ON COMPUTING THE NON-ABELIAN TENSOR SQUARES OF THE FREE 2-ENGEL GROUPS

Proceedngs of the Ednburgh Mathematcal Socety 004) 47, 305 33 c DOI:10.1017/S001309150000998 Prnted n the Unted Kngdom ON COMPUTING THE NON-ABELIAN TENSOR SQUARES OF THE FREE -ENGEL GROUPS RUSSELL D. BLYTH 1, ROBERT FITZGERALD MORSE AND JOANNE L. REDDEN 3 1 Department of Mathematcs and Computer Scence, Sant Lous Unversty, St. Lous, MO 63103, USA Department of Electrcal Engneerng and Computer Scence, Unversty of Evansvlle, Evansvlle, IN 477, USA rfmorse@evansvlle.edu) 3 Department of Mathematcs, Unversty of Evansvlle, Evansvlle, IN 477, USA Receved 1 November 00) Abstract In ths paper we compute the non-abelan tensor square for the free -Engel group of rank n>3. The non-abelan tensor square for ths group s a drect product of a free abelan group and a nlpotent group of class whose derved subgroup has exponent 3. We also compute the non-abelan tensor square for one of the group s fnte homomorphc mages, namely, the Burnsde group of rank n and exponent 3. Keywords: non-abelan tensor square; free -Engel groups; Burnsde groups 000 Mathematcs subject classfcaton: Prmary 0F05; 0F45 Secondary 0F18 1. Introducton The non-abelan tensor square G G of a group G s the group generated by the symbols g h, where g, h G, subject to the relatons gg h = g g g h)g h) and g hh =g h) h g h h ), for all g, g,h,h G, where g g denotes the conjugaton acton gg g 1. A group s -Engel f t satsfes the left normed commutator law [x, y, y]. The purpose of ths paper s to compute the non-abelan tensor square of the free -Engel group of rank n for each n>3. By computng the tensor square of a group G, we mean fndng a smplfed and standard presentaton for G G. The non-abelan tensor square has ts roots n algebrac K-theory [9] as well as n topology [6, 7]. Ths group constructon was frst nvestgated from a purely group theoretc perspectve n the semnal paper by Brown et al. [8], n whch they compute G G 305

306 R. D. Blyth, R. F. Morse and J. L. Redden for all groups of order up to thrty. Subsequent papers nvestgate explct descrptons of the non-abelan tensor square for partcular groups; for nlpotent of class groups see [1 3,13,16], for metacyclc groups see [5,14], and for lnear groups see [1]. A recent survey artcle on the non-abelan tensor squares and the more general non-abelan tensor products of groups can be found n [15]. In [1,], the non-abelan tensor squares of nlpotent of class groups were nvestgated. The non-abelan tensor square of the free nlpotent group of class and rank, whch s exactly the free -Engel group of rank, was shown to be free abelan of rank 6. The non-abelan tensor square of the free -Engel group of rank 3 was computed n [4]; t s the drect product of a free abelan group of rank 11 and a nlpotent group of class. Snce -Engel groups are metabelan, the followng proposton from [4] shows that ther tensor squares are abelan or nlpotent of class. Proposton 1.1. Let G be a group. If the derved subgroup G s nlpotent of class clg ), then G G s nlpotent wth class equal to clg ) or clg )+1. In lght of Proposton 1.1 and the fact that the tensor square of the free -Engel group of rank 3 s not abelan [4], we conclude that the tensor squares of the free -Engel groups of rank n>3 are nlpotent of class. Our method for computng non-abelan tensor squares uses the concept of a crossed parng. Defnton 1.. Let G and L be groups. A functon Φ : G G L s called a crossed parng f for all g, g,g G. Φgg,g )=Φ g g, g g )Φg, g ), 1.1) Φg, g g )=Φg, g )Φ g g, g g ) 1.) The followng proposton allows us to determne homomorphc mages of G G. Proposton 1.3 see [7]). A crossed parng Φ : G G L determnes a unque homomorphsm of groups Φ : G G L such that Φ g g )=Φg, g ) for all g, g n G. To compute the non-abelan tensor square of a group G, we frst conjecture a group L and fnd a crossed parng Φ : G G L. We then show that the homomorphsm nduced by the crossed parng s actually an somorphsm and hence G G = L. In the case of computng the tensor square for the free -Engel groups, we were guded n our conjecture of L by computng the tensor square of a fnte homomorphc mage of these groups, namely, the Burnsde groups of rank n and exponent 3 for n = 4, 5 and 6. These computatons were performed usng methods developed n [10] and mplemented n Gap [11]. We provde a more detaled outlne of our computatonal method n 3.

On computng the non-abelan tensor squares of the free -Engel groups 307 Our man result s the followng. Theorem 1.4. The non-abelan tensor square of the free -Engel group of rank n> s a drect product of a free abelan group of rank 1 3 nn +)and an nn 1)-generated nlpotent group of class whose derved subgroup has exponent 3. Corollary 1.5. The non-abelan tensor square of the Burnsde group of rank n> and exponent 3 s a drect product of an elementary abelan 3-group of rank 1 3 nn +) and an nn 1)-generated nlpotent group of class havng exponent 3. In the next secton we record varous results needed to compute the non-abelan tensor square for -Engel groups. We prove Theorem 1.4 and Corollary 1.5 n 3.. Prelmnary results The followng famlar denttes for -Engel groups are stated here wthout proof see, for example, [18]). Recall that any nlpotent group of class s a -Engel group. Lemma.1. Let G be a -Engel group. For x, y, z, w G and n Z we have [x, y, z, w] =[x, y, z] 3 =1,.1) [x, y, z] =[z, x, y] =[x, z, y] 1,.) [x, y n ]=[x n,y]=[x, y] n..3) Throughout the paper, let E = En, ) denote the free -Engel group of rank n> wth a fxed orderng on ts generators, whch are labelled g 1,g,...,g n. It follows from Lemma.1 that any element g n E can be wrtten unquely wth respect to the orderng of the generators) as the product g = g α 1 j<k n [g j,g k ] β j,k 1 r<s<t n [g r,g s,g t ] γr,s,t,.4) where each α and β j,k s an nteger and by Equaton.1)) each γ r,s,t s an nteger modulo 3. Snce E/E 3 s somorphc to Bn, 3), the Burnsde group of rank n and exponent 3, we can also express each element of Bn, 3) smlarly, where each α and β j,k s now also an nteger modulo 3. The followng lemma provdes formulae for multplcaton and conjugaton of arbtrary elements n E. Lemma.. Let g and g be elements of E, where g s defned n.4) and g = g α 1 j<k n [g j,g k ] β j,k 1 r<s<t n [g r,g s,g t ] γ r,s,t,.5) where each α and β j,k s an nteger and each γ r,s,t s an nteger modulo 3. Then the product g g can be wrtten n the form g g = [g j,g k ] β j,k [g r,g s,g t ] γ r,s,t,.6) g α 1 j<k n 1 r<s<t n

308 R. D. Blyth, R. F. Morse and J. L. Redden where α = α + α, β j,k = β j,k + β j,k α j α k and γ r,s,t γ r,s,t + γ r,s,t + β s,t α r β r,t α s + β r,s α t α rα s α t + α rα sα t α rα s α t mod 3. Left conjugaton of g by g s the product where g g = g α 1 j<k n [g j,g k ] β j,k 1 r<s<t n α = α, β j,k = β j,k α jα k + α j α k, γ r,s,t γ r,s,t + β s,t α r β r,t α s + β r,s α t α rα s α t + α rα sα t [g r,g s,g t ] γ r,s,t,.7) α rα s α t β s,tα r + β r,tα s β r,sα t α r α s α t + α r α sα t + α r α sα t mod 3. The followng lemma states general propertes for the non-abelan tensor square of any group. Lemma.3 see [8]). Let G be any group and x, v, y, z G. Then n G G we have where 1 denotes the dentty element of G G. x 1=1 x =1,.8) [x v, y z] =[x, v] [y, z],.9) Let each of g and g be ether a generator or commutator of E. We defne the weght of an element g g of E E as the sum of the commutator weghts of g and g, where a generator has weght one. For example, the three varable element x [y, z] has weght three whle the three varable element x [x, y, z] has weght four. We lst several denttes used to express the elements of E E n terms of the generators g 1,g,...,g n of E. We start wth two- and three-varable denttes that hold n the tensor square of a -Engel group and ts subgroups. The proofs of these denttes can be found n [3, 4]. Lemma.4. Let G be a nlpotent group of class at most. Then, for x, y G and m, n Z, the followng dentty holds n G G: x m y n =x y) mn y [x, y]) mn ) x [x, y]) n m )..10) Lemma.5. Let G be a -Engel group. Then, for any x, y, z G, the followng denttes hold n G G: z [y, x])y [x, z])x [z,y]) = [y, z] [x, z])[y, z] [x, y])[x, z] [x, y]),.11) [x, y] z) 1 =[y, x] z) =z [y, x]) 1 =z [x, y])..1)

On computng the non-abelan tensor squares of the free -Engel groups 309 The followng lemma lsts four-varable denttes found n [4]. Lemma.6. Let G be a -Engel group. For x, y, z, v G, the followng denttes hold n G G: [x, v] [y, z] =[x, v, y] z)y [x, v, z]),.13) [x, v, y] z)y [x, v, z])[y, z, x] v)x [y, z, v])=1..14) The next lemma explctly shows that all generators of E E nvolvng fve or more varables equal the dentty. Moreover, all generators of E E expressed n four varables can be wrtten as a product of weght four generators of a common form and have exponent 3. Lemma.7. Let G be a -Engel group. Then, for any u, v, x, y, z G, the followng hold n G G: [u, v] [x, y, z] =[x, y, z] [u, v] =1,.15) [x, v] [y, z]) 3 =1,.16) v [x, y, z] =[v, x] [y, z])[v, y] [x, z]) 1 [v, z] [x, y])..17) Proof. Substtutng [u, v] for x, x for y, y for z and z for v n.14) yelds.15). We note that, by.10) and.1), for any v, x, y, z n G, Hence, by.13), we have v [x, y, z]) 3 = v [x, y, z] 3 = v 1=1. [x, v] [y, z]) 3 =[x, v, y] z) 3 y [x, v, z]) 3 =1, and thus.16) holds. To show.17) we frst rearrange.13) usng.1) as follows: z [x, v, y] =y [x, v, z])[v, x] [y, z])..18) By nterchangng v wth y n.18), and usng.) and.1), we obtan z [v, x, y] =v [x, y, z])[v, z] [x, y])..19) Now nterchange respectvely) y wth z and x wth z n.19) to get and Also, from.13), we have y [v, x, z] =v [x, y, z]) 1 [v, y] [x, z]).0) x [v, y, z] =v [x, y, z])[v, x] [y, z])..1) [x, v] [y, z] =z [v, x, y])y [x, v, z])..)

310 R. D. Blyth, R. F. Morse and J. L. Redden Interchange x wth z and y wth v n.), and use.) and.1) to obtan [x, v] [y, z] =x [z,y,v])v [y, z, x])..3) Note that the left-hand sdes of.) and.3) are equal, so we may equate the rghthand sdes, and, after applyng.) as needed, substtute the expressons on the rghthand sdes of.19),.0) and.1). Ths gves v [x, y, z])[v, z] [x, y])v [x, y, z])[v, y] [x, z]) 1 =v [x, y, z]) 1 [v, x] [y, z]) 1 v [x, y, z]). Hence v [x, y, z]) =[v, x] [y, z]) 1 [v, y] [x, z])[v, z] [x, y]) 1. Snce v [x, y, z]) 3 =1, we have v [x, y, z]) =v [x, y, z]) 1 and we arrve at Equaton.17). The followng proposton s from [4]. Proposton.8. For a -Engel group G, the defnng relatons of G G reduce to xy z =x [y, z])y z)x z),.4) x yz =[z,x] y)x y)x z)..5) We generalze Proposton.8 n two steps. We frst show the followng. Proposton.9. Let G be a -Engel group. Let x 1,...,x n G. Let b be an element of the derved subgroup of G. Then, for n, and n ) x b = n ) b x = k 1 x j [x k,b]) x b).6) k= k= k 1 [x k,b] x j ) b x )..7) Proof. We nduct on n. Consder Equaton.6). Let x 1,x G and let b le n the derved subgroup of G. By Proposton.8 and Equaton.9) we have x 1 x b =x 1 [x,b])x b)x 1 b) =x 1 [x,b])x 1 b)x b), whch has the form of.6) for n =. Suppose now that Equaton.6) holds for some n. For =1,...,n + 1, let x G. Let b be an element of the derved subgroup of G. Then, by Proposton.8, we have n+1 ) n x b = ) ) n x [x n+1,b] x n+1 b) ) ) x b..8)

On computng the non-abelan tensor squares of the free -Engel groups 311 Snce [x n+1,b] s n the centre of G, the factor n x ) [x n+1,b] of.8) expands lnearly by Proposton.8). By the nductve hypothess, the last factor on the rghthand sde of.8) s n ) x b = k 1 x j [x k,b]) x b). k= These facts put together gve us the followng: n+1 ) n ax b = ) ) n x [x n+1,b] x n+1 b) k= ) ) x b n ) k 1 = x [x n+1,b]) x n+1 b) x j [x k,b]) x b) = n+1 k 1 n+1 x j [x k,b]) x b). k= The proof of Equaton.7) s smlar. Proposton.9 s used to show the followng result, whch has a smlar nductve proof. The proof can be found n [17]. Proposton.10. Let G be a -Engel group. For u =1,...,n, let x u, y u, x and y be elements of G. Then, for n 3, n ) x u y = u=1 and n ) x y u = u=1 l 1 l=3 k= l 1 l=3 k= k 1 s 1 x j [x k, [x l,y]]) x r [x s,y]) x n m y) s= r=1 k 1 s 1 [y k, [y l,x]] y j ) [y s,x] y r ) s= r=1 m=0 x y u ). Approprate substtutons n Proposton.10 show that some expansons are lnear, such as n the followng corollary. Corollary.11. Let G be a -Engel group. Let x G and suppose that b s an element of the derved subgroup of G for =1,...,n. Then, for n, n ) n ) b x = b x) and x b = x b ). u=1

31 R. D. Blyth, R. F. Morse and J. L. Redden 3. Computng the tensor square In ths secton we prove Theorem 1.4, whch provdes a smplfed presentaton for E E. The followng s an outlne of our method for makng the computatons needed n provng the theorem. Usng the denttes for E Edeveloped n, we frst express an arbtrary generator g g of E E as a product of a fxed set of elements of E E whose exponent expressons depend on the exponents n the representatons of g and g found n.4) and.5). Then we construct a group L n and defne a multplcaton formula for t n terms of the exponents of ts generators see Example 3.3). To prove Theorem 1.4, we defne a functon Φ : E E L n that we show s a crossed parng. Verfyng that Φ s a crossed parng nvolves multplyng and conjugatng elements of E usng Lemma. and multplyng elements n L n as defned n Example 3.3. These operatons are descrbed by formulae on the exponents of the generators of these groups, respectvely. We consder these formulae as functons. Verfcaton of the denttes 1.1) and 1.) for Φ s completed by composng these exponent functons. Snce some of these compostons nvolve manpulatng hundreds of terms, the computatons were performed usng Maple [19]. We complete the proof of the theorem by showng the homomorphsm nduced by the crossed parng Φ s an somorphsm. We express the element g of E, defned n.4), as the product abc, where a = g α, b = 1 j<k n [g j,g k ] β j,k, c = 1 r<s<t n [g r,g s,g t ] γr,s,t. 3.1) Smlarly, we express g E, defned n.5), as the product a b c, where a = g α, b = [g j,g k ] β j,k, c = [g r,g s,g t ] γ r,s,t. 3.) 1 j<k n 1 r<s<t n As a frst step, we use Proposton.8 repeatedly to compute an expanson formula for g g n terms of the factors a, b, c, a, b and c. We use the fact that E s nlpotent of class 3 to smplfy the expanson, g g = a b c a b c =a b [c, a b c ])c a b c )a b a b c ) =c a b c )a b a b c ) =[c,c] a b )c a b )c c )[c,a b] a b )a b a b )a b c ) =[b,c] a )c a )c b )c c )a b a b )a [b, c ])b c )a c ) =c a )c b )c c )a b a b )b c )a c ). 3.3) By Corollary.11, the factors c c ), c b ) and b c ) n Equaton 3.3) expand lnearly nto products of elements of E Eof weght at least fve, whch are all equal to the dentty by.15). Hence our expanson becomes a b c a b c =c a )a b a b )a c ). 3.4)

On computng the non-abelan tensor squares of the free -Engel groups 313 We expand a b a b ) usng Proposton.8, obtanng a b a b )=a [b, a b ])b a b )a a b ) =a [b, a ])[b,b] a )b a )b b )[b,a] a )a a )a b ) =a [b, a ])b a )b b )[b,a] a )a a )a b ). 3.5) Puttng the expansons 3.4) and 3.5) together we get a b c a b c =c a )a [b, a ])b a )b b )[b,a] a )a a )a b )a c ). 3.6) We next show that every generator of E E can be expressed as a product of a prescrbed set of elements n E En a fxed orderng. We accomplsh ths goal by expandng each of the eght factors n 3.6) as products nvolvng the generators g 1,...,g n of E. For the sake of notatonal convenence n the sequel, we set I 1 = {, j, k) 1, j, k n; max{j, k}; j<k}, I = {, j, k, l) 1 <j n; k<l n;, j) < k, l) lexcographcally}. Lemma 3.1. Let g and g be arbtrary elements of E. Then g g = g g ) ρ g [g j,g k ]) σ,j,k,j,k) I 1 [g,g j ] [g k,g l ]) τ,j,k,l,j,k,l) I =0 j n g n g j ) νn,j, 3.7) where each ρ, σ,j,k and ν,j s an nteger and each τ,j,k,l s an nteger modulo 3. Proof. We contnue to consder g and g as wrtten n.4) and.5), respectvely. The eght sets of forms lsted below n 3.8) 3.15) are the possble element forms that arse when the eght terms of 3.6) are expanded usng Propostons.8.10: g αr r g αu u [g αs s [g r,g s,g t ] γr,s,t g α u u, 3.8) g αu u [[g r,g s ] βr,s,g α v v ], 3.9) [g r,g s ] βr,s g α u u, [g α q q, [g r,g s ] βr,s ] g α p p, 3.10) [g r,g s ] βr,s [g p,g q ] β p,q, 3.11) g α v v, g αr r,g α p p,g α q q ], g α r r [[g r,g s ] β r,s,g α v v ] g α u [gs αs,g α u u ], gr αr [g α t t u, 3.1) [g αs s,gu αu,g α s s ], g α r r,gt αt,g α u u ], [g α s s,g αu u ], 3.13) gu αu [g r,g s ] β r,s, g α p p [gq αq, [g r,g s ] β r,s ], 3.14) gu αu [g r,g s,g t ] γ r,s,t, 3.15) where 1 r<s<t n, 1 p<q n and 1 u, v n.

314 R. D. Blyth, R. F. Morse and J. L. Redden For example, by Corollary.11 and Proposton.9, b a = 1 r<s n = 1 r<s n = 1 r<s n [g r,g s ] βr,s [g r,g s ] βr,s n q 1 q= p=1 [g α q q u=1 g α u u u=1 g α u u ), [g r,g s ] βr,s ] g α p p ) u=1 ) [g r,g s ] βr,s g α u u ). Hence we have the element forms shown n 3.10). We extract the exponents for each of the forms found n 3.8) 3.15) usng.3) and.10): [g r,g s,g t ] g u ) α u γr,s,t, 3.16) g u [g r,g s,g v ]) αuβr,sα v, 3.17) [g r,g s ] g u ) βr,sα u, gu [g r,g s,g u ]) βr,sα u ), [gq, [g r,g s ]] g p ) α q βr,sα p, 3.18) [g r,g s ] [g p,g q ]) β p,q βr,s, 3.19) [g r,g s,g v ] g u ) β r,s αvα u,, 3.0) g u g v ) αuα v, gv [g u,g v ]) αuα v ), g u [g u,g v ]) α v αu ), gr [g s,g u ]) αrαsα u, g r [g r, [g s,g u ]]) αsα u αr ), gr [g s, [g t,g u ]]) αrαsαtα u, g r [g s,g p,g q ]) αrαsα p α q, gr [g t,g u,g s ]) α r α t αuα s, g r [g u,g s ]) α r α s αu, g r [g r, [g s,g u ]]) α s αuα r ), 3.1) g u [g r,g s ]) αuβ r,s, gu [g u, [g r,g s ]]) β r,s α ), gp [g q, [g r,g s ]]) αpαqβ r,s, 3.) g u [g r,g s,g t ]) γ r,s,t αu, 3.3) where 1 r<s<t n, 1 p<q n and 1 u, v n. Hence an arbtrary generator of E E can be wrtten as a product of the element forms lsted n 3.16) 3.3). The only non-central elements n ths lst are the powers of elements of the form g g j for j. Hence the exponents have no effect n rewrtng the other element forms to match the factors lsted n the statement of the lemma. There are only two basc form types that need to be rewrtten. For 3.16), we have, usng Equatons.1) and.17), [g r,g s,g t ] g u )=g u [g r,g s,g t ]) 1 =[g u,g r ] [g s,g t ]) 1 [g u,g s ] [g u,g t ])[g u,g t ] [g s,g r ]) 1.

On computng the non-abelan tensor squares of the free -Engel groups 315 We then reorder the subscrpts as needed by repeatedly usng.1) and keepng track of the sgn changes of the exponents as necessary. We rewrte the other element forms that nvolve a weght three commutator n E smlarly. Elements of the form g k [g,g j ] for whch k>max{, j} can be rewrtten usng.11), g k [g,g j ]) =g [g j,g ]) 1 g j [g j,g k ]) 1 [g,g k ] [g j,g k ])[g,g k ] [g j,g ])[g j,g k ] [g j,g ]). Agan, we reorder the subscrpts, as needed, usng.1). We can order the non-central factors g g j, for j, as specfed n the lemma, notng that the commutators [g g j,g k g l ]=[g,g j ] [g k,g l ] where k l) le n the centre of E E. Hence the reorderng of the non-central factors does not add any new element forms. An arbtrary element of E E s a product of generators of the form g g.by Lemma 3.1, each of these generators can be wrtten n a common form 3.7). Hence E E s fntely generated by the factors n the product 3.7). Remark 3.. We note that Lemma 3.1 also holds for the generators of Bn, 3) Bn, 3), where the exponents are taken to be ntegers modulo 3. Moreover, an arbtrary element of Bn, 3) Bn, 3) can be expressed as the product 3.7), where all the exponents are agan consdered to be ntegers modulo 3. The exponents ρ, σ,j,k, ν,j and τ,j,k,l specfed n Lemma 3.1 are used later n the proof of Theorem 1.4 to defne a mappng Φ : E E L n where L n s defned n Example 3.3 below). An explct expresson for these exponents s determned by a careful analyss of how the exponents arse durng the expanson and collecton process of g g as a product of the generators ndcated n Lemma 3.1. Ths analyss, whch was performed completely n [17], reles on trackng the ndces of each of the products n the expanson process descrbed n the proof of the lemma to form the exponent expressons lsted below. When all of the forms n the expanson are collected, we note that each basc form must be consdered n several cases dependent only on the relatve orderng of ts ndces. We arrve at the followng descrptons of the exponents of the generators of E E. For generators of weght two, the exponents are ρ = α α and ν,j = α α j j). 3.4) For generators of weght three, we have four cases to consder, wth ndces ordered <j<k. The exponent for each generator g [g j,g k ]s σ,j,k = α β j,k + α β j,k α α jα k + α α j α k + α α j α k α α jα k + α kβ,j α k β,j. 3.5)

316 R. D. Blyth, R. F. Morse and J. L. Redden For each generator g [g,g j ], the exponent s σ,,j = α β,j + α β,j α α α j + α α α j α j α The exponent for each generator g j [g,g k ]) s ) + α j ) α. σ j,,k = α jβ,k + α j β,k α α j α k + α α jα k α kβ,j + α k β,j. For each generator g j [g,g j ]), the exponent s σ j,,j = α jβ,j + α j β,j + α α j ) α For generators of weght four, we have sx cases to consder, wth ndces ordered <j<k<l. The exponent for each generator [g,g j ] [g,g k ]s αj ). τ,j,,k = α γ,j,k + α γ,j,k + α α β j,k α α jβ,k + α α kβ,j α α β j,k + α α j β,k ) α α α k β,j + β,j β,k β,jβ,k + β j,k α α jβ,k + α α kβ,j + α α j β,k ) α α k β,j β j,k α α α α j α k + α α α jα k α α α j α k + α α α jα k ) ) ) ) + α jα α α k α j α k α j α k α α + α jα k α kβ,j + α k β,j. For each generator [g,g j ] [g j,g k ], the exponent s τ,j,j,k = α jγ,j,k + α j γ,j,k + α α j β j,k α j α jβ,k + α j α kβ,j α α jβ j,k + α j α jβ,k ) α α jα k β,j + β,j β j,k β,jβ j,k β j,k + α jα kβ,j α j α k β,j ) ) ) + β,k αj + α α j α jα k α α αj α k + α α k j α kβ,j + α k β,j. The exponent for each generator [g,g k ] [g j,g k ]s τ,k,j,k = α kγ,j,k + α k γ,j,k + α α k β j,k α jα k β,k + α k α kβ,j α α kβ j,k + α j α kβ,k ) ) α α k α kβ,j + β,k β j,k β,kβ j,k + β k,j β,j αk α kβ,j + α k β,j. For each generator [g,g j ] [g k,g l ], the exponent s τ,j,k,l = α γ j,k,l + α jγ,k,l + α kγ,j,l α lγ,j,k + α γ j,k,l α j γ,k,l α k γ,j,l + α l γ,j,k + α α jβ k,l α α j β k,l α α k β j,l + α α l β j,k α α kβ j,l + α j α kβ,l + α jα k β,l α jα l β,k + α α lβ j,k α j α lβ,k α k α lβ,j + α kα l β,j α α j β k,l + α α jβ k,l + α α kβ j,l α α lβ j,k + α α k β j,l α jα k β,l α j α kβ,l + α j α lβ,k

On computng the non-abelan tensor squares of the free -Engel groups 317 α α l β j,k + α jα l β,k + α kα l β,j α k α lβ,j + β,j β k,l β,jβ k,l + α α jβ k,l α α kβ j,l + α jα kβ,l + α α lβ j,k α jα lβ,k α kα lβ,j α α j β k,l + α α k β j,l α j α k β,l α α l β j,k + α j α l β,k + α k α l β,j + α α j α k α l α α jα kα l α α jα k α l + α α j α kα l + α α j α kα l α α jα k α l α α j α k α l + α α jα kα l + α α j α kα l α α jα k α l + α α j α k α l + α α jα kα l α α j α kα l α α jα k α l. The exponent for each generator [g,g k ] [g j,g l ]s τ,k,j,l = α γ j,k,l + α jγ,k,l + α kγ,j,l + α lγ,j,k α γ j,k,l α j γ,k,l α k γ,j,l α l γ,j,k α α jβ k,l α α j β k,l α α k β j,l α α l β j,k + α α kβ j,l + α j α kβ,l + α jα k β,l + α jα l β,k α α lβ j,k α j α lβ,k α k α lβ,j α kα l β,j + α α j β k,l + α α jβ k,l + α α kβ j,l + α α lβ j,k α α k β j,l α jα k β,l α j α kβ,l α j α lβ,k + α α l β j,k + α jα l β,k + α kα l β,j + α k α lβ,j + β,k β j,l β,kβ j,l α α jβ k,l + α α kβ j,l + α jα kβ,l α α lβ j,k α jα lβ,k α kα lβ,j + α α j β k,l α α k β j,l α j α k β,l + α α l β j,k + α j α l β,k + α k α l β,j + α α j α k α l α α jα kα l + α α jα k α l α α j α kα l α α j α kα l + α α jα k α l + α α j α k α l α α jα kα l α α j α kα l + α α jα k α l + α α j α k α l α α jα kα l α α j α kα l α α jα k α l. Fnally, the exponent of each generator [g,g l ] [g j,g k ]s τ,l,j,k = α γ j,k,l α jγ,k,l + α kγ,j,l + α lγ,j,k + α γ j,k,l + α j γ,k,l α k γ,j,l α l γ,j,k + α α jβ k,l + α α j β k,l α α k β j,l α α l β j,k α α kβ j,l α j α kβ,l + α jα k β,l + α jα l β,k + α α lβ j,k + α j α lβ,k α k α lβ,j α kα l β,j α α j β k,l α α jβ k,l + α α kβ j,l + α α lβ j,k + α α k β j,l + α jα k β,l α j α kβ,l α j α lβ,k α α l β j,k α jα l β,k + α kα l β,j + α k α lβ,j + β,l β j,k β,lβ j,k + α α jβ k,l α α kβ j,l α jα kβ,l + α α lβ j,k + α jα lβ,k α kα lβ,j α α j β k,l + α α k β j,l + α j α k β,l α α l β j,k α j α l β,k + α k α l β,j α α j α k α l + α α jα kα l α α jα k α l + α α j α kα l + α α j α kα l α α jα k α l α α j α k α l + α α jα kα l + α α j α kα l α α jα k α l α α j α k α l + α α jα kα l + α α j α kα l α α jα k α l. We now construct the group L n that we show s somorphc to E E. Example 3.3. Let F be the free group of rank nn 1) and set N to be F/γ 3 F )= y,j 1, j n; j, the free nlpotent group of class and rank nn 1). Set N = [y,j,y j, ], [y,j,y k,l ][y,j,y l,k ], [y,j,y k,l ][y j,,y k,l ], [y,j,y k,l ][y l,k,y j, ], [y,j,y k,l ] 3 1, j, k, l n; j; k l. Snce N N ZN ), the subgroup N s normal n N. Set W n = N /N = w,j 1, j n; j,

318 R. D. Blyth, R. F. Morse and J. L. Redden where w,j = y,j N for 1, j n and j. Set U n to be the free abelan group on the generatng set {x 1 n} {u,j,k 1, j, k n; maxj, k); j<k}. The allowable subscrpt trples for u,j,k match the four cases of subscrpt trples that arse n the generators of the tensor square E E for factors of the form g [g j,g k ]. There are n ) + n ) 3 = n+1 ) 3 such trples, and thus Un has rank n + ) n+1 3 = 1 3 nn + ). We set L n to be the drect product U n W n. Denote the commutator [w,j,w k,l ]byz,j,k,l. We represent an arbtrary element h of L n as h = x κ u λ,j,k,j,k,j,k) I 1,j,k,l) I z µ,j,k,l,j,k,l =0 j n w ηn,j n,j, 3.6) where each κ, λ,j,k and η n,j s an nteger and each µ,j,k,l s an nteger modulo 3. Let h = x κ,j,k,j,k,l w η n,j n,j 3.7) u λ,j,k,j,k) I 1 z µ,j,k,l,j,k,l) I =0 j n be another element of L n, where each κ, λ,j,k and η n,j s an nteger and each µ,j,k,l s an nteger modulo 3. Then the product hh s where hh = x κ u λ,j,k,j,k,j,k) I 1 z µ,j,k,l,j,k,l,j,k,l) I =0 j n w η n,j n,j, and computng modulo 3) κ = κ + κ, η,j = η,j + η,j, λ,j,k = λ,j,k + λ,j,k 3.8) µ,j,,k = µ,j,,k + µ,j,,k + η j, η k, η,j η k, + η,k η j, η,k η,j, <j<k, µ,j,j,k = µ,j,j,k + µ,j,j,k + η j, η k,j η,j η k,j + η j,k η j, + η,j η j,k, <j<k, µ,k,j,k = µ,k,j,k + µ,k,j,k η k,j η k, + η j,k η k, η,k η k,j + η,k η j,k, <j<k, µ,j,k,l = µ,j,k,l + µ,j,k,l + η j, η l,k η,j η l,k η j, η k,l + η,j η k,l, <j<k<l, µ,k,j,l = µ,k,j,l + µ,k,j,l + η k, η l,j η,k η l,j + η j,l η k, + η,k η j,l, <j<k<l, µ,l,j,k = µ,l,j,k + µ,l,j,k η k,j η l, + η j,k η l, η,l η k,j + η,l η j,k, <j<k<l. Proof of Theorem 1.4. Let g, g, g be arbtrary elements of E, where g and g are defned n.4) and.5), respectvely, and g = g α 1 j<k n [g j,g k ] β j,k 1 r<s<t n [g r,g s,g t ] γ r,s,t, 3.9)

On computng the non-abelan tensor squares of the free -Engel groups 319 where each α and β j,k s an nteger and each γ r,s,t s an nteger modulo 3. We defne the mappng Φ : E E L n by Φg, g )= x ρ u σ,j,k,j,k,j,k) I 1,j,k,l) I z τ,j,k,l,j,k,l =0 j n w νn,j n,j, where the exponents ρ, σ,j,k, τ,j,k,l and ν n,j are as specfed n Lemma 3.1. To show that Φ s a crossed parng, we must show that Equatons 1.1) and 1.) hold. Consder Equaton 1.1) and let and Φgg,g )= Φ g g, g g )Φg, g )= x ρ x ρ u σ,j,k,j,k,j,k) I 1 u σ,j,k,j,k,j,k) I 1 z τ,j,k,l,j,k,l,j,k,l) I z τ,j,k,l,j,k,l,j,k,l) I =0 =0 j n j n w ν n,j n,j 3.30) w ν n,j n,j. 3.31) The expressons 3.30) and 3.31) are equal f the exponents of the correspondng generators are equal. Hence we must show that ρ = ρ, σ,j,k = σ,j,k, τ,j,k,l τ,j,k,l mod 3, and ν,j = ν,j. We start by showng that ρ = ρ. For each, where 1 n, consder α, α, ρ, κ defned n.6),.7), 3.4) and 3.8), respectvely) as functons of two varables. Thus α α,α )=α + α, α α,α )=α, ρ α,α )=α α and κ κ,κ )=κ + κ. By the defnton of Φ and the formulae for computng n E and n L n, we obtan and ρ = ρ α α,α ),α ) = ρ α + α,α ) =α + α )α = α α + α α ρ = κ ρ α α,α ),α α,α )),ρ α,α )) = κ ρ α,α ),ρ α,α )) = ρ α,α )+ρ α,α ) = α α + α α. By nspecton, ρ = ρ, as needed. A smlar argument shows that ν,j = ν,j for each par, j), where 1, j n and j.

30 R. D. Blyth, R. F. Morse and J. L. Redden We next show σ,j,k = σ,j,k for two of the four possble orderngs of the ndces, j and k. Frst we wrte β,j, β,j and λ,j,k defned n.6),.7) and 3.8), respectvely) as functons: β,jα,α j,β,j,α,α j,β,j) =β,j + β,j α α j, β,j α,α j,β,j,α,α j,β,j) =β,j α α k + α α j, λ,j,kλ,j,k,λ,j,k) =λ,j,k + λ,j,k. Consder the case when the ndces are ordered < j < k. In functonal notaton, Equaton 3.5) becomes σ,j,k α,α j,α k,β,j,β,k,β j,k,α,α j,α k,β,j,β,k,β j,k) = α β j,k + α β j,k α α jα k + α α j α k + α α j α k α α jα k + α kβ,j α k β,j. Usng the defnton of Φ and the formulae for computng n E and n L n, we express σ,j,k = σ,j,kα α,α ),α j α j,α j),α kα k,α k),β,jα,α j,β,j,α,α j,β,j), β,kα,α k,β,k,α,α k,β,k),β j,kα j,α k,β j,k,α j,α k,β j,k), α,α j,α k,β,j,β,k,β j,k) 3.3) and σ,j,k = λ,j,kσ,j,k α α,α ),α j α j,α j),α k α k,α k),β,j α,α j,β,j,α,α j,β,j), β,k α,α k,β,k,α,α k,β,k),β j,k α j,α k,β j,k,α j,α k,β j,k), α α,α ),α j α j,α j ),α k α k,α k),β,j α,α j,β,j,α,α j,β,j), β,k α,α k,β,k,α,α k,β,k),β j,k α j,α k,β j,k,α j,α k,β j,k)), σ,j,k α,α j,α k,β,j,β,k,β j,k,α,α j,α k,β,j,β,k,β j,k)). 3.33) Maple [19] was then used to compose these functons and to smplfy the resultng expressons. All subsequent calculatons were performed wth Maple and ndependently checked usng Gap [11]. In partcular, these computatons mmedately show that σ,j,k σ,j,k =0. The correspondng computaton of the case σ,,j σ,,j, where <j, results n the expresson α j ) α α j ) α α α α j + α j Snce, for any two ntegers m and n, the dentty ) ) ) m + n m n = + α + α ). + mn 3.34)

On computng the non-abelan tensor squares of the free -Engel groups 31 holds, ths expresson also evaluates to zero. By smlar calculatons and applcatons of 3.34), for each of the remanng two cases, σ,j,k σ,j,k s also zero. The exponent of z,j,k,l has sx possble orderngs of the ndces, j, k and l. The case τ,j,,k τ,j,,k mod 3 for <j<kyelds the expresson ) ) β j,k α α α α + + α + α )) ) α + α jα k α α α α + α + α )) )) ) ) + α j α k α α α + α j α k α α α α + + α + α )) + α j α + α ) ) )) α k α α α α ) α + α j α k α α +α α α + α )) +. By the dentty 3.34) and the fact that m ) m m mod 3, we conclude that τ,j,,k τ,j,,k 0 mod 3. For each of the other fve possble orderngs of the ndces, j, k and l, the expresson τ,j,k,l τ,j,k,l mod 3 was smlarly computed usng Maple and verfed to be congruent to zero modulo three. These computatons show that Equaton 1.1) holds for Φ. Equaton 1.) holds by a smlar analyss. Thus Φ s a crossed parng. By Proposton 1.3, Φ determnes a unque homomorphsm Φ : E E L n wth Φ g g )=Φg, g ) for all g, g E. It follows that Φ g g )=Φg,g )=x, Φ g g j )=Φg,g j )=w,j, Φ g [g j,g k ]) = Φg, [g j,g k ]) = u,j,k, Φ [g,g j ] [g k,g l ]) = Φ[g,g j ], [g k,g l ]) = z,j,k,l. It remans to be shown that Φ s one-to-one and onto. Let h, represented as n 3.6), be an arbtrary element of L n and defne V n E Eto be V = 1 n g g ) κ g [g j,g k ]) λ,j,k,j,k) I 1 [g,g j ] [g k,g l ]) µ,j,k,l,j,k,l) I =0 j n g n g j ) ηn,j. 3.35)

3 R. D. Blyth, R. F. Morse and J. L. Redden Then Φ V )= = 1 n = h. Φ g g ) κ x κ,j,k) I 1 Φ g [g j,g k ]) λ,j,k,j,k,l) I Φ [g,g j ] [g k,g l ]) µ,j,k,l u λ,j,k,j,k,j,k) I 1,j,k,l) I z µ,j,k,l,j,k,l =0 =0 j n j n Φ g n g j ) ηn,j w ηn,j n,j Hence Φ s onto. Suppose Φ V )=1 Ln, where V s defned as n 3.35). It s mmedate from the defntons of L n and Φ that V =1 E E. We conclude that Φ : E E L n s one-to-one, and therefore s an somorphsm. Proof of Corollary 1.5. Let B denote the Burnsde group of rank n and exponent 3 and let ) ) n +1 n +1 fn) =n + +3 + nn 1), 3 4 whch enumerates both the generators lsted n Equaton 3.7) and the generators x, u,j,k, z,j,k,l and w n,j of L n. As we noted n Remark 3., our analyss holds for B B by replacng nteger wth nteger modulo 3 throughout. It follows that B B 3 fn). Let ḡ, ḡ, ḡ Bhave representatons analogous to.4),.5) and 3.9), where all of the exponents are taken modulo 3. We defne Ψ : B B L n /L 3 n by Ψḡ, ḡ )=ΨgE 3,g E 3 )=Φg, g )L 3 n. Snce Φ s a crossed parng, a smlar argument shows that Ψ s a crossed parng. Thus, by Proposton 1.3 and the fact that Ψ s onto, L n /L 3 n s a homomorphc mage of B B. Thus 3 fn) B B L n /L 3 n =3 fn), provng our clam. Acknowledgements. R.F.M. ntated ths research durng a vst at the Unversty of St Andrews, whch was supported by a grant from the Centenary Fund of the Ednburgh Mathematcal Socety and an ARSAF grant from the Unversty of Evansvlle. He thanks S. A. Lnton for obtanng the grant from the Ednburgh Mathematcal Socety and hs hosptalty whle vstng there. Ths paper s part of J.L.R. s PhD dssertaton [17] at Sant Lous Unversty. References 1. R. Aboughaz, Produt tensorel du group d Hesenberg, Bull. Soc. Math. France 115 1987), 95 106.. M. R. Bacon, The nonabelan tensor square of a nlpotent group of class, Glasgow Math. J. 36 1994), 91 96.

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