l + d x (w) d x (ρ) = (s + 1) l. By Lemma 1 each w k is a roduc of commuaors of ye (i) or (ii). A commuaor of ye (ii) jus increases he degree in y. Le[y j1,...,x jr ] be a commuaor of ye (i). So all j 1,..., j r, exce ossibly one, are equal o 0. This imlies ha [y j1,...,x jr ] G j+i( j jr ( l 1 1)). Hence, w G d, where d = μj + id x (w) μi( l 1 1) j + (μ 1)i l + (μ 1)i l + id x (w) μi( l 1 1) j + id x (ρ) + i[(μ 1) l μ( l 1 1)]. Now i remains o noice ha (μ 1) l μ( l 1 1) >0. We showed ha d > j + id x (ρ), which imlies he lemma. Lemma 4 The Lie ring L(Ɣ) is weakly graded nil, i.e., for arbirary homogeneous elemens a, b L(Ɣ) here exiss n(a, b) 1 such ha bad(a) n(a,b) = 0. Proof Le a Ɣ i, a l = 1. Le n(a) = 2i l. By Lemma 3, for an arbirary elemen b Ɣ j, j n(a), here exiss an ineger n(a, b) 1 such ha [b, a, a,...,a] n(a,b) G j+in(a,b)+1. Since Ɣ is a orsion grou i follows ha for an arbirary k 1 he subgrou Ɣ k has finie index in Ɣ, hence Ɣ k is oen in Ɣ. The subgrou G k is he comleion of Ɣ k. Hence Ɣ G k = Ɣ k. We roved ha bad(a) n(a,b) = 0inL(Ɣ).Nowleb be an arbirary homogeneous elemen from L(Ɣ). Then he degree of he elemen b = bad(a) n(a) is greaer han n(a). Hence, bad(a) n(a)+n(a,b ) = b ad(a) n(a,b ) = 0, which finishes he roof of he lemma. Lemma 5 Le L be a Lie algebra over a field Z/Z generaed by elemens x 1,...,x m. Le a L be an elemen such ha x i ad(a) k = 0, 1 i m. Then Lad(a) k = (0). Proof The algebra L is embeddable in is universal associaive enveloing algebra U(L). Lea k be he ower of he elemen a in U(L). For an arbirary elemen b L we have bad(a) k =[b, a k ]. If he elemen a k commues wih all generaors x 1,...,x m hen [L, a k ]=Lad(a) k = (0), which finishes he roof of he lemma.

7 On Lie rings of orsion grous 377 Lemma 6 Le L be a Lie ring generaed by elemens x 1,...,x m. Suose ha l L = (0). Le a L be an elemen such ha x i ad(a) k = 0, 1 i m. Then Lad(a) kl = (0). Proof By Lemma 5 we have Lad(a) k L. Hence L(ad(a) k ) l l L = (0), which roves he lemma. Proof of Theorem 1 Le x 1,...,x m be generaors of he grou Ɣ. Then he elemens x i Ɣ 2,1 i m, generae he Lie ring L(Ɣ). Le l be he maximum of orders of he elemens x 1,...,x m,sox l i = 1, 1 i m. Then l (x i Ɣ 2 ) = 0inheLiering L(Ɣ). Hence l L(Ɣ) = (0). Le a be a homogeneous elemen of L(Ɣ). By Lemma 4 here exiss k 1 such ha (x i Ɣ 2 )ad(a) k = 0fori = 1,...,m. Now Lemma 6 imlies ha L(Ɣ)ad(a) k l = (0), which finishes he roof of Theorem 1. Proof of Theorem 2 Wihou loss of generaliy we assume ha i Ɣ i = (1). Weview he subgrous {Ɣ i i 1} as a basis of neighborhoods of 1 hus making Ɣ a oological grou. Le G be a comleion of Ɣ in his oology. Le G i be he closure of Ɣ i in G. Then G i Ɣ = Ɣ i and G = G 1 > G 2 > is a cenral series of he grou G. Arguing as in Lemmas 3, 4 we conclude ha he Lie ring L(Ɣ) = i 1 Ɣ i /Ɣ i+1 is weakly graded nil. Choose homogeneous elemens a 1,...,a m L(Ɣ). Since Ɣ is a -grou i follows ha here exiss l 1 such ha l a i = 0, 1 i m. Consider he subring L of L(Ɣ) generaed by a 1,...,a m, l L = (0).Ifa is a homogeneous elemen from L and a i ad(a) k = 0, 1 i m, hen by Lemma 6 we have L ad(a) k l = (0), which finishes he roof of Theorem 2. Oen Access This aricle is disribued under he erms of he Creaive Commons Aribuion 4.0 Inernaional License (h://creaivecommons.org/licenses/by/4.0/), which ermis unresriced use, disribuion, and reroducion in any medium, rovided you give aroriae credi o he original auhor(s) and he source, rovide a link o he Creaive Commons license, and indicae if changes were made. References 1. Dixon, J.M., du Sauoy, M.P.F., Mann, A., Segal, D.: Analyic ro- grous, London Mah. Soc. Lecure Noes Series, vol Cambridge Universiy Press, Cambridge (1991) 2. Ershov, M.: Golod-Shafarevich grous: a survey. In. J. Algebra Comu. 22, Aricle ID0001 (2012). doi: /s Golod, E.S.: On nil algebras and residually finie -grous. Izv. Akad. Nauk SSSR Ser. Ma. 28, (1964) 4. Grigorchuk, R.I.: Degrees of growh of finiely generaed grous and he heory of invarian means. Izv. Akad. Nauk SSSR Ser. Ma. 48, (1984) 5. Gua, N., Sidki, N.: On he Burnside roblem for eriodic grous. Mah. Z. 182, (1983) 6. Higman, G.: Lie ring mehods in he heory of finie niloen grous. In: Proc. Inerna. Congress Mah. 1958, Cambridge Univ. Press, New York (1960) 7. Vaughan-Lee, M.: The resriced Burnside roblem, 2nd edn. London Mahemaical Sociey Monograhs, New Series 8. The Clarendon Press. Oxford Universiy Presss, New York (1993) 8. Zelmanov, E.: Nil rings and eriodic grous, KMS Lecure Noes in Mahemaics. Korean Mahemaical Sociey, Seoul (1992) 9. Zelmanov, E.: Lie ring mehods in he heory of niloen grous, Grous 93 Galway/S. Andrews, v.2, London Mah. Soc. Lecure Noe Ser., vol. 212, Cambridge Univ. Press, Cambridge (1995)