The Formanek-Procesi group with base a right-angled Artin group: Residual nilpotence and Lie algebra
aa r X i v : . [ m a t h . R A ] F e b The Formanek-Procesi group with base a right-angled Artingroup: Residual nilpotence and Lie algebra
V. Metaftsis and A.I. Papistas
Abstract
In the present work, we investigate the Lie algebra of the Formanek-Procesi group
F P ( H ) with base group H a right-angled Artin group. We show that the Lie algebragr(FP(H)) has a presentation that is dictated by the group presentation. As a result, weare able to show that F P ( H ) is residually nilpotent. Contents Z -modulevia the adjoint representation 6 Z -module I/γ ( I ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 FP( H ) J P Y and L ( E ≀
Ω) . . . . . . . . . . . . . . . . . . . . . 184.2 The decomposition of J ∩ L (Ω) . . . . . . . . . . . . . . . . . . . . . . . . . . 214.3 A decomposition of J . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224.4 The case θ n = ∅ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 gr( F n +1 /N ) J Y in gr( F n +1 ) . . . . . . . . . . . . . . . . . . . . . . . . . . . 286.2 Recognition of L ( E ≀
Ω) in gr( F n +1 ) . . . . . . . . . . . . . . . . . . . . . . . . 296.3 Recognition of I Ω in gr( F n +1 ) . . . . . . . . . . . . . . . . . . . . . . . . . . . 306.4 Proof of Theorem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336.4.1 A presentation of gr( F P ( H )) . . . . . . . . . . . . . . . . . . . . . . . 336.4.2 F n +1 /N is residually nilpotent . . . . . . . . . . . . . . . . . . . . . . 35 The Formanek-Procesi group was introduced in [9] as a counterexample to the linearity ofAut( F n ), the group of automorphisms of a free group F n of rank n , with n ≥
3. For any1roup G , the Formanek-Procesi group is the HNN-extension with presentation F P ( G ) = h t, G × G | t − ( g, g ) t = (1 , g ) , g ∈ G i . It was shown in [9] that
F P ( F n ) is not linear. On the other hand, it was shown in [15] that anHNN-extension with base a finitely generated abelian group is linear if and only if is residuallyfinite. Hence, if A is a finitely generated abelian group, then F P ( A ) is an HNN-extensionwith base an abelian group and, by results of [1], it is residually finite and hence, it is linear.A right-angled Artin group (raag) is a finitely generated group with a standard presenta-tion of the form H = h a , . . . , a k | [ a i , a j ] = 1 for certain pairs ( i, j ) with i, j ∈ { , . . . , k }i .The generating set { a , . . . , a k } is the standard generating set for H . The defining graph Γof H is the graph whose vertices correspond to infinite cyclic groups generated by h a i i andwhose edges connect a pair of vertices a i , a j if and only if [ a i , a j ] = 1. For an extensive surveyconcerning raags, one may consult [6] and the references therein. It is obvious that if H isnot abelian, then H contains free subgroups and so, F P ( H ) is not linear.In this work, we investigate the associated Lie algebra of F P ( H ) by providing an Elim-ination method for gr(FP(H)). The Elimination method was introduced by Lazard in [13]for the study of nilpotent groups and N -series. In [7], Duchamp and Krob study the freepartially commutative Lie algebra and its universal enveloping algebra and prove a partiallycommutative version of Lazard’s elimination, thus providing a presentation of the Lie algebraof a partially commutative set.In the present work, we prove a Lazard’s Elimination version of the free Lie algebraby decomposing it into a summand of free Lie algebras and ideals that originate by thevarious relators that appear in the Formanek-Procesi group with base a raag. To achievethis decomposition, besides the classical Lazard’s Elimination, we also use results of Labute[11, 12] and of Duchamp and Krob [7]. As a result, we are able to get a presentation of theLie algebra of the Formanek-Procesi group with basis a raag and to show that the Formanek-Procesi group with basis a raag is residually nilpotent. A first attempt in that direction wasmade in [16]. In the process, we recover results of [ ? ] concerning the Lie algebra of a raag.To be more specific, for a positive integer n , with n ≥
2, let [ n ] = { , . . . , n } and ∆ n =[ n ] × [ n ] \ { ( a, a ) | a ∈ [ n ] } . Let θ n be a non-empty subset of ∆ n such that θ n containsboth ( a, b ) and ( b, a ) for a, b ∈ [ n ]. We call θ n a partial commutation relation on [ n ]. Define θ n,δ = { ( a, b ) ∈ θ n | a > b } . Notice that θ n,δ is the edge set of a defining graph of a raag H .Now, let H and H two copies of a raag with presentations H i = h a i, , . . . , a i,n | [ a i,k , a i,l ] = 1 for all ( k, l ) ∈ θ n,δ i , i = 1 , F P ( H ) to be the group with generating set t, a , , . . . , a ,n , a , , . . . , a ,n and defin-ing relations ta ,k a ,k t − = a ,k , k ∈ [ n ] , [ a i,r , a i,s ] = 1 , i = 1 , , ( r, s ) ∈ θ n,δ and[ a ,k , a ,l ] = 1 , k, l ∈ [ n ] . Using the first set of the above relation, we may solve for a ,k , k ∈ [ n ] and replace into theother two sets of relations to get a presentation with generators t, a ,k , k ∈ [ n ] and relations[ t, a ,k , a ,l ] = 1 , [ a ,r , a ,s ] = 1 , [[ t, a ,r ] , [ t, a ,s ]] = 1 , for k, l ∈ [ n ] , ( r, s ) ∈ θ n,δ y i = a ,i , we finally have F P ( H ) = h t, y , . . . , y n | [ t, y k , y l ] = [ y r , y s ] = [[ t, y r ] , [ t, y s ]] = 1 , k, l ∈ [ n ] , ( r, s ) ∈ θ n,δ i For a positive integer n , with n ≥
2, let gr(F n+1 ) be the free Lie algebra (over Z ) withfree generating set { t, y , . . . , y n } and J be the ideal of gr(F n+1 ) generated by the set J = { [ t, y k , y l ] , [ y r , y s ] , [[ t, y r ] , [ t, y s ]] , k, l ∈ [ n ] , ( r, s ) ∈ θ n,δ } . Our main theorem is the following.
Theorem 1
With the previous notations, let G = F P ( H ) . Then,1. gr( G ) ∼ = gr( F n +1 ) /J as Lie algebras.2. G is residually nilpotent. In particular, G is a Magnus group. The structure of the paper is the following. In section 2, we introduce some notationand we give some known results concerning Lazard’s Elimination method and a method onchanging basis on a Lie algebra using the free associative Z -algebra. In section 3, we give analgorithm that inductively describes the basis of an ideal I of the free Lie algebra generatedby a set described by a partial commutative relation. Moreover, we describe the Z -module I/γ ( I ) via the adjoint representation. In section 4, we analyze the ideal generated by theFormanek-Procesi relations with base group a raag into various pieces. Section 5 containssome group theoretic observations and their connection with the Lie algebra. Finally, inSection 6, we put things together by showing that the various pieces of the ideal generatedby the Formanek-Procesi relations with base group a raag are direct summands of the freeLie algebra. That way we are able to describe the presentation of the Lie algebra of theFormanek-Procesi group and prove that the group is residually nilpotent. Let G be any group. For non-empty subsets X and Y of G , we write [ X, Y ] for the subgroupof G generated by the (group) commutators [ x, y ] = x − y − xy with x ∈ X and y ∈ Y . Fora positive integer c , let γ c ( G ) be the c -th term of the lower central series of G . That is, γ ( G ) = G and, for c ≥ γ c ( G ) = [ γ c − ( G ) , G ]. We write G ′ = γ ( G ) for the commutatorsubgroup of G . The associated graded abelian group gr( G ) = L c ≥ gr c ( G ), where gr c ( G ) = γ c ( G ) /γ c +1 ( G ), has the structure of a graded Lie algebra over Z , the Lie bracket operationin gr( G ) being induced by the commutator operation in G (see, for example, [3], [13], [14]).A group G is called residually nilpotent if to each g = 1 in G there corresponds a normalsubgroup K g such that G/K g is nilpotent and g / ∈ K g . An equivalent assertion is that G is residually nilpotent if T c ≥ γ c ( G ) = { } . A group G is called a Magnus group if G isresidually nilpotent and each quotient group γ c ( G ) /γ c +1 ( G ) is torsion-free.For a positive integer n , with n ≥
2, let F n be a free group of rank n with a free generatingset { x , . . . , x n } . For each c ∈ N , the quotient group gr c ( F n ) is a free abelian group offinite rank. Furthermore, gr( F n ) is a free Lie algebra over Z with a free generating set { x F ′ n , . . . , x n F ′ n } (see [14]). Let H be a subgroup of F n . For d ∈ N , let H d = H ∩ γ d ( F n )3nd so, for d = 1, we get H = H . For d ≥
1, we have H d +1 = H d ∩ γ d +1 ( F n ). Since H is asubgroup of F n , we obtain a descending series of subgroups of H ( H d ) d ≥ : H = H ⊇ H ⊇ H ⊇ · · · . Since [ H κ , H λ ] ⊆ H κ + λ for all κ, λ ∈ N , we get the series ( H d ) d ≥ is a strongly central seriesof H . For d ∈ N , we define I d ( H ) = H d γ d +1 ( F n ) /γ d +1 ( F n ) ≤ gr d ( F n ). Form the (restricted)direct sum L ( H ) = L d ≥ I d ( H ) of the abelian groups I d ( H ). Since the series ( H d ) d ≥ isstrongly central, we have L ( H ) is a Lie subalgebra of gr( F n ). In the case H = N is a normalsubgroup of F n , the Lie subalgebra L ( N ) is an ideal of gr( F n ) (see [13]).By a Lie algebra L , we mean a Lie algebra over the ring Z of integers and by U ( L )we denote the universal enveloping algebra of L (see [3]). For a positive integer n , with n ≥
2, let V n be a free Z -module with a Z -basis the set V n = { v , . . . , v n } . We write T ( V n ) for the free associative Z -algebra (with identity element) on V n (that is, T ( V n ) isa free associative Z -algebra such that V n is a free Z -submodule of T ( V n ) and every basisof V n is a free generating set of T ( V n )). Hence, we may write T ( V n ) = T ( V n ) and so, T ( V n ) = L r ≥ T r ( V n ), where T ( V n ) = Z , T ( V n ) = V n and, for all r ≥ T r ( V n ) is the Z -submodule of T ( V n ) freely spanned by all elements of the form v i · · · v i r with v i , . . . , v i r ∈ V n .For r = 0, the trivial product v i · · · v i r is interpreted as the neutral element 1 in T ( V n ). Give T ( V n ) the structure of a Lie algebra by defining the Lie product [ u, v ] = uv − vu for all u, v ∈ T ( V n ). The Lie subalgebra of T ( V n ) generated by the set V n is denoted by L ( V n ) (orsimply L n ). It is well known that L n is a free Lie algebra on the set V n (see [3]). For apositive integer c , let L cn = T c ( V n ) ∩ L n . That is, L cn is the Z -submodule of L n generated byall Lie commutators [ v i , . . . , v i c ], i , . . . , i c ∈ [ n ]. The Z -submodule of L cn is called the c -thhomogeneous component of L n . Thus, L n = L c ≥ L cn . It is well known that L n and gr( F n )are naturally isomorphic as Lie algebras by an isomorphism χ n : L n → gr( F n ) satisfying theconditions χ n ( v i ) = x i F ′ n , i ∈ [ n ]. In the light of the isomorphism χ n , we identify L n withgr( F n ) and write, for any c ≥
2, [ v i , . . . , v i c ] = [ x i , . . . , x i c ] γ c +1 ( F n ) for all i , . . . , i c ∈ [ n ].Furthermore, for any c ≥ L cn = gr c ( F n ) and so, I d ( N ) ≤ L dn for all d ≥ L n be a free Lie algebra of rank n , with n ≥
2, freely generated by the set { v , . . . , v n } .Then, v , . . . , v n are the Lie monomials of degree 1. If u is a Lie monomial of degree r and v is a Lie monomial of degree s , with u = v , then [ u, v ] is a Lie monomial of degree r + s .M. Hall proved in [10, Theorem 3.1] that the basic Lie monomials form a Z -basis of L n . Inthe free group F n freely generated by the set { x , . . . , x n } , as observed in [10], we may definebasic group commutators in terms of x , . . . , x n . In [10, Theorem 4.1] proved that, for each c ∈ N , the quotient group gr c ( F n ), being a free abelian group of finite rank n c , has a Z -basis { w γ c +1 ( F n ) , . . . , w n c γ c +1 ( F n ) } where each w i is a basic group commutator of weight c (see,also, [14, Theorem 5.12, Corollary 5.12 (iv)]).Throughout this paper, we use the left-normed convention for Lie commutators. Forsubsets B and C of any Lie algebra, we write C ≀ B = { [ c, b , . . . , b k ] : c ∈ C , b , . . . , b k ∈ B , k ≥ } . The following result is a version of Lazard’s “Elimination Theorem” (see [3, Chapter 2,Section 2.9, Proposition 10]). Theorem 2 (Elimination)
Let A = B ∪ C be the disjoint union of its proper non-emptysubsets B and C and consider the free Lie algebra L ( A ) . Then, B and C ≀ B freely generate Liealgebras L ( B ) and L ( C ≀ B ) , and there is a Z -module decomposition L ( A ) = L ( B ) ⊕ L ( C ≀ B ) .Furthermore, L ( C ≀ B ) is the ideal of L ( A ) generated by C . L ( A ) be the free Lie algebra on A . For all non-empty subsets V , U of L ( A ), we write S ( V , U ) = { [ v, u ] : v ∈ V , u ∈ U } . We inductively define the subsets ( A m ) m ≥ of L ( A ) by A = A and, for all m ≥ , A m = S p + q = m ( S ( A p , A q ) \ { } ). For m ≥
1, an element of A m is called a simple Lie commutator of degree m . The following result has been proved in [7,Lemma II.5]. Lemma 1
Let A = B ∪ C be the disjoint union of its proper subsets B and C and considerthe free Lie algebra L ( A ) . Let u be a simple Lie commutator of L ( A ) . Then, either u ∈ L ( B ) or u ∈ L ( C ≀ B ) . Let V be a free Z -module with a Z -basis a finite non-empty set A . Write A = B ∪ C forthe disjoint union of its non-empty proper subsets B and C . Let T ( V ) be the free associative Z -algebra on V . Thus, T ( V ) is a free associative Z -algebra subject to any Z -basis of V and T ( V ) = T ( A ). Let L ( A ) be the Lie subalgebra of T ( A ) generated by the set A , which is afree Lie algebra on A . By Theorem 2, L ( A ) = L ( B ) ⊕ L ( C ≀ B ).The aim of this section is to construct a Z -basis of L ( C ≀ B ), by means of the Poincare-Birkhoff-Witt theorem (see [3]), suitable to our purposes. Let B = { b , . . . , b m } and B ∗ be thefree monoid over the set B . That is, B ∗ = { ε } ∪ ( S t ≥ B t ), where B t = { b i · · · b i t : i , . . . , i t ∈ [ m ] } . Since, for t ≥ B t ∼ = B × · · · × B | {z } t , we have |B t | = |B| t . For any c ∈ C and b i · · · b i t ∈ B t ,we write [ c ; b i · · · b i t ] = [ c, b i , . . . , b i t ] ∈ L t +1 ( A ). For a positive integer r , with r ≥
2, let U r = { [ c ; b i · · · b i r − ] : c ∈ C ; b i · · · b i r − ∈ B r − } and so, C ≀ B = S r ≥ U r with U = C . For r ≥
1, let U r be the Z -module spanned by the set U r . Since L ( C ≀ B ) is a free Lie algebra on
C ≀ B , we obtain the set U r is a Z -basis of U r for all r .Let T ( B ) be the free associative Z -algebra on B being a subalgebra of T ( A ). Hence, T ( B ) = L r ≥ T r ( B ). The set B r is a Z -basis of T r ( B ). Furthermore, T ( B ) has the structureof a Lie algebra. The Lie subalgebra of T ( B ) spanned by the set B is the free Lie algebra L ( B ). We denote by V C the free Z -module with a Z -basis C . By Theorem 2, it follows that,for any r ≥
1, there exists a Z -module isomorphism ξ r : U r → V C ⊗ Z T r − ( B ) subject to ξ r ([ c, b j , . . . , b j r − ]) = c ⊗ b j · · · b j r − for all c ∈ C , b j , . . . , b j r − ∈ B and so, there exists a Z -module isomorphism ξ : L r ≥ U r → V C ⊗ Z T ( B ) such that ξ ( u + · · · + u κ ) = ξ ( u )+ · · · + ξ κ ( u κ )for all u i ∈ U i , i = 1 , . . . , κ .The following result is well known (and it can be easily proved). Lemma 2
1. In T ( B ) , for any positive integer n , with n ≥ , and κ ∈ { , . . . , n − } , let S n,κ be the set of all permutations σ ∈ S n satisfying σ (1) > · · · > σ ( κ ) > σ ( κ + 1) < σ ( κ + 2) < · · · < σ ( n ) . Then, for a , . . . , a n ∈ T ( B ) , [ a , . . . , a n ] = n − X κ =0 X σ ∈ S n,κ ( − κ a σ (1) · · · a σ ( n ) .
2. For all c ∈ C and b , . . . , b n , . . . , b λ , . . . , b λm λ ∈ B , ξ − ( c ⊗ [ b , . . . , b m ] · · · [ b λ , . . . , b λm λ ]) = [ c, [ b , . . . , b m ] , . . . , [ b λ , . . . , b λm λ ]] . M be any Z -basis of L ( B ) consisting of simple Lie commutators. For any positiveinteger r , the set M r = M ∩ L r ( B ) is a Z -basis of L r ( B ). So, M = S r ≥ M r . For r ≥
1, wearbitrarily order the elements of M r and extend it, by increasing the degree, to the elementsof M . We denote by M < the set of all elements of T ( B ) of the form v v · · · v κ , (2 . κ ≥ v , . . . , v κ ∈ M and v ≤ v ≤ · · · ≤ v κ . Note that the elements (2 .
1) aredistinct as written. A consequence of Poincare-Birkhoff-Witt Theorem (see [3]) gives that theset M < is a Z -basis of T ( B ). Thus, the set G = { c ⊗ v · · · v κ : c ∈ C , v · · · v κ ∈ M < , κ ≥ } is a Z -basis of the free Z -module V C ⊗ Z T ( B ). For any u = v v · · · v λ , where any v i ∈ M ν i ,we write [ c ; u ] = [ c, v , . . . , v λ ], with c ∈ V C . Since ξ − is a Z -module isomorphism, it follows,by Lemma 2 (2), that the set ξ − ( G ) = P = { [ c ; u ] : c ∈ C , u ∈ M < } is a Z -basis of L r ≥ U r .For a non-negative integer r , let M 1) of degree r . Thus,for r = 0, M < = { } . It is clear enough that the set M With the above notation, L ( A ) = L ( B ) ⊕ L ( P ) . Furthermore, L ( C ≀ B ) = L ( P ) .Proof. Since the set P is a Z -basis of L r ≥ U r , we obtain L ( C ≀ B ) = L ( P ) and so, L ( A ) = L ( B ) ⊕ L ( P ). (cid:3) Let v , . . . , v κ ∈ M and consider the product v v · · · v κ . For any permutation π of { , , . . . , κ } , the product v π (1) v π (2) · · · v π ( κ ) is called a re-arrangement of v v · · · v κ . For aproof of the following result, we refer to [4, Lemma 1]. Lemma 3 Let M be an ordered basis of L ( B ) . For each element u of M < , let u ′ be are-arrangement of u . Then, the set { u ′ : u ∈ M < } is a basis of T ( B ) . By the above discussion and Lemma 3, we may replace P with the set P ′ = { [ c ; u ′ ] : c ∈C ; u ∈ M < } . Hence, we restate Proposition 1 as follows. Proposition 2 With the previous notations, L ( A ) = L ( B ) ⊕ L ( P ′ ) . Furthermore, L ( C ≀ B ) = L ( P ) = L ( P ′ ) . Z -module via the adjoint representation Let L be any Lie algebra. For each v ∈ L , the Z -linear mapping ad v : L → L is definedby u (ad v ) = [ u, v ] for all u ∈ L . In particular, ad v is a derivation of L . The set of allderivations of L is denoted by Der L and is regarded as a Lie algebra in a natural way. Themap ad : L → Der L sending v to ad v for all v ∈ L is called the adjoint representation of L .Let U ( L ) be the universal enveloping Z -algebra of L and we regard L is contained in U ( L ).Any element of U ( L ) \ { } is a Z -linear combination of elements of the form u u · · · u m where u , . . . , u m ∈ L . 6et L be a free Lie algebra and J be an ideal of L generated by a non-empty set consistingof simple Lie commutators. We write γ ( J ) = [ J, J ] for the Z -submodule of J spanned byall [ u, v ], with u, v ∈ J , and ˜ v = v + J for all v ∈ L . Note that, since J is an ideal of L , γ ( J ) is an ideal of L . The Z -module J/γ ( J ) becomes a (right) L/J -module via the adjointrepresentation of L/J . Namely,( u + γ ( J ))˜ v = u (ad v ) + γ ( J )= [ u, v ] + γ ( J )for all u ∈ I and v ∈ L . Hence, the Z -module J/γ ( J ) becomes a (right) U ( L/J )-module bydefining ( u + γ ( J ))˜ v ˜ v · · · ˜ v κ = u (ad v )(ad v ) · · · (ad v κ ) + γ ( J )= [ u, v , v , . . . , v κ ] + γ ( J )for all u ∈ I and v , v , . . . , v κ ∈ L . For a positive integer m , with m ≥ 2, let Y = { y , y , . . . , y m } and θ m be a partial commu-tation relation in [ m ]. Let L ( Y ) be the free Lie algebra on Y and I be the ideal of L ( Y )generated by the set I = { [ y i , y j ] : ( i, j ) ∈ θ m } . For any Lie algebra L , we write γ ( L ) for thederived algebra of L .One of our aims in this section is to show that I is a direct summand of L ( Y ). Provingthis, we have L ( Y ) /I is a free Z -module. Since I is a homogeneous ideal, we get, by a resultof Witt (see, for example, [2, Section 2.4, Theorem 5. See, also, Theorem 3]), I is a freeLie algebra. Next, we present a decomposition of I where each of its summand is a free Liealgebra by explicitly giving a free generating set. Moreover, if I = { } , we show that the Z -module I/γ ( I ) is a free (right) U ( L ( Y ) /I )-module with a free generating set I + γ ( I ).If θ m = ∅ , then I = { } and so, there is nothing to prove. Without loss of generality,we may assume that (1 , ∈ θ m . Order the elements of Y as y < y < . . . < y n and define θ m,δ = { ( a, b ) ∈ θ m | a > b } . Let T ( Y ) be the universal enveloping Z -algebra of L ( Y ) and I be the ideal of L ( Y ) generated by the set I = { [ y i , y j ] : ( i, j ) ∈ θ m,δ } . For i ∈ [ m ], let Y i = { y , y , . . . , y i } . Note that Y m = Y . Furthermore, for i ∈ [ m ] \ { } , we write I Y i = { [ y a , y b ] : y a , y b ∈ Y i ; ( a, b ) ∈ θ m,δ } and I Y i for the ideal of L ( Y i ) generated by the set I Y i . Inparticular, I = I Y . Since Y ⊂ Y ⊂ · · · ⊂ Y m = Y , we have I Y ⊆ I Y ⊆ · · · ⊆ I Y m = I . Byour choice, I Y = { [ y , y ] } . Hence, I Y j = ∅ for all j ∈ [ m ] \ { } . For i ∈ [ m − \ { } , byTheorem 2, L ( Y i +1 ) = L ( Y i ) ⊕ L ( { y i +1 } ≀ Y i ) . (3 . i ∈ [ m − \ { } , I Y i +1 = ( I Y i +1 ∩ L ( Y i )) ⊕ ( I Y i +1 ∩ L ( { y i +1 } ≀ Y i )) . (3 . I Y i +1 ∩ L ( Y i ) = I Y i (3 . i ∈ [ m − \ { } . By (3 . 2) and (3 . I Y i +1 = I Y i ⊕ ( I Y i +1 ∩ L ( { y i +1 } ≀ Y i )) . (3 . heorem 3 With the previous notations, for any i ∈ [ m − , there are explicitly describedsubsets B , . . . , B i +1 of L ( { y } ≀ Y ) , . . . , L ( { y i +1 } ≀ Y i ) , respectively, consisting of simple Liecommutators of degree at least in terms of elements of Y , . . . , Y i +1 , respectively, such that L ( Y i +1 ) = hY i +1 i ⊕ L ( B ) ⊕ · · · ⊕ L ( B i +1 ) ⊕ I Y i +1 , where hY i +1 i is the free Z -module with a Z -basis the set Y i +1 .Moreover, for any i , with i ∈ [ m − \ { } , the ideal I Y i +1 is a free Lie algebra, thereare explicitly described subsets D , . . . , D i +1 of L ( { y } ≀ Y ) , . . . , L ( { y i +1 } ≀ Y i ) , respectively,consisting of simple Lie commutators of degree at least in terms of elements of Y , . . . , Y i +1 ,respectively, and I Y i +1 = γ ( L ( Y )) ⊕ L ( D ) ⊕ · · · ⊕ L ( D i +1 ) .Proof. We induct on i and let i = 1. Then, B = ∅ and, by Theorem 2, L ( { y } ≀ { y } ) = h y i ⊕ γ ( L ( Y )), where h y i is the Z -module spanned by y . Since the derived algebra γ ( L ( Y )) of L ( Y ) is generated as an ideal by the set I Y , we have I Y = γ ( L ( Y )). Thus, L ( Y ) = hY i⊕ I Y . It is well known that I Y is a free Lie algebra with a free generating set con-sisting of simple Lie commutators. Namely, I Y = L ( Y (1)2 ), where Y (1)2 = { [ y , y , α y , β y ] : α + β ≥ } . Thus, i ≥ 2. Let κ ∈ [ m − \ { } and assume that our claim is valid for κ .Therefore, there are explicitly described subsets B , . . . , B κ of L ( { y } ≀ Y ) , . . . , L ( { y κ } ≀ Y κ − ),respectively, consisting of simple Lie commutators of degree at least 2 in terms of elements of Y , . . . , Y κ , respectively, such that L ( Y κ ) = hY κ i ⊕ L ( B ) ⊕ · · · ⊕ L ( B κ ) ⊕ I Y κ . (3 . I Y κ is a free Lie algebra, there are explicitly described subsets D , . . . , D κ of L ( { y }≀Y ) , . . . , L ( { y κ } ≀ Y κ − ), respectively, consisting of simple Lie commutators of degree at least2 in terms of elements of Y , . . . , Y κ , respectively, and I Y κ = γ ( L ( Y )) ⊕ L ( D ) ⊕ · · · ⊕ L ( D κ ) . (3 . j ∈ [ κ ] \ { , } , let M D j be the ordered Z -basis of L ( D j ) consisting of the basic Liecommutators which are formed by the elements of D j . Write M Y κ ,I = M ′Y ∪M D ∪· · ·∪M D κ ,where M ′Y denotes the ordered Z -basis of γ ( L ( Y )) consisting of the basic Lie commutatorsin the elements of the standard free generating set Y (1)2 of γ ( L ( Y )) = L ( Y (1)2 ). We give theset M Y κ ,I a total ordering such that u < u < · · · < u κ for all u ∈ M ′Y and u j ∈ M D j ( j ∈ [ κ ] \ { , } ). The set M Y κ ,I is a totally ordered Z -basis of I Y κ . By (3 . 1) (for i = κ ), L ( Y κ +1 ) = L ( Y κ ) ⊕ L ( { y κ +1 } ≀ Y κ ). For each j ∈ [ κ ] \ { , } , let M B j be the ordered Z -basisof L ( B j ) consisting of the basic Lie commutators which are formed by the elements of B j .Write M Y κ = Y κ ∪ M B ∪ · · · ∪ M B κ ∪ M Y κ ,I . We give the set M Y κ a total ordering suchthat w < w < · · · < w κ < w I for all w ∈ Y κ , w j ∈ M B j ( j ∈ [ κ ] \ { , } ) and w I ∈ M Y κ ,I .By (3 . 5) and (3 . M Y κ is an ordered Z -basis of L ( Y κ ).For any positive integer r , the set M Y κ ,r = M Y κ ∩ L r ( Y κ ) is a Z -basis of L r ( Y κ ). So, M Y κ = ∪ r ≥ M Y κ ,r . Note that M Y κ , = Y κ . Let M < Y κ be the Z -basis of T ( Y κ ) consisting ofall elements of the form u · · · u µ , where µ ≥ u , . . . , u µ ∈ M Y κ and u ≤ u ≤ · · · ≤ u µ . Atypical element of M < Y κ \ { } has the form u = y α · · · y α κ κ v v · · · v ν , (3 . α , . . . , α κ ≥ 0, each v i ∈ ( S κj =3 M B j ) ∪ M Y κ ,I ( i ∈ [ ν ]) and v ≤ · · · ≤ v ν . The set P Y κ = { [ y κ +1 ; u ] : u ∈ M < Y κ } is a free generating set of L ( { y κ +1 } ≀ Y κ ). Thus, L ( { y κ +1 } ≀ Y κ ) =8 ( P Y κ ). For any κ -tuple of non negative integers a κ = ( α , . . . , α κ ) ∈ N κ , where N is the setof non negative integers, and r ∈ N , we write y ( a κ ; r ) = y α · · · y α κ κ = y · · · y | {z } α y · · · y | {z } α · · · y κ · · · y κ | {z } α κ with α + · · · + α κ = r . For r = 0, we interpret y ( a κ ; 0) as 1. Let S Y κ ,r = { y ( a κ ; r ) : a κ ∈ N κ } with S Y κ , = { } . So, (3 . 7) becomes u = y ( a κ ; r ) v v · · · v ν , (3 . y ( a κ ; r ) ∈ S Y κ ,r , α , . . . , α κ ≥ α + · · · + α κ = r ∈ N , each v i ∈ ( S κj =3 M B j ) ∪ M Y κ ,I ( i ∈ [ ν ]) (when it occurs in the expression) and v ≤ · · · ≤ v ν .Fix r ∈ N , with r ≥ 2. We replace each element y ( a κ ; r ) of S Y κ ,r by its re-arrangementdenoted by y ( a κ ; r ; π ), where π is a permutation of { , , . . . , r } , as follows: Suppose that theelements y j , . . . , y j λ of Y κ , with j < · · · < j λ , subject to ( κ +1 , j a ) ∈ θ m,δ for all a ∈ [ λ ], occurin y ( a κ ; r ). Then, by a suitable permutation π of { , , . . . , r } , we rewrite (a re-arrangement) y ( a κ ; r ) as y ( a κ ; r ; π ) = y a j j . . . y a jλ j λ · y a jλ +1 j λ +1 . . . y a jκ j k = y ( a κ ; r ) · y ( a κ − κ ; r − r ) , (3 . y ( a κ ; r ) = y a j j . . . y a jλ j λ , y ( a κ − κ ; r − r ) = y a jλ +1 j λ +1 . . . y a jκ j k , r = a j + · · · + a j λ . Oth-erwise, we leave y ( a κ ; r ) unchanged. Hence, any element of M < Y κ \ { } of the form (3 . 8) isreplaced, by means of (3 . u ′ = y ( a κ ; r ; π ) v v · · · v ν , (3 . v i ∈ ( S κj =3 M B j ) ∪ M Y κ ,I ( i ∈ [ ν ]), when it occurs in (3 . v ≤ · · · ≤ v ν .By Lemma 3, the set M < Y κ ,θ m = { u ′ : u ∈ M < Y κ } is a Z -basis of T ( Y κ ). Moreover, the set P ′Y κ = { [ y κ +1 ; u ′ ] : u ′ ∈ M < Y κ ,θ m } is a free generating set of the free Lie algebra L ( { y κ +1 } ≀ Y κ ).By Theorem 2 and by Proposition 2, L ( Y κ +1 ) = L ( Y κ ) ⊕ L ( P ′Y κ ) . (3 . . 10) (the writing of the elements of M < Y κ ,θ m ), a typical element of P ′Y κ \ { y κ +1 } has theform [ y κ +1 ; u ′ ] = [ y κ +1 ; y ( a κ ; r ; π ) v v · · · v ν ] , where each v i ∈ ( S κj =3 M B j ) ∪ M Y κ ,I ( i ∈ [ ν ]), when it occurs in (3 . v ≤ · · · ≤ v ν .We decompose the set P ′Y κ into its subsets, P ′Y κ = P ′Y κ , , ∪ P ′′Y κ , as follows. Let P ′Y κ , bethe subset of P ′Y κ consisting of all elements of the form [ y κ +1 ; u ′ ], with u ′ is element of theform (3 . 10) and y ( a κ ; r ) = 1, and P ′Y κ , is the subset of P ′Y κ consisting of all elements ofthe form [ y κ +1 ; u ′ ], with u ′ of the form (3 . y ( a κ ; r ) = 1 and at least one of v , . . . , v ν belongs to M Y κ ,I . Write P ′Y κ , , = P ′Y κ , ∪ P ′Y κ , and let P ′′Y κ = P ′Y κ \ P ′Y κ , , . That is, P ′′Y κ is the union of { y κ +1 } and the set P ′′Y κ ,y κ +1 of all elements of the form [ y κ +1 ; u ′ ], where u ′ isthe form (3 . 10) with y ( a κ ; r ) = 1 and each v i ∈ S κj =3 M B j ( i ∈ [ ν ]) (if they occur in (3 . v ≤ . . . ≤ v ν . Thus, P ′′Y κ = { y κ +1 } ∪ P ′′Y κ ,y κ +1 . By Theorem 2, L ( P ′Y κ ) = L ( P ′′Y κ ) ⊕ L ( P ′Y κ , , ≀ P ′′Y κ ) (3 . L ( P ′Y κ , , ≀ P ′′Y κ ) is the ideal of L ( P ′Y κ ) generated by the set P ′Y κ , , . By (3 . 4) (for i = κ ), I Y κ +1 = I Y κ ⊕ ( I Y κ +1 ∩ L ( { y κ +1 } ≀ Y κ )) . (3 . Claim. We claim that I Y κ +1 ∩ L ( { y κ +1 } ≀ Y κ ) = L ( P ′Y κ , , ≀ P ′′Y κ ). It is clear enough that P ′Y κ , and P ′Y κ , are contained in I Y κ +1 ∩ L ( { y κ +1 } ≀ Y κ ) and since I Y κ +1 ∩ L ( { y κ +1 } ≀ Y κ ) is anideal in L ( Y κ +1 ), we have L ( P ′Y κ , , ≀ P ′′Y κ ) ⊆ I Y κ +1 ∩ L ( { y κ +1 } ≀ Y κ ) . (3 . T Y κ , = { [ y κ +1 ; zy p · · · y p s ] : z = y ν ∈ Y κ ; y p · · · y p s ∈ Y ∗ κ ; ( κ + 1 , ν ) ∈ θ m,δ } and T Y κ , = { [ y κ +1 ; z · · · z µ [ y a , y b ] z ′ · · · z ′ ν ] : z · · · z µ , z ′ · · · z ′ ν ∈ Y ∗ κ ; y a , y b ∈ Y κ ; ( a, b ) ∈ θ m } , where Y ∗ κ is the free monoid on Y κ . By a result of Duchamp and Krob [7, Theorem II.7], I Y κ +1 ∩ L ( { y κ +1 }≀Y κ ) is the ideal of L ( { y κ +1 }≀Y κ ) = L ( P ′Y κ ) generated by the set T Y κ , ∪T Y κ , .A natural Z -basis of T ( Y κ ) is the free monoid Y ∗ κ of Y κ . We observe that by using the identity ab = ba + [ a, b ] for all a, b ∈ T ( Y κ ), the Jacobi identity, as many times as needed and the factthat the set M Y κ is a Z -basis of L ( Y κ ), we have, for any p , . . . , p s ∈ [ κ ], with s ≥ y p · · · y p r = y α · · · y α κ κ + X finite c w w, (3 . α , . . . , α κ ≥ α + · · · + α κ = r , c w ∈ Z and w is of the form (3 . L ( Y κ +1 ), by using the Jacobi identity in the form [ x, y, z ] = [ x, z, y ] + [ x, [ y, z ]], wehave [ y κ +1 ; zy p · · · y p r ] = [ y κ +1 ; zy α · · · y α κ κ ] + X finite c w [ y κ +1 ; zw ] . (3 . τ ∈ [ r ], ( κ + 1 , p τ ) / ∈ θ m,δ . By (3 . y κ +1 ; zy p · · · y p r ] is a Z -linear combination of elements of P ′Y κ , .2. Suppose that the elements y j , . . . , y i λ of Y κ , with j < · · · < j λ , subject to ( κ +1 , j a ) ∈ θ m,δ for all a ∈ [ λ ], occur in y p · · · y p r . Then, as before, by using the identity ab = ba + [ a, b ] for all a, b ∈ T ( Y κ ) and the Jacobi identity, as many times as needed,and the fact that the set M Y κ is a Z -basis of L ( Y κ ), the element zy p · · · y p r is writtenas a sum of an element of the form (3 . 9) and P finite a w ′ w ′ , where a w ′ ∈ Z and w ′ has theform (3 . . y κ +1 ; zy p · · · y p r ] is a Z -linear combination of elementsof P ′Y κ , .Thus, in any case, any element of T Y κ , belongs to L ( P ′Y κ , , ≀ P ′′Y κ ). Hence, T Y κ , is a subsetof L ( P ′Y κ , , ≀ P ′′Y κ ). Next, we need the following identity which is recorded as [5, (2.7)]. If a, b, c , . . . , c r are elements of an arbitrary Lie algebra, then[ a, b, c , . . . , c r ] = r X i =0 X σ ∈ K i [[ a, c σ (1) , . . . , c σ ( i ) ] , [ b, c σ ( i +1) , . . . , c σ ( r ) ]] , (3 . i = 0 , , . . . , r , K i is the set of all permutations of { , . . . , r } such that σ (1) < · · · <σ ( i ) and σ ( i + 1) < · · · < σ ( r ). We put a = [ y κ +1 , z , . . . , z µ ] and b = [ y a , y b ]. Then,[ y κ +1 , z , . . . , z µ , [ y a , y b ] , z ′ . . . , z ′ ν ] = [ a, b, z ′ . . . , z ′ ν ](by (3 . P νi =0 P σ ∈ K i [[ a, z ′ σ (1) , . . . , z ′ σ ( i ) ] , [ b, z ′ σ ( i +1) , . . . , z ′ σ ( ν ) ]]= [ a, z ′ , . . . , z ′ ν , b ] + [ a, [ b, z ′ , . . . , z ′ ν ]]+ P ν − i =1 P σ ∈ K i [[ a, z ′ σ (1) , . . . , z ′ σ ( i ) ] , [ b, z ′ σ ( i +1) , . . . , z ′ σ ( ν ) ]] . Since I Y κ is the ideal of L ( Y κ ) generated by the set I Y κ = { [ y a , y b ] : y a , y b ∈ Y κ ; ( a, b ) ∈ θ m } , we obtain each [ b, z ′ σ ( i +1) , . . . , z ′ σ ( ν ) ] ∈ I Y κ . Thus, each [ b, z ′ σ ( i +1) , . . . , z ′ σ ( ν ) ] is a Z -linearcombination of elements of M Y κ ,I . Having in mind (3 . 17) (and the above discussion), (3 . x, y, z ] = [ x, z, y ] + [ x, [ y, z ]] as many times asneeded, we have each element of T Y κ , as a Z -linear combination of elements of L ( P ′Y κ , , ≀P ′′Y κ ).Thus, T Y κ , is a subset of L ( P ′Y κ , , ≀ P ′′Y κ ). Since I Y κ +1 ∩ L ( { y κ +1 } ≀ Y κ ) is the ideal in L ( P ′Y κ )generated by the set T Y κ , ∪ T Y κ , and L ( P ′Y κ , , ≀ P ′′Y κ ) is an ideal of L ( P ′Y κ ), we have I Y κ +1 ∩ L ( { y κ +1 } ≀ Y κ ) ⊆ L ( P ′Y κ , , ≀ P ′′Y κ ) . (3 . . 14) and (3 . I Y κ +1 ∩ L ( { y κ +1 } ≀ Y κ ) = L ( P ′Y κ , , ≀ P ′′Y κ ) (3 . . 12) and the Claim, L ( P ′Y κ ) = L ( P ′′Y κ ) ⊕ ( I Y κ +1 ∩ L ( { y κ +1 } ≀ Y κ )) . (3 . . . . . 4) (for i = κ ) and (3 . L ( Y κ +1 ) = hY κ +1 i ⊕ L ( B ) ⊕ · · · ⊕ L ( B κ ) ⊕ L ( B κ +1 ) ⊕ I Y κ +1 , (3 . I Y κ +1 = I Y κ ⊕ L ( D κ +1 ), B κ +1 = P ′′Y κ ,y κ +1 ≀{ y κ +1 } and D κ +1 = P ′Y κ , , ≀P ′′Y κ . In particular,for κ = m − 1, by (3 . 20) and (3 . L ( Y ) = hYi ⊕ L ( B ) ⊕ · · · ⊕ L ( B m ) ⊕ I , where I = γ ( L ( Y )) ⊕ L ( D ) ⊕ · · · ⊕ L ( D m ). (cid:3) Corollary 1 With the previous notations, L ( Y ) = h y , y i ⊕ L ( P ′′Y ) ⊕ · · · ⊕ L ( P ′′Y m − ) ⊕ I ,where h y , y i is the free Z -module of rank and I = γ ( L ( Y )) ⊕ L ( P ′Y , , ≀ P ′′Y ) ⊕ · · · ⊕ L ( P ′Y m − , , ≀ P ′′Y m − ) . Z -module I/γ ( I ) In this section, we prove the following essential result. Theorem 4 If θ m = ∅ , then, for all κ ∈ [ m − , the free Z -module I Y κ +1 /γ ( I Y κ +1 ) is a free(right) U ( L ( Y κ +1 ) /I Y κ +1 ) -module with a free generating set I Y κ +1 + γ ( I Y κ +1 ) = { [ y a , y b ] + γ ( I Y κ +1 ) : y a , y b ∈ Y κ +1 ; ( a, b ) ∈ θ m,δ } . In particular, for κ = m − , the Z -module I/γ ( I ) is a free (right) U ( L ( Y ) /I ) -module with a free generating set I + γ ( I ) = { [ y a , y b ] + γ ( I ) : y a , y b ∈ Y ; ( a, b ) ∈ θ m,δ } . roof. Let m = 2. Since I Y = γ ( L ( Y )) = L ( Y (1)2 ), we have I Y /γ ( I Y ) is a free Z -module with a free generating set Y (1)2 = { u + γ ( I Y ) : u ∈ Y (1)2 } . It is well knownthat I Y /γ ( I Y ) is a free (right) U ( L ( Y ) /I Y )-module with a free generating set I Y = { [ y , y ] + γ ( I Y ) } .Thus, we assume that m ≥ 3. We induct on κ . For κ = 1, the proof is shown above. So,we may assume that our claim is valid for all κ ∈ [ m − I Y κ /γ ( I Y κ ) is a free (right) U ( L ( Y κ ) /I Y κ )-module with a free generating set I Y κ + γ ( I Y κ ) = { [ y a , y b ] + γ ( I Y κ ) : y a , y b ∈Y κ ; ( a, b ) ∈ θ m,δ } . The set M Y κ ,I = M ′Y ∪ M D ∪ · · · ∪ M D κ is a totally ordered Z -basis of I Y κ . Furthermore, the set M Y κ = Y κ ∪ M B ∪ · · · ∪ M B κ ∪ M Y κ ,I is a totally ordered Z -basisof L ( Y κ ). By (3 . 1) (for i = κ ), L ( Y κ +1 ) = L ( Y κ ) ⊕ L ( { y κ +1 } ≀ Y κ ). The set M < Y κ ,θ m = { u ′ : u ∈ M < Y κ } is a Z -basis of T ( Y κ ). Moreover, the set P ′Y κ = { [ y κ +1 ; u ′ ] : u ′ ∈ M < Y κ ,θ m } is afree generating set of the free Lie algebra L ( { y κ +1 } ≀ Y κ ). Thus, L ( { y κ +1 } ≀ Y κ ) = L ( P ′Y κ ). Atypical element of P ′Y κ \ { y κ +1 } has the form [ y κ +1 ; u ′ ] = [ y κ +1 ; y ( a κ ; r ; π ) v v · · · v ν ], whereeach v i ( ∈ M Y κ , i ∈ [ ν ]), in the above expression, is a simple Lie commutator of degree atleast 2 and v ≤ v ≤ · · · ≤ v ν . The set P ′Y κ is decomposed into its subsets P ′Y κ , , and P ′′Y κ where P ′Y κ , is the subset of P ′Y κ consisting of all elements of the form [ y κ +1 ; u ′ ] with u ′ iselement of the form (3 . 10) with y ( a κ ; r ) = 1, and P ′Y κ , is the subset of P ′Y κ consisting of allelements of the form [ y κ +1 ; u ′ ], with u ′ is of the form (3 . y ( a κ ; r ) = 1 and at least oneof v , . . . , v ν belongs to M Y κ ,I . Write P ′Y κ , , = P ′Y κ , ∪ P ′Y κ , and let P ′′Y κ = P ′Y κ \ P ′Y κ , , .That is, P ′′Y κ is the union of { y κ +1 } and the set P ′′Y κ ,y κ +1 of all elements of the form [ y κ +1 ; u ′ ],where u ′ is the form (3 . 10) with y ( a κ ; r ) = 1, y ( a κ − κ ; r − r ) = 1 and each v i ( i ∈ [ ν ]) (ifit occurs in (3 . S κj =3 M B j . Thus, P ′′Y κ = { y κ +1 } ∪ P ′′Y κ ,y κ +1 .For any non-empty subset X of L ( Y κ +1 ), we write e X = X + I Y κ +1 I Y κ +1 . The Lie subalgebra ] L ( Y κ ) of L ( Y κ +1 ) /I Y κ +1 is naturally isomorphic as a Lie algebra to L ( Y κ ) L ( Y κ ) ∩ I Y κ +1 . Since I Y κ +1 = I Y κ ⊕ L ( D κ +1 ) and L ( Y κ +1 ) = L ( Y κ ) ⊕ L ( { y κ +1 } ≀ Y κ ), we have, by the modular law, L ( Y κ ) ∩ I Y κ +1 = I Y κ and so, ] L ( Y κ ) ∼ = L ( Y κ ) /I Y κ as Lie algebras in a natural way. Hence, U ( ] L ( Y κ )) ∼ = U ( L ( Y κ ) /I Y κ ) as Z -algebras in a natural way. Since L ( Y κ ) is a free Lie algebra and I Y κ isa direct summand of L ( Y κ ), we have both I Y κ and L ( Y κ ) /I Y κ are free Z -modules. Thus, ] L ( Y κ ) is a free Z -module. The set e L Y κ is a totally ordered Z -basis of ] L ( Y κ ), where L Y κ = Y κ ∪ M B ∪ · · · ∪ M B κ with a total ordering such that w < w < · · · < w κ for all w ∈ Y κ , w j ∈ M B j , j ∈ [ κ ] \ { , } . It is clear enough that ] L ( Y κ ) = g hY κ i ⊕ ^ L ( B ) ⊕ · · · ⊕ ^ L ( B κ ) . (3 . L ( Y κ +1 ) /I Y κ +1 ^ L ( Y κ ) ∼ = L ( Y κ +1 ) L ( Y κ )+ I Y κ +1 as Z -modules in a natural way. Since L ( Y κ ) + I Y κ +1 = hY κ i ⊕ L ( B ) ⊕ · · · ⊕ L ( B κ ) ⊕ I Y κ +1 , we get L ( Y κ +1 ) L ( Y κ )+ I Y κ +1 ∼ = h y κ +1 i⊕ L ( B κ +1 ) as Z -modules in a natural way. Therefore, L ( Y κ +1 ) /I Y κ +1 ^ L ( Y κ ) is a free Z -module. Observe that ^ L ( Y κ +1 ) = L ( Y κ +1 ) /I Y κ +1 = ^ hY κ +1 i ⊕ ^ L ( B ) ⊕ · · · ⊕ ^ L ( B κ +1 ) , (3 . ^ hY κ +1 i ∼ = hY κ +1 i as abelian Lie algebras and ^ L ( B j ) ∼ = L ( B j ) as Lie algebras for any j ∈ [ κ + 1] \ { , } . Since the above Lie algebras are free Z -modules, we obtain the elements12 y α · · · y α κ +1 κ +1 u u · · · u µ ) + I Y κ +1 , where α , . . . , α κ +1 ≥ u , . . . , u µ ∈ S κ +1 j =3 M B j and u ≤ u ≤ · · · ≤ u µ , form a Z -basis for U ( ^ L ( Y κ +1 )) (see, [3, Chapter 1, § U ( ^ L ( Y κ +1 )) is a free Z -module. By (3 . 23) and (3 . ^ L ( Y κ +1 ) = ^ h y κ +1 i ⊕ ^ L ( B κ +1 ) ⊕ ] L ( Y κ ) . (3 . ] L κ +1 is a totally ordered Z -basis of ^ h y κ +1 i ⊕ ^ L ( B κ +1 ), where L κ +1 = { y κ +1 } ∪ M B κ +1 with y κ +1 < w κ +1 for all w κ +1 ∈ M B κ +1 . We give g L Y κ ∪ ] L κ +1 a total ordering such thatevery element of g L Y κ is less than every element of ] L κ +1 . The elements ˜ u ˜ u · · · ˜ u λ with u , . . . , u λ ∈ L Y κ and u ≤ u ≤ · · · ≤ u λ , form a Z -basis of U ( ^ L ( Y κ )) (see, [3, Chapter 1, § y α · · · y α κ κ u · · · u µ y α κ +1 κ +1 v · · · v λ ) + I Y κ +1 , with α , . . . , α κ +1 ≥ u , . . . , u µ ∈ S κj =3 M B j , v , . . . , v λ ∈ M B κ +1 and u ≤ u ≤ · · · ≤ u µ < y κ +1 < v ≤ · · · ≤ v λ form a Z -basis of U ( ^ L ( Y κ +1 )). Hence, the canonical homomorphism of U ( ^ L ( Y κ )) into U ( ^ L ( Y κ +1 )) maps the elements of a Z -basis of U ( ^ L ( Y κ )) to Z -linearly independent elementsof U ( ^ L ( Y κ +1 ) and so, we may regard U ( ^ L ( Y κ )) as Z -subalgebra of U ( ^ L ( Y κ +1 )).For any non-empty subset Y of L ( Y κ +1 ), we write Y = Y + γ ( I Y κ +1 ) γ ( I Y κ +1 ) . Since I Y κ +1 is anideal of L ( Y κ +1 ), we have I Y κ +1 /γ ( I Y κ +1 ) = I Y κ +1 . Now, γ ( I Y κ +1 ) = γ ( I Y κ ) ⊕ ([ L ( D κ +1 ) , I Y κ ] + γ ( L ( D κ +1 )) . (3 . I Y κ +1 , L ( P ′Y κ ) are ideals in L ( Y κ +1 ), (3 . 13) and the Claim, wehave [ L ( D κ +1 ) , I Y κ ] ⊆ I Y κ +1 ∩ L ( P ′Y κ ) = L ( D κ +1 ). We claim that I Y κ +1 = I Y κ ⊕ L ( D κ +1 ).Indeed, we point out that I Y κ +1 = I Y κ + L ( D κ +1 ). Let u + γ ( I Y κ +1 ) = v + γ ( I Y κ +1 ) for u ∈ I Y κ \ γ ( I Y κ +1 ) and v ∈ L ( D κ +1 ) \ γ ( I Y κ +1 ). Let ϕ be the Lie algebra endomorphism of L ( Y κ +1 ) with ϕ ( y κ +1 ) = 0 and ϕ ( y j ) = y j for all j ∈ [ κ ]. By (3 . u − v = w + w ,where w ∈ γ ( I Y κ ) and w ∈ [ L ( D κ +1 ) , I Y κ ]+ γ ( L ( D κ +1 )). By applying ϕ on u − v = w + w ,we get u = ϕ ( u ) = ϕ ( w ) = w ∈ γ ( I Y κ ) a contradiction. Therefore, I Y κ +1 = I Y κ ⊕ L ( D κ +1 ) . (3 . . I Y κ ∩ L ( D κ +1 ) = { } , we have I Y κ ∼ = I Y κ /γ ( I Y κ ) as Z -modules in a natural way. By our hypothesis, I Y κ /γ ( I Y κ ) is a free (right) U ( L ( Y κ ) /I Y κ )-module with a free generating set I Y κ + γ ( I Y κ ). Since I Y κ ∼ = I Y κ /γ ( I Y κ ) as Z -modules (ina natural way) and by (3 . I Y κ as a free (right) U ( ^ L ( Y κ ))-module with afree generating set I Y κ = I Y κ + γ ( I Y κ +1 ). In other words, I Y κ = M ya,yb ∈Y κ ( a,b ) ∈ θ m,δ (cid:16) [ y a , y b ] + γ ( I Y κ +1 )) U ( ^ L ( Y κ ) (cid:17) . (3 . . 25) and since L ( D κ +1 ) = hD κ +1 i ⊕ γ ( L ( D κ +1 )), we have L ( D κ +1 ) = hD κ +1 i and so,(3 . 26) is rewritten as I Y κ +1 = I Y κ ⊕ hD κ +1 i . (3 . j ∈ [ κ ], ( κ + 1 , j ) / ∈ θ m,δ . Then, P ′Y κ , = ∅ . The set P ′Y κ , consists of the elementsof the form [ y κ +1 ; y α · · · y α κ κ u u · · · u ν v · · · v µ ], where α , . . . , α κ ≥ u i ∈ S κj =3 M B j (if u , . . . , u ν occur in the expression), i ∈ [ ν ], v j ∈ M Y κ ,I (at least one of v j occursin the above expression), j ∈ [ µ ] and u ≤ u ≤ · · · ≤ u ν ≤ v ≤ · · · ≤ v µ . The set M Y κ ,I is a Z -basis of I Y κ . The set [ P ′Y κ , ≀ P ′′Y κ , M Y κ ,I ] consists of all elements of theform [ u, v , . . . , v s , w ], where u ∈ P ′Y κ , , v , . . . , v s ∈ P ′′Y κ , s ≥ w ∈ M Y κ ,I . By thedefinition of M Y κ ,I and (3 . P ′Y κ , ≀ P ′′Y κ , M Y κ ,I ] ⊂ γ ( I Y k +1 ) . (3 . . y κ +1 ; y α · · · y α κ κ u u · · · u ν v · · · v µ ]+ γ ( I Y k +1 ) = [ y κ +1 ; y α · · · y α κ κ u u · · · u ν v ]+ γ ( I Y k +1 )= − [ v, [ y κ +1 ; y α · · · y α κ κ u u · · · u ν ]] + γ ( I Y k +1 ) . In particular, v ∈ Y (1)2 ∪ D ∪ · · · ∪ D κ . The element f = [ y κ +1 ; y α · · · y α κ κ u u · · · u ν ] ∈P ′′Y κ for all u j ∈ S κj =3 M B j and u ≤ · · · ≤ u ν (if u , . . . , u ν occur in the expression). By(3 . v + γ ( I Y k +1 ) is uniquely written as a Z -linear combination of elements of the form([ y a , y b ])ad( y α ) · · · ad( y α κ κ )ad( u ) · · · ad( u ν )+ γ ( I Y κ +1 ), where y a , y b ∈ Y κ , ( a, b ) ∈ θ m,δ , α , . . . , α κ ≥ u j ∈ S κj =3 M B j and u ≤ · · · ≤ u ν . Let P ′Y κ , , be the subset of P ′Y κ , consisting of all elements of the form [ y κ +1 ; y α · · · y α κ κ u u · · · u ν v ], where α , . . . , α κ ≥ u i ∈ S κj =3 M B j , i ∈ { , . . . , ν } (if u , . . . , u ν occur in the expression) and v ∈ Y (1)2 ∪D ∪ · · · ∪ D κ . Since L ( D κ +1 ) is a free Lie algebra on D κ +1 = P ′Y κ , ≀ P ′′Y κ , M Y κ ,I is a Z -basis of I Y κ and by (3 . P ′Y κ , = P ′Y κ , , is Z -linearly independent. Theset P ′′Y κ consists of y κ +1 and elements of the form f , where α , . . . , α κ ≥ α + · · · + α κ ≥ u i ∈ S κj =3 M B j , i ∈ [ µ ] and u ≤ u ≤ · · · ≤ u µ . Since D κ +1 = P ′Y κ , ≀ P ′′Y κ , we have hD κ +1 i = hP ′Y κ , ≀ P ′′Y κ i = hP ′Y κ , , ≀ P ′′Y κ i . It is clear enough that the set P ′Y κ , , ≀ P ′′Y κ isa Z -basis of hD κ +1 i . By (3 . G = I Y κ ∪ ( P ′Y κ , , ≀ P ′′Y κ ) isa Z -basis of I Y κ +1 . Recall that the free Z -module I Y κ +1 becomes a (right) U ( ^ L ( Y κ +1 ))-module by defining ( u + γ ( I Y κ +1 ))˜ v · · · ˜ v r = u (ad v ) · · · (ad v r ) + γ ( I Y κ +1 ) for all u ∈ I Y κ +1 and v , . . . , v r ∈ L ( Y κ +1 ). By (3 . x, [ y, z ]] = [ x, y, z ] − [ x, z, y ], we may assumethat v , . . . , v r ∈ Y κ +1 ∪ ( S κ +1 j =3 M B j ). Let M be the U ( ^ L ( Y κ +1 ))-submodule of I Y κ +1 generated by the set M = { [ y a , y b ] + γ ( I Y κ +1 ) : y a , y b ∈ Y κ ; ( a, b ) ∈ θ m,δ } . Since U ( ^ L ( Y κ )) is regarded in a canonical way as a Z -subalgebra of U ( ^ L ( Y κ +1 )), we have,by (3 . I Y κ ⊆ M . By the decomposition of P ′Y κ , = P ′Y κ , , and the fact that P ′Y κ , , ≀ P ′′Y κ is a Z -basis of hD κ +1 i , we get hD κ +1 i is contained in M . By (3 . M = I Y κ +1 . Since G is a Z -basis of I Y κ +1 , we obtain I Y κ +1 = I Y κ +1 /γ ( I Y κ +1 ) = M ya,yb ∈Y κ ( a,b ) ∈ θm,δ (cid:16) [ x, y ] + γ ( I Y κ +1 )) U ( ^ L ( Y κ +1 ) (cid:17) and so, I Y κ +1 is a free (right) U ( ^ L ( Y κ +1 ))-module with a free generating set M .14. For all j ∈ [ κ ], ( κ + 1 , j ) ∈ θ m,δ . The set P ′Y κ , consists of the elements of the form[ y κ +1 ; y α · · · y α κ κ u u · · · u ν ] + γ ( I Y k +1 ), where α , . . . , α κ ≥ α + · · · + α κ ≥ u i ∈ S κj =3 M B j (if u , . . . , u ν occur), i ∈ [ ν ] and u ≤ u ≤ · · · ≤ u ν . The set P ′Y κ , consists ofthe elements of the form [ y κ +1 ; u u · · · u ν v · · · v µ ], where u i ∈ S κj =3 M B j (if u , . . . , u ν occur in the expression), i ∈ [ ν ], v j ∈ M Y κ ,I (at least one of v j occurs in the aboveexpression), j ∈ [ µ ] and u ≤ u ≤ · · · ≤ u ν ≤ v ≤ · · · ≤ v µ . The set M Y κ ,I is a Z -basisof I Y κ . The set [ P ′Y κ , , ≀P ′′Y κ , M Y κ ,I ] consists of all elements of the form [ u, v , . . . , v s , w ],where u ∈ P ′Y κ , , , v , . . . , v s ∈ P ′′Y κ , s ≥ w ∈ M Y κ ,I . By the definition of M Y κ ,I and (3 . P ′Y κ , , ≀ P ′′Y κ , M Y κ ,I ] ⊂ γ ( I Y k +1 ) . (3 . . y κ +1 ; u u · · · u ν v · · · v µ ] + γ ( I Y k +1 ) = [ y κ +1 ; u u · · · u ν v ] + γ ( I Y k +1 )= − [ v, [ y κ +1 ; u u · · · u ν ]] + γ ( I Y k +1 ) . In particular, v ∈ Y (1)2 ∪ D ∪ · · · ∪ D κ . The element [ y κ +1 ; u u · · · u ν ] ∈ P ′′Y κ for all u j ∈ S κj =3 M B j and u ≤ · · · ≤ u ν (if u , . . . , u ν occur in the expression). By (3 . v is uniquely written as a Z -linear combination of elements of the form([ y a , y b ])ad( y α ) · · · ad( y α κ κ )ad( u ) · · · ad( u ν ) + γ ( I Y κ +1 ) , where y a , y b ∈ Y κ , ( a, b ) ∈ θ m,δ , α , . . . , α κ ≥ u j ∈ S κj =3 M B j and u ≤ · · · ≤ u ν . Theset P ′′Y κ consists of y κ +1 and elements of the form [ y κ +1 ; u u · · · u ν ] with u i ∈ S κj =3 M B j (at least one of u , . . . , u ν occurs), i ∈ [ ν ] and u ≤ · · · ≤ u ν . Since D κ +1 = ( P ′Y κ , ∪P ′Y κ , ) ≀ P ′′Y κ , we have hD κ +1 i = h ( P ′Y κ , ∪ P ′Y κ , ) ≀ P ′′Y κ i = h ( P ′Y κ , ∪ P ′Y κ , ) ≀ P ′′Y κ i . It isclear enough that the set ( P ′Y κ , ∪ P ′Y κ , ) ≀ P ′′Y κ is a Z -basis of hD κ +1 i . By (3 . G = I Y κ ∪ (( P ′Y κ , ∪ P ′Y κ , ) ≀ P ′′Y κ ) is a Z -basis of I Y κ +1 . Let M bethe U ( ^ L ( Y κ +1 ))-submodule of I Y κ +1 generated by the set M = { [ y a , y b ] + γ ( I Y κ +1 ) : y a , y b ∈ Y κ +1 ; ( a, b ) ∈ θ m,δ } . Since U ( ^ L ( Y κ )) is regarded in a canonical way as a Z -subalgebra of U ( ^ L ( Y κ +1 )), we have, by (3 . I Y κ ⊆ M . By the decomposition of( P ′Y κ , ∪ P ′Y κ , ) and the fact that ( P ′Y κ , ∪ P ′Y κ , ) ≀ P ′′Y κ is a Z -basis of hD κ +1 i , we get hD κ +1 i is contained in M . By (3 . M = I Y κ +1 . Since G is a Z -basis of I Y κ +1 , we obtain I Y κ +1 = I Y κ +1 /γ ( I Y κ +1 ) = M ya,yb ∈Y κ +1( a,b ) ∈ θm,δ (cid:16) [ y a , y b ] + γ ( I Y κ +1 )) U ( ^ L ( Y κ +1 ) (cid:17) and so, I Y κ +1 is a free (right) U ( ^ L ( Y κ +1 ))-module with a free generating set M .3. There are j ∈ [ κ ] such that ( κ + 1 , j ) ∈ θ m,δ . The set P ′Y κ , consists of the elements ofthe form [ y κ +1 ; y ( a κ ; r ) y ( a κ − κ ; r − r ) u u · · · u ν ] + γ ( I Y k +1 )where y ( a κ ; r ) = 1, u i ∈ S κj =3 M B j (if u , . . . , u ν occur), i ∈ [ ν ] and u ≤ u ≤· · · ≤ u ν . The set P ′Y κ , consists of the elements of the form [ y κ +1 ; y ( a κ − κ ; r − ) u u · · · u ν v · · · v µ ], with u i ∈ S κj =3 M B j (if u , . . . , u ν occur), i ∈ [ ν ], v j ∈ M Y κ ,I (at least one of v j occurs in the above expression), j ∈ [ µ ] and u ≤ u ≤ · · · ≤ u ν ≤ v ≤ · · · ≤ v µ . The set M Y κ ,I is a Z -basis of I Y κ . By the definition of M Y κ ,I , (3 . . y κ +1 ; y ( a κ − κ ; r − r ) u u · · · u ν v · · · v µ ] + γ ( I Y k +1 )= [ y κ +1 ; y ( a κ − κ ; r − r ) u u · · · u ν v ] + γ ( I Y k +1 )= − [ v, [ y κ +1 ; y ( a κ − κ ; r − r ) u u · · · u ν ]] + γ ( I Y k +1 ) . In particular, v ∈ Y (1)2 ∪ D ∪ · · · ∪ D κ . Note that the element[ y κ +1 ; y ( a κ − κ ; r − r ) u u · · · u ν ] ∈ P ′′Y κ for all u j ∈ S κj =3 M B j and u ≤ · · · ≤ u ν (if u , . . . , u ν occur in the expression). By(3 . v + γ ( I Y k +1 ) is uniquely written as a Z -linear combination of elements of the form([ y a , y b ])ad( y α ) · · · ad( y α κ κ )ad( u ) · · · ad( u ν )+ γ ( I Y κ +1 ), where y a , y b ∈ Y κ , ( a, b ) ∈ θ m,δ , α , . . . , α κ ≥ u j ∈ S κj =3 M B j and u ≤ · · · ≤ u ν . The set P ′′Y κ consists of y κ +1 and elements of the form [ y κ +1 ; y ( a κ − κ ; r − r ) u u · · · u ν ] with y ( a κ − κ ; r − r ) = 1, u i ∈ S κj =3 M B j (if u , . . . , u ν occur in the expression), i ∈ [ ν ] and u ≤ · · · ≤ u ν . Since D κ +1 = ( P ′Y κ , ∪ P ′Y κ , ) ≀ P ′′Y κ , we have hD κ +1 i = h ( P ′Y κ , ∪ P ′Y κ , ) ≀ P ′′Y κ i = h ( P ′Y κ , ∪ P ′Y κ , ) ≀ P ′′Y κ i . It is clear enough that the set ( P ′Y κ , ∪ P ′Y κ , ) ≀ P ′′Y κ is a Z -basis of hD κ +1 i . By (3 . G = I Y κ ∪ (( P ′Y κ , ∪ P ′Y κ , ) ≀P ′′Y κ ) is a Z -basis of I Y κ +1 . Let M bethe U ( ^ L ( Y κ +1 ))-submodule of I Y κ +1 generated by the set M = { [ y a , y b ] + γ ( I Y κ +1 ) : y a , y b ∈ Y κ +1 ; ( a, b ) ∈ θ m,δ } . Since U ( ^ L ( Y κ )) is regarded in a canonical way as a Z -subalgebra of U ( ^ L ( Y κ +1 )), we have, by (3 . I Y κ ⊆ M . By the decomposition of( P ′Y κ , ∪ P ′Y κ , ) and the fact that ( P ′Y κ , ∪ P ′Y κ , ) ≀ P ′′Y κ is a Z -basis of hD κ +1 i , we get hD κ +1 i is contained in M . By (3 . M = I Y κ +1 . Since G is a Z -basis of I Y κ +1 , we obtain I Y κ +1 = I Y κ +1 /γ ( I Y κ +1 ) = M x,y ∈Y κ +1( x,y ) ∈ θm,δ (cid:16) [ x, y ] + γ ( I Y κ +1 )) U ( ^ L ( Y κ +1 ) (cid:17) and so, I Y κ +1 is a free (right) U ( ^ L ( Y κ +1 ))-module with a free generating set M .So, in all cases, I Y κ +1 is a free (right) U ( ^ L ( Y κ +1 ))-module with a free generating set I Y κ +1 . For κ = m − 1, the free Z -module I/γ ( I ) is a free (right) U ( L ( Y ) /I )-modulewith a free generating set I = { [ y a , y b ] + γ ( I ) : y a , y b ∈ Y ; ( a, b ) ∈ θ m,δ } . (cid:3) FP( H ) For a positive integer n , with n ≥ 2, let A = { y , y , . . . , y n +1 } . Write Y = { y , . . . , y n } and C = { y n +1 } . Let θ n be a partial commutation relation on [ n ]. As before, without lossof generality, we may order the elements of A as y < . . . < y n < y n +1 and assume that162 , ∈ θ n . Define θ n,δ = { ( k, l ) ∈ θ n | k > l } and so θ n,δ = ∅ . By Theorem 2, we have L ( A ) = L ( Y ) ⊕ L ( C ≀ Y ), where C ≀ Y = { [ y n +1 , b , . . . , b k ] : b , . . . , b k ∈ Y , k ≥ } . Let J bethe subset of L ( A ), J = { [ y n +1 , y i , y j ] , [ y a , y b ] , [[ y n +1 , y a ] , [ y n +1 , y b ]] | i, j ∈ [ n ] , ( a, b ) ∈ θ n,δ } and J be the ideal of L ( A ) generated by the set J . Our main purpose in this section is1. to show that J is a direct summand of L ( A ),2. to give a Z -module decomposition of J as a direct sum of explicitly described Liesubalgebras of L ( A ) by giving their free generating sets and3. to describe the quotient Lie algebra L ( A ) /J .The following result is immediate. Lemma 4 The ideal J Y = J ∩ L ( Y ) of L ( Y ) is generated by the set { [ y a , y b ] | ( a, b ) ∈ θ n,δ } . By Corollary 1 (see, also, proof of Theorem 3), L ( Y ) = h y , y i ⊕ L ( P ′′Y ) ⊕ · · · ⊕ L ( P ′′Y n − ) ⊕ J Y , (4 . h y , y i is the free Z -module of rank 2 and J Y = γ ( L ( Y )) ⊕ L ( P ′Y , , ≀ P ′′Y ) ⊕ · · · ⊕ L ( P ′Y n − , , ≀ P ′′Y n − ) . (4 . i ∈ [ n ] \ { , } , put D i = P ′Y i − , , ≀ P ′′Y i − . By (4 . i ∈ [ n ] \ { , } , let M D i be the totally ordered Z -basis of L ( D i ) consisting of the basic Lie commutators in terms ofelements of D i . Write M J Y = M ′Y ∪ M D ∪ · · · ∪ M D n , where M ′Y denotes the totallyordered Z -basis of γ ( L ( Y )) consisting of the basic Lie commutators in the elements of thestandard free generating set of γ ( L ( Y )). The set M J Y is a totally ordered Z -basis of J Y subject to u < u < · · · < u n for all u ∈ M ′ and u i ∈ M D i ( i ∈ [ n ] \ { , } ). By Theorem 2,for i ∈ [ n ] \{ , } , L ( P ′′Y i − ) = h y i i⊕ L ( P ′′Y i − ,y i ≀{ y i } ) = h y i i⊕ L ( B i ), where B i = P ′′Y i − ,y i ≀{ y i } .Hence, (4 . 1) becomes L ( Y ) = hYi ⊕ L ( B ) ⊕ · · · ⊕ L ( B n ) ⊕ J Y . (4 . i ∈ [ n ] \ { , } , let M B j be the totally ordered Z -basis of L ( B j ) consisting of basicLie commutators in terms of elements of B j . Write M Y = Y ∪ M B ∪ · · · ∪ M B n ∪ M J Y . By(4 . 3) and (4 . M Y is a totally ordered Z -basis of L ( Y ) subject to w < w < · · · Ω) is the ideal in L ( P Y ) generated by the set E . Since E ⊆ J and J is an idealin L ( A ), we have L ( E ≀ Ω) ⊆ J . By (4 . 4) and the modular law, we get J P Y = ( J ∩ L (Ω)) ⊕ L ( E ≀ Ω) . (4 . . J ∩ L (Ω) ⊆ L (Ω) and the modular law, we have J P Y ∩ L (Ω) = ( J ∩ L (Ω)) ⊕ ( L ( E ≀ Ω) ∩ L (Ω))(by (4 . J ∩ L (Ω) . So, we obtain the following decomposition of J , J = J Y ⊕ J P Y = J Y ⊕ ( J ∩ L (Ω)) ⊕ L ( E ≀ Ω) . (4 . J P Y and L ( E ≀ Ω) Let J be the ideal of L ( A ) generated by the set J = { [ y a , y b ] : ( a, b ) ∈ θ n,δ } . By Lemma 1,we have J = J , Y ⊕ J , P Y , (4 . J , Y = J ∩ L ( Y ) and J , P Y = J ∩ L ( P Y ). By a result of Duchamp and Krob [7,Theorem II.7], we have J , Y is the ideal of L ( Y ) generated by the set J . By Lemma 4, J Y = J , Y . Let P ′Y be the subset of P Y consisting of all elements of the form [ y n +1 ; u ] where u ∈ M < Y with at least one of v , . . . , v ν belongs to M J Y , and P ′′Y = P Y \ P ′Y . Proposition 3 With the previous notations,1. J , P Y = L ( P ′Y ≀ P ′′Y ) and L ( P Y ) = L ( P ′′Y ) ⊕ J , P Y .2. J P Y is the ideal of L ( A ) generated by the set { [ y n +1 , y i , y j ] , [[ y n +1 , y a ] , [ y n +1 , y b ]] : i, j ∈ [ n ]; ( a, b ) ∈ θ n,δ } .3. L ( E ≀ Ω) is an ideal in L ( A ) .4. J ∩ L (Ω) is the ideal of L (Ω) generated by the set { [[ y n +1 , y a ] , [ y n +1 , y b ]] : ( a, b ) ∈ θ n,δ } .Proof. 1. By Theorem 2, L ( P Y ) = L ( P ′′Y ) ⊕ L ( P ′Y ≀ P ′′Y ) . (4 . J , P Y is the ideal of L ( C≀Y ) = L ( P Y )generated by the set T = { [ y n +1 , z , . . . , z µ , [ y a , y b ] , z ′ , . . . , z ′ ν ] : z · · · z µ , z ′ · · · z ′ ν ∈ Y ∗ , ( a, b ) ∈ θ n,δ } , where Y ∗ is the free monoid on Y . We point out that in [7] right-normed Lie commu-tators are used. We put a = [ y n +1 , z , . . . , z µ ] and b = [ y a , y b ]. Then,[ y κ +1 , z , . . . , z µ , [ y a , y b ] , z ′ . . . , z ′ ν ] = [ a, b, z ′ . . . , z ′ ν ]18 . = P νi =0 P σ ∈ K i [[ a, z ′ σ (1) , . . . , z ′ σ ( i ) ] , [ b, z ′ σ ( i +1) , . . . , z ′ σ ( ν ) ]]= [ a, z ′ , . . . , z ′ ν , b ] + [ a, [ b, z ′ , . . . , z ′ ν ]]+ P ν − i =1 P σ ∈ K i [[ a, z ′ σ (1) , . . . , z ′ σ ( i ) ] , [ b, z ′ σ ( i +1) , . . . , z ′ σ ( ν ) ]] . We proceed as in the proof of Theorem 3. Since J , Y = J Y is the ideal of L ( Y ) generatedby the set J , we obtain each [ b, z ′ σ ( i +1) , . . . , z ′ σ ( ν ) ] ∈ J Y . Thus, each [ b, z ′ σ ( i +1) , . . . , z ′ σ ( ν ) ]is a Z -linear combination of elements of M J Y . Having in mind (3 . 17) (and the abovediscussion), (3 . 15) and using the Jacobi identity in the form [ x, y, z ] = [ x, z, y ]+ [ x, [ y, z ]]as many times as needed, we may write each element of T as a Z -linear combination ofelements of L ( P ′Y ≀ P ′′Y ). Thus, T is a subset of L ( P ′Y ≀ P ′′Y ). Since J , P Y is the ideal in L ( P Y ) generated by the set T and L ( P ′Y ≀ P ′′Y ) is an ideal of L ( P Y ), we have J , P Y ⊆ L ( P ′Y ≀ P ′′Y ) . (4 . P ′Y is contained in J , P Y and since J , P Y is an ideal in L ( A ), wehave L ( P ′Y ≀ P ′′Y ) ⊆ J , P Y . (4 . . 9) and (4 . J , P Y = L ( P ′Y ≀ P ′′Y ) and so, by (4 . L ( P Y ) = L ( P ′′Y ) ⊕ J , P Y .2. Let J and J be the ideals of L ( A ) generated by the sets J = { [ y a , y b ] : ( a, b ) ∈ θ n,δ } and J = { [ y n +1 , y i , y j ] , [[ y n +1 , y a ] , [ y n +1 , y b ]] : i, j ∈ [ n ] , ( a, b ) ∈ θ n,δ } , respectively. Clearly, J = J + J . Since L ( P Y ) is an ideal in L ( A ) and J ⊆ L ( P Y ),we conclude that J ⊆ L ( P Y ) and so, J ⊆ J P Y . But J P Y = J ∩ L ( P Y )= ( J + J ) ∩ L ( P Y )(modular law) = ( J ∩ L ( P Y )) + J ( J , P Y = J ∩ L ( P Y )) = J , P Y + J . In order to show that J P Y ⊆ J , it is enough to prove that J , P Y ⊆ J . Let Y ∗ be thefree monoid on Y . By a result of Duchamp and Krob [7, Theorem II.7], we have J , P Y is the ideal of L ( C ≀ Y ) = L ( P Y ) generated by the set T = { [ y n +1 , z , . . . , z µ , [ y a , y b ] , z ′ , . . . , z ′ ν ] : z · · · z µ , z ′ · · · z ′ ν ∈ Y ∗ , ( a, b ) ∈ θ n,δ } . Since T ⊆ J and J is an ideal in L ( A ), we get J , P Y ⊆ J and so, J P Y = J . That is, J P Y is the ideal in L ( A ) generated by the set J .3. For the next few lines, we write S = L ( E ≀ Ω). To show that S is an ideal of L ( A ), it isenough to show that [ u, v ] ∈ S for all u ∈ S and v ∈ L ( A ). Since L ( A ) = L ( Y ) ⊕ L ( C ≀Y )and S is an ideal of L ( C ≀ Y ), it is enough to show that [ u, v ] ∈ S for all u ∈ S and19 ∈ L ( Y ). Since L ( Y ) is generated by the set Y , every element v of L ( Y ) is a Z -linearcombination of Lie commutators of the form [ y i , . . . , y i κ ], κ ≥ y i , . . . , y i κ ∈ Y . Byusing the Jacobi identity in the form [ x, [ y, z ]] = [ x, y, z ] − [ x, z, y ] as many times asneeded, it is enough to show that [ u, y i , . . . , y i κ ] ∈ S for all u ∈ S and y i , . . . , y i κ ∈ Y , κ ≥ 1. Since E ≀ Ω = E ∪ ( [ α ≥ [ E , Ω , . . . , Ω | {z } α ])and S is a free Lie algebra on E ≀ Ω, it is enough to show that [ w, y i , . . . , y i κ ] ∈ S forall w ∈ E ≀ Ω and y i , . . . , y i κ ∈ Y , κ ≥ 1. Since E ≀ Ω is a free generating set of S andthe linearity of Lie commutator, we may assume that w ∈ [ E , Ω , . . . , Ω | {z } α ≥ ](= [ E , α Ω]) with α ≥ 1. For x, y ∈ L ( A ) and a non-negative integer m , we write [ x, m y ] = [ x, y, . . . , y ]with m copies of y (this product is interpreted as x if m = 0). Since S is an ideal of L ( C ≀ Y ) and by using the Jacobi identity in the form[ x, y, z ] = [ x, z, y ] + [ x, [ y, z ]] (4 . w to have the following form w = [ u, m y n +1 , µ [ y n +1 , y ] , . . . , µ n [ y n +1 , y n ]] , where u ∈ E , m, µ , . . . , µ n ≥ m + µ + · · · + µ n ≥ 1. By using (4 . 11) as many asit is needed and since S is an ideal of L ( C ≀ Y ), we get[ w, y i , . . . , y i κ ] = [ u, y i , . . . , y i κ , m y n +1 , µ [ y n +1 , y ] , . . . , µ n [ y n +1 , y n ]] + v ′ , where v ′ ∈ S . Since [ u, m y n +1 , µ [ y n +1 , y ] , . . . , µ n [ y n +1 , y n ]] ∈ S , we conclude that[ w, y i , . . . , y i κ ] ∈ S . Therefore, L ( E ≀ Ω) is an ideal of L ( A ).4. For non-empty subsets X, Y of L ( A ) and a non-negative integer m , we write [ X, m Y ] =[ X, Y, . . . , Y ] with m copies of Y (this product is interpreted as X if m = 0). Write V and V for the Z -modules spanned by the sets { [ y n +1 , y i , y j ] : i, j ∈ [ n ] } and { [[ y n +1 , y a ] , [ y n +1 , y b ]] : ( a, b ) ∈ θ n,δ } , respectively. By Proposition 3 (2), we have J P Y = V ⊕ ([ V , L ( A )] + V ) ⊕ ( M m ≥ ([ V , m L (cid:0) A )] + [ V , ( m − L ( A )] (cid:1) . (4 . J P Y ⊆ L m ≥ L m ( A ). Let φ be the natural mapping from L ( C ≀ Y ) onto L (Ω).Thus, φ ( y n +1 ) = y n +1 , φ ([ y n +1 , y i ]) = [ y n +1 , y i ], i ∈ [ n ], and φ ([ y n +1 , b , . . . , b k ]) = 0for all b , . . . , b k ∈ Y , with k ≥ 2. By using the Jacobi identity and since φ is a Liealgebra epimorphism, we have, by (4 . 12) and since L ( A ) = hYi ⊕ h y n +1 i , φ ( J P Y ) isthe ideal of L (Ω) generated by the set { [[ y n +1 , y a ] , [ y n +1 , y b ]] : ( a, b ) ∈ θ n,δ } . It is clearenough that φ ( J P Y ) ⊆ J P Y ∩ L (Ω) = J ∩ L (Ω). Let y ∈ J P Y ∩ L (Ω). Then, there exists x ∈ L ( C ≀ Y ) = L ( P Y ) such that φ ( x ) = y . By (4 . x = u + v , where u ∈ L (Ω)and v ∈ L ( E ≀ Ω). Since φ ( E ≀ Ω) = 0 and L ( E ≀ Ω) is a free Lie algebra on E ≀ Ω, we have φ ( x ) = φ ( u ) + φ ( v ) = u = y and so, u ∈ J P Y ∩ L (Ω) = J ∩ L (Ω). By (4 . x ∈ J P Y . Hence, y ∈ φ ( J P Y ) and so, φ ( J P Y ) = J P Y ∩ L (Ω) = J ∩ L (Ω) . Therefore, the ideal J ∩ L (Ω) of L (Ω) is generated by the set { [[ y n +1 , y a ] , [ y n +1 , y b ]] :( a, b ) ∈ θ n,δ } . (cid:3) .2 The decomposition of J ∩ L (Ω) Write I Ω = φ ( J P Y ) = J ∩ L (Ω) and Ω = { [ y n +1 , y i ] : i ∈ [ n ] } ⊂ L ( A ) and so, Ω = { y n +1 } ∪ Ω = C ∪ Ω with C = { y n +1 } . By Theorem 2, L (Ω) = L (Ω ) ⊕ L ( C ≀ Ω ) . (4 . . I Ω = φ ( J P Y ) = J ∩ L (Ω) = I Ω ⊕ I C≀ Ω , (4 . I Ω = I Ω ∩ L (Ω ) and I C≀ Ω = I Ω ∩ L ( C ≀ Ω ). Let ψ be the (natural) Lie algebra iso-morphism from L ( A ) onto L (Ω) satisfying the conditions ψ ( y n +1 ) = y n +1 , ψ ( y i ) = [ y n +1 , y i ], i ∈ [ n ]. By (4 . 13) and since ψ is a Lie algebra isomorphism, we have ψ ( L ( Y )) = L (Ω ) and ψ ( L ( C ≀ Y )) = L ( C ≀ Ω ) . (4 . ψ is a Lie algebra isomorphism and by (4 . ψ ( J ) = I Ω . Furthermore, ψ ( J , Y ) = ψ ( J Y ) = I Ω and ψ ( J , P Y ) = I C≀ Ω . For i ∈ [ n ] } , let ω i = [ y n +1 , y i ] andΩ i = { ω , . . . , ω i } . Note that Ω n = Ω . By (4 . 15) and (4 . ψ is a Lie alge-bra isomorphism, we have L (Ω ) = h Ω i ⊕ L ( ψ ( B )) ⊕ · · · ⊕ L ( ψ ( B n )) ⊕ I Ω , (4 . κ ∈ [ n − \ { } , ψ ( B κ +1 ) = P ′′ Ω κ ,ω κ +1 ≀ { ω κ +1 } . By (4 . 2) and since ψ is a Lie algebraisomorphism, we have I Ω = γ ( L (Ω )) ⊕ L ( ψ ( D )) ⊕ · · · ⊕ L ( ψ ( D n )) , (4 . κ ∈ [ n − \ { } , ψ ( D κ +1 ) = P ′ Ω κ , , ≀ P ′′ Ω κ . We point out that the set M Ω = M ψ ( Y ) = Ω ∪ M ψ ( B ) ∪ · · · ∪ M ψ ( B n ) ∪ M I Ω0 , where M I Ω0 = ψ ( M J Y ) = ψ ( M ′Y ) ∪ M ψ ( D ) ∪ · · · ∪ M ψ ( D n ) , is a Z -basis of L (Ω ). Fur-thermore, let M < Ω be the Z -basis of T (Ω ) consisting of the elements of the form u · · · u µ ,where µ ≥ u , . . . , u µ ∈ M Ω and u ≤ u ≤ · · · ≤ u µ . A typical element of M < Ω \ { } hasthe form w = ω α · · · ω α n n v v · · · v ν , where α , . . . , α n ≥ 0, each v i ∈ ( S nj =3 M ψ ( B j ) ) ∪ M I Ω0 ( i ∈ [ ν ]) and v ≤ · · · ≤ v ν . The set P Ω = { [ y n +1 ; w ] : w ∈ M < Ω } is a free generating set for L ( { y n +1 } ≀ Ω ). Thus, L ( { y n +1 } ≀ Ω ) = L ( P Ω ). Let P ′ Ω be the subset of P Ω consisting ofall elements of the form [ y n +1 ; w ] where w as above with at least one of v , . . . , v ν belongs to M I Ω0 , and P ′′ Ω = P Ω \ P ′ Ω . Thus, P ′′ Ω = { y n +1 } ∪ { [ y n +1 ; ω α · · · ω α n n v · · · v µ ] : α + · · · + α n ≥ v , . . . , v n ∈ n [ j =3 M ψ ( B j ) } . By (4 . L ( C ≀ Y ) = L ( P Y ), L ( C ≀ Ω ) = L ( P Ω ), ψ ( J , P Y ) = I C≀ Ω and Proposition 3 (1), wehave I C≀ Ω = L ( P ′ Ω ≀ P ′′ Ω ) and L ( P Ω ) = L ( P ′′ Ω ) ⊕ I C≀ Ω . Thus, I Ω = J ∩ L (Ω) = γ ( L (Ω )) ⊕ L ( ψ ( D )) ⊕ · · · ⊕ L ( ψ ( D n )) ⊕ L ( P ′ Ω ≀ P ′′ Ω ) . (4 . .3 A decomposition of J By (4 . 6) and (4 . J = J Y ⊕ I Ω ⊕ I C≀ Ω ⊕ L ( E ≀ Ω). By (4 . 3) and L ( A ) = L ( Y ) ⊕ L ( C ≀ Y ), we get L ( A ) = hYi ⊕ L ( B ) ⊕ · · · ⊕ L ( B n ) ⊕ J Y ⊕ L ( C ≀ Y ) . (4 . L ( C ≀ Y ) = L ( P Y ) and by (4 . . 19) becomes L ( A ) = hYi ⊕ L ( B ) ⊕ · · · ⊕ L ( B n ) ⊕ J Y ⊕ L (Ω) ⊕ L ( E ≀ Ω) . or L ( A ) = hYi ⊕ L ( B ) ⊕ · · · ⊕ L ( B n ) ⊕ J Y ⊕ L ( E ≀ Ω) ⊕ L (Ω ) ⊕ L ( C ≀ Ω ) = hYi ⊕ n M j =3 L ( B j ) ⊕ J Y ⊕ L ( E ≀ Ω) ⊕ h Ω i ⊕ n M j =3 L ( ψ ( B j )) ⊕ I Ω ⊕ L ( P ′′ Ω ) ⊕ I C≀ Ω , where J = J Y ⊕ I Ω ⊕ I C≀ Ω ⊕ L ( E ≀ Ω)= γ ( L ( Y )) ⊕ γ ( L (Ω )) ⊕ n M j =3 L ( D j ) ⊕ n M j =3 L ( ψ ( D j )) ⊕ L ( P ′ Ω ≀ P ′′ Ω ) ⊕ L ( E ≀ Ω) . Thus, we summarize in the following result. Theorem 5 With the previous notations, L ( A ) = hYi ⊕ h Ω i ⊕ n M j =3 L ( B j ) ⊕ n M j =3 L ( ψ ( B j )) ⊕ L ( P ′′ Ω ) ⊕ J, where hYi and h Ω i are the free Z -modules on the sets Y and Ω , respectively. Moreover, J = γ ( L ( Y )) ⊕ γ ( L (Ω )) ⊕ n M j =3 L ( D j ) ⊕ n M j =3 L ( ψ ( D j )) ⊕ L ( P ′ Ω ≀ P ′′ Ω ) ⊕ L ( E ≀ Ω) . θ n = ∅ In this section, we assume that θ n = ∅ . The ideal J of L ( A ) is generated by the set { [ y n +1 , y i , y j ] : i, j ∈ [ n ] } . Since L ( C ≀ Y ) = L ( P Y ), we have L ( A ) = L ( Y ) ⊕ L ( P Y ). ByLemma 4, J Y = { } and so, J = J P Y ⊆ L ( P Y ). By (4 . J = ( J ∩ L (Ω)) ⊕ L ( E ≀ Ω).By Proposition 3 (4), J ∩ L (Ω) = { } and so, J = L ( E ≀ Ω) (Alternatively, by Theorem 5, J = L ( E ≀ Ω)). By (4 . L ( A ) = L ( Y ) ⊕ L (Ω) ⊕ L ( E ≀ Ω) and so, we obtain the followingresult. Corollary 2 If θ n = ∅ , then L ( A ) = L ( Y ) ⊕ L (Ω) ⊕ J , where J = L ( E ≀ Ω) . Since E ≀ Ω = { [ y n +1 , y i , . . . , y i κ , b j , . . . , b j λ ] : κ ≥ λ ≥ y i · · · y i κ ∈ Y ∗ ; b j · · · b j λ ∈ Ω ∗ } , we have the Z -module J/γ ( J ) is free on E ≀ Ω modulo γ ( J ).22 heorem 6 In case θ n = ∅ , the Z -module J/γ ( J ) is a free (right) U ( L ( A ) /J ) -module withfree generating set { [ y n +1 , y i , y j ] + γ ( J ) : i, j ∈ [ n ] } .Proof. For the next few lines, we write e ij = [ y n +1 , y i , y j ] for all i, j ∈ [ n ]. We point outthat e ij = [ ω i , y j ] with ω i = [ y n +1 , y i ] for all i, j ∈ [ n ]. Moreover, we let L J ( Y ) = L ( Y ) + JJ and L J (Ω) = L (Ω) + JJ . By Corollary 2, we obtain L ( A ) /J = L J ( Y ) ⊕ L J (Ω). Thus, the free Z -module L ( A ) /J isa direct sum, as Z -modules, of its free Lie subalgebras L J ( Y ) and L J (Ω). We point outthat L J ( Y ) ∼ = L ( Y ) and L J (Ω) ∼ = L (Ω) as Lie algebras in a natural way. For κ ∈ [ r ], let w κ = w κ + J ∈ U ( L ( A ) /J ) \ { } and w , . . . , w r ∈ L ( A ). Write each w i = u i + v i + t i with u i ∈ L ( Y ), v i ∈ L (Ω) and t i ∈ J . Furthermore, we assume that each u i , v i is a simple Liecommutator of degree ≥ Y and Ω, respectively. Then,for all i, j ∈ [ n ],( e ij + γ ( J )) w · · · w r = e ij (ad w ) · · · (ad w r ) + γ ( J )= [ e ij , w , . . . , w r ] + γ ( J )= [ e ij , u + v + t , . . . , u r + v r + t r ] + γ ( J ) . By the multi-linearity of the Lie commutator, e ij , t , . . . , t r ∈ J and by using the Jacobiidentity in the form [ a, [ b, c ]] = [ a, b, c ] − [ a, c, b ], we can write the element ( e ij + γ ( J )) ¯ w · · · ¯ w r as a Z -linear combination of elements of the form [ e ij , z , . . . , z µ ] + γ ( J ), where z , . . . , z µ ∈Y ∪ Ω. For a moment, we consider the product z · · · z µ as a µ -tuple ( z , . . . , z µ ) and assumethat, for some j ∈ [ n ], y j occurs in the k -th position ( k ≥ 2) of ( z , . . . , z µ ). If in the ( k − z , . . . , z µ ) is ω τ , for some τ ∈ [ n ], then[ e ij , z , . . . , ω τ , y j , . . . , z µ ] + γ ( J ) = [ e ij , z , . . . , y j , ω τ , . . . , z µ ] + γ ( J ) , (4 . e ij and [ ω τ , y j ] = e τj ∈ J . If in the ( k − z , . . . , z µ ) is y n +1 , then[ e ij , z , . . . , y n +1 , y j , . . . , z µ ] − ([ e ij , z , . . . , y j , y n +1 , . . . , z µ ] + [ e ij , z , . . . , ω j , . . . , z µ ]) ∈ γ ( J ) . (4 . . 20) and (4 . e ij , z , . . . , z µ ] + γ ( J ), where z , . . . , z µ ∈ Y ∪ Ω, as a Z -linear combination of elements of the form e ij (ad y i ) · · · (ad y i κ )(ad b j ) · · · (ad b j λ ) + γ ( J ) = e ij ¯ y i · · · ¯ y i κ ¯ b j · · · ¯ b j λ , where y i · · · y i κ ∈ Y ∗ and b j · · · b j λ ∈ Ω ∗ . Hence, since the set E ≀ Ω is a free generating setfor J , we have the required result. (cid:3) In this section, we give some auxiliary results that we shall use in the next section. Forsubgroups X , . . . , X m of G , with m ≥ 2, we write [ X , X , . . . , X m ] = [[ X , . . . , X m − ] , X m ]for the subgroup of G generated by all group commutators [ u, x ] where u ∈ [ X , . . . , X m − ]and x ∈ X m . The proof of the following result is elementary and is left to the reader.23 emma 5 1. Let G be a group and H be a subgroup of G . Then, for positive integers m and r ,(a) [ G ] H,m = [ H, G, . . . , G | {z } m ] is a normal subgroup of G and it is generated by the groupcommutators [ h, a , . . . , a m ] for all h ∈ H and a , . . . , a m ∈ G ,(b) [ H, γ m ( G )] ≤ [ G ] H,m and(c) [ γ r ( G ) , H, γ m ( G )] ≤ [ γ r ( G ) , H, G, . . . , G | {z } m ] ≤ [ G ] H,m + r .(d) (Jacobi identity) For any a ∈ H and b, c ∈ G , [ a, [ b, c ]] = [ a, b, c ][ a, c, b ] − v , where v ∈ [ G ] H, .(e) For a positive integer m , with m ≥ , and κ ∈ { , . . . , m − } , let S m,κ be the theset of all permutations of { , . . . , m } satisfying the conditions σ (1) > · · · > σ ( κ ) >σ ( κ + 1) < σ ( κ + 2) < · · · < σ ( m ) and S ∗ m = S m − κ =0 S m,κ . If a ∈ H , g , . . . , g m ∈ G ,then [ a, [ g , . . . , g m ]] γ m +2 ( G ) = m − Y κ =0 Y σ ∈ S m,κ [ a, g σ (1) , . . . , g σ ( m ) ] ( − κ γ m +2 ( G ) . 2. Let G be a finitely generated group with a generating set G , S be a non-empty subset of G \ { } and N S be the normal closure of S in G . If H is the subgroup of G generatedby the set S , then N S = H [ H, G ] . In particular, N S is the subgroup of G generated bythe set { [ s, z ε , . . . , z ε κ κ ] : κ ≥ s ∈ S ; ε j = ± , j ∈ [ κ ]; z ε , . . . , z ε κ κ ∈ G} . By a graded Z -module, we mean a Z -module V with a distinguished Z -module decompo-sition V = V ⊕ V ⊕ · · · , where each V m is a free Z -module of finite rank. For each positiveinteger d , let L d grad ( V ) be the Z -submodule of L ( V ) spanned by all left-normed Lie commu-tators [ v , v , . . . , v κ ], with κ ≥ 1, such that, for i ∈ [ κ ], v i ∈ V m ( i ) for some m ( i ) ≥ m (1) + m (2) + · · · + m ( κ ) = d . Then, L ( V ) is a graded Z -module L ( V ) = L ( V ) ⊕ L ( V ) ⊕ · · · . If V is a free Z -module of finite rank, then we may regard V as a graded Z -module: V = V ⊕ ⊕ ⊕ · · · . Then, the Z -module L d ( V ) is the d -th homogeneous component of L ( V ) asdefined in Section 2.For a positive integer m , with m ≥ 2, let F m be the free group on the m letters y , . . . , y m .The Lie algebra gr( F m ), associated to the lower central series ( γ c ( F m )) c ≥ of F m , is a free Liealgebra with a free generating set { ζ = y F ′ m , . . . , ζ m = y m F ′ m } . Write Z m = { ζ , . . . , ζ m } .Let T ( Z m ) be the free associative Z -algebra on Z m . The Lie subalgebra L ( Z m ) of T ( Z m )generated by the set Z m is a free Lie algebra on Z m . Then, L ( Z m ) = L s ≥ L s ( Z m ). It is clearenough that we may regard L ( Z m ) = gr( F m ). Moreover, for all s ∈ N , L s ( Z m ) = gr s ( F m ) = γ s ( F m ) /γ s +1 ( F m ). For a positive integer e , with e ≥ 2, let r , . . . , r t ∈ γ e ( F m ) \ γ e +1 ( F m ) andlet ρ = r γ e +1 ( F m ) , . . . , ρ t = r t γ e +1 ( F m ). Write R = { r , . . . , r t } and let N = R F m be thenormal closure of R in F m . For a positive integer d , let N d = N ∩ γ d ( F m ). Note that N d = N for all d ∈ [ e ]. Let L ( N ) = L d ≥ e I d ( N ), where I d ( N ) = N d γ d +1 ( F m ) /γ d +1 ( F m ) ≤ gr d ( F m ).Since N is a normal subgroup of F m , the Lie algebra L ( N ) is an ideal of L ( Z m ). Let I be the24deal of L ( Z m ) generated by the set R L = { ρ , . . . , ρ t } . Since I is a homogeneous ideal, wehave I = L d ≥ e I d , where I d = I ∩ gr d ( F m ). It is clear enough that R L ⊆ I e ( N ). Since L ( N ) isan ideal in L ( Z m ), we obtain I ⊆ L ( N ). In particular, for any d ≥ e , I d ⊆ I d ( N ) ≤ gr d ( F m ).By using an inductive argument on d , we may show the following result. Lemma 6 (Labute) With the previous notations, I = L ( N ) if and only if the induced ho-momorphism ξ : I/ [ I, I ] → L ( N ) / [ L ( N ) , L ( N )] is onto. Lemma 6 has been observed in [11, Proof of the Theorem, page 19] (see, also, [12, Proofof Theorem 1, page 53, for a general setting]).For the rest of this section, we assume that m ≥ 3. Let Z m = Z m − ∪{ ζ m } , where Z m − = { ζ , . . . , ζ m − } . Let N = N y m be the normal closure of { y m } in F m . The Lie subalgebra L ( N ) = L d ≥ I d ( N ), where I d ( N ) = N d γ d +1 ( F m ) /γ d +1 ( F m ), of gr( F m ) is an ideal of gr( F m ).Let F m − be the free group on the set { y , . . . , y m − } and form the restricted direct sum L ( F m − ) = L c ≥ L ,c ( F m − ) of the abelian groups L ,c ( F m − ) = γ c ( F m − ) γ c +1 ( F m ) /γ c +1 ( F m ).Since [ γ µ ( F m − ) , γ ν ( F m − )] ⊆ γ µ + ν ( F m − ) for all µ, ν ∈ N , L ( F m − ) is a Lie subalgebraof gr( F m ). By Theorem 2, gr( F m ) = L ( Z m − ) ⊕ L ( { ζ m } ≀ Z m − ). The free Lie algebra L ( { ζ m } ≀ Z m − ) is the ideal of gr( F m ) generated by the element ζ m . For positive integer r , wewrite U r for the Z -module spanned by the set U r = { [ ζ m , z , . . . , z r − ] : z , . . . , z r − ∈ Z m − } .We point out that U = { ζ m } and |U r | = |Z m − | r − for all r ≥ 2. Thus, { ζ m }≀Z m − = S r ≥ U r and L ( { ζ m } ≀ Z m − ) = L ( U ) where U = L r ≥ U r .In the following result, we realize both L ( Z m − ) and L ( { ζ m } ≀ Z m − ) as Lie subalgebrasof gr( F m ) by means of certain subgroups of F m . Namely, we show that L ( F m − ) = L ( Z m − )and L ( N ) = L ( { ζ m } ≀ Z m − ). Proposition 4 With the previous notations,1. The free group F m , with m ≥ , is the semidirect product of N = { y m } F m by F m − .Moreover, N = X [ X, F m ] , where X is the cyclic subgroup of F m generated by theelement y m .2. For any positive integer d ≥ , N d = N ∩ γ d ( F m ) = [ F m ] X,d − . For d ∈ N , I ( N ) equalsthe Z -module h ζ m i spanned by the element ζ m and, for d ≥ , I d ( N ) = L d grad ( { ζ m } ≀Z m − ) .3. L ( N ) = L ( { ζ m } ≀ Z m − ) .4. L ( F m − ) = L ( Z m − ) and gr( F m ) = L ( F m − ) ⊕ L ( N ) .Proof. 1. This is straightforward.2. We induct on d and let d = 2. By Proposition 4 (1), N = X [ X, F m ]. Since [ X, F m ] ⊆ F ′ m ,we have I ( N ) = h ζ m i . By the modular law, N ∩ F ′ m = [ X, F m ]( X ∩ F ′ m ). Since X is an infinite cyclic group, we get X ∩ F ′ m = { } and so, N = N ∩ F ′ m = [ X, F m ].It is easily verified that I ( N ) = U . Assume that our claim is valid for some d ≥ N d = [ F m ] X,d − and I d ( N ) = L d grad ( U ). Let N (1 ,d ) be the subgroup of N d generated by the set { [ y m , y i , . . . , y i d − ] : i , . . . , i d − ∈ [ m ] } and let N (2 ,d ) = [ F m ] X,d .25y Lemma 5 (1)( a ), N (2 ,d ) is a normal subgroup of F m and so, N (1 ,d ) N (2 ,d ) ≤ F m . Weclaim that N d = N (1 ,d ) N (2 ,d ) . By Lemma 5 (1)( a ), it is enough to show that, for all g , . . . , g d − ∈ F m , [ y m , g , . . . , g d − ] ∈ N (1 ,d ) N (2 ,d ) . Write g j = y ε ,j i (1 ,j ) · · · y ε λ,j i ( λ,j ) with ε µ,j = ± i ( µ, j ) ∈ [ m ] and µ ∈ [ λ ]. By using a well known commutator identities andby working modulo N (2 ,d ) , we get[ y m , g , . . . , g d − ] N (2 ,d ) = Y finite [ y m , y j , . . . , y j d − ] α ( j ,...,jd − N (2 ,d ) , where α ( j ,...,j d − ) ∈ Z and j , . . . , j d − ∈ [ m ]. Thus, N d = N (1 ,d ) N (2 ,d ) . Since N (2 ,d ) ⊆ γ d +1 ( F m ), we have I d ( N ) = N (1 ,d ) γ d +1 ( F m ) /γ d +1 ( F m ) = L d grad ( U ). Write V d = U ⊕· · · ⊕ U d and W d = L t ≥ d +1 U t . Thus, L ( { ζ m } ≀ Z m − ) = L ( U ) = L ( V d ⊕ W d ). ByTheorem 2, we have L ( { ζ m } ≀ Z m − ) = L ( V d ) ⊕ L ( W d ≀ V d ) and so, L d grad ( { ζ m } ≀ Z m − ) =gr d ( F m ) ∩ ( L ( V d ) ⊕ L ( W d ≀ V d )). Since L ( V d ) = ⊕ s ≥ L s grad ( V d ), we get L d grad ( { ζ m } ≀ Z m − ) = gr d ( F m ) ∩ M s ≥ L s grad ( V d ) ⊕ L ( W d ≀ V d ) . Since, for all s ≥ L s grad ( V d ) ⊆ gr s ( F m ) and L ( W d ≀ V d ) ⊆ γ d +1 (gr( F m )) = M t ≥ d +1 gr t ( F m ) , we obtain L d grad ( { ζ m } ≀ Z m − ) = L d grad ( U ⊕ · · · ⊕ U d ). Therefore, L d grad ( { ζ m } ≀ Z m − )is the Z -submodule of gr d ( F m ) spanned by all Lie commutators of the form [ v , . . . , v µ ]with µ ≥ v i ∈ U d ( i ) , d ( i ) ≥ i ∈ [ µ ] and d (1) + · · · + d ( µ ) = d . Order the elementsof V d = U ∪ U ∪ · · · ∪ U d , which is a Z -basis of V d , by increasing homogeneous degrees ζ m < [ ζ m , ζ ] < · · · < [ ζ m , ζ m − ] < [ ζ m , ζ , ζ ] < [ ζ m , ζ , ζ ] < · · · . Form the set of basic Lie commutators B ( V , L ) on the elements of the set V d by regardingthe elements of V d having length 1. We point out that the elements of V d have degreeas elements in gr( F m ) as well. Thus, for all i ∈ [ d ], the elements of U i have length 1and degree i . Let B ( d, V , L ) be the subset of B ( V , L ) of degree d . Therefore, the set B ( d, V , L ) is a Z -basis of L d ( { ζ m } ≀ Z m − ). Note that every element of B ( d, V , L ) may beconsidered as an element of N d in a natural way. For example, the basic Lie commutator[[ ζ m , ζ , ζ , ζ ] , ( d − ζ m ] of length d − d may be regarded as an element of N d as [[ y m , y , y , y ] , ( d − y m ]. Write B ( d, G , F ) for the subset of N d constructed by theset B ( d, V , L ). By applying Lemma 5 (1)( d ) (Jacobi identity), we get, for y i , . . . , y i d − ,with i , . . . , i d − ∈ [ m ], the element [ y m , y i , . . . , y i d − ] N (2 ,d ) written as a product ofelements of the form vN (2 ,d ) , where v ∈ B ( d, G , F ). Now, N d +1 = N d ∩ γ d +1 ( F m )= ( N (1 ,d ) N (2 ,d ) ) ∩ γ d +1 ( F m )(modular law) = N (2 ,d ) ( N (1 ,d ) ∩ γ d +1 ( F m )) . Since, for any y i , . . . , y i d − , with i , . . . , i d − ∈ [ m ], the element [ y m , y i , . . . , y i d − ] N (2 ,d ) is written as a product of elements of the form vN (2 ,d ) , where v ∈ B ( d, G , F ), we have,26or all y i , . . . , y i d − , with i , . . . , i d − ∈ [ m ],[ y m , y i , . . . , y i d − ] N (2 ,d ) = Y v ∈B ( d, G ,F ) ( vN (2 ,d ) ) α v , where α v ∈ Z . Since I d ( N d ) = L d grad ( { ζ m } ≀ Z m − and B ( d, V , L ) is a Z -basis of L d grad ( { ζ m } ≀ Z m − ), we get N (1 ,d ) ∩ γ d +1 ( F m ) ⊆ N (2 ,d ) . Therefore, N d +1 = N (2 ,d ) .It is clear enough that I d +1 ( F m ) = [ F m ] X,d γ d +2 ( F m ) /γ d +2 ( F m ) = L d +1grad ( { ζ m } ≀ Z m − ).3. It follows by Proposition 4 (2).4. Recall that gr( F m ) = L ( Z m − ) ⊕ L ( { ζ m } ≀ Z m − ). Since γ d +1 ( F m − ) = γ d ( F m − ) ∩ γ d +1 ( F m ) for all d ∈ N , we obtain L ( F m − ) ∼ = gr( F m − ) as Lie algebras in a naturalway. Since L ( Z m − ) is a free Lie algebra on Z m − and it is a Lie subalgebra of L ( F m − ),we get L ( F m − ) = L ( Z m − ). By Proposition 4 (3), L ( { ζ m }≀Z m − ) = L ( N ). Therefore,gr( F m ) = L ( F m − ) ⊕ L ( N ). (cid:3) gr( F n +1 /N ) For a positive integer n , with n ≥ 2, let F = { x , . . . , x n , t } . Write X = { x , . . . , x n } ,order the elements of X as x < . . . < x n and let C = { t } . Let θ n be a non-empty partialcommutation relation on [ n ] and θ n,δ = { ( a, b ) ∈ θ n | a > b } 6 = ∅ . Without loss of generality,we may assume that (2 , ∈ θ n,δ . Let F n +1 be the free group of rank n + 1 with a free generating set F and R = { [ x a , x b ] , [ t, x i , x j ] , [[ t, x a ] , [ t, x b ]] : i, j ∈ [ n ]; ( a, b ) ∈ θ n,δ } and let N = R F n +1 be the normal closure of R in F n +1 . Namely, F n +1 /N ∼ = F P ( H ) for H the raag with defining graph Γ, vertex set X and edge set { ( x a , x b ) | ( a, b ) ∈ θ n,δ } .Write A = { y , . . . , y n , s } , where y i = x i F ′ n +1 and s = tF ′ n +1 , Y = { y , . . . , y n } with order y < . . . < y n and C = { s } . Let ρ be the natural 1 − X onto Y , thatis, ρ ( x i ) = y i for all i ∈ [ n ]. As in Section 4, the Lie subalgebra L ( A ) of T ( A ) generated bythe set A is a free Lie algebra on A and L ( A ) = gr( F n +1 ). We denote by J N the ideal ofgr( F n +1 ) generated by the set R L = { [ y a , y b ] , [ s, y i , y j ] , [[ s, y a ] , [ s, y b ]] : i, j ∈ [ n ]; ( a, b ) ∈ θ n,δ } .Our main purpose in this section is to prove gr( F n +1 /N ) ∼ = gr( F n +1 ) /J N as Lie algebrasover Z . By Theorem 2, gr( F n +1 ) = L ( Y ) ⊕ L ( { s } ≀ Y ) . (6 . L ( { s } ≀ Y ) is the ideal of gr( F n +1 ) generated by s . Let N t = { t } F n +1 be thenormal closure of { t } in F n +1 . By Proposition 4 (for m = n + 1), we have L ( F n ) = L ( Y ) and L ( N t ) = L ( { s } ≀ Y ). Define J Y = J N ∩ L ( Y ). By Lemma 4, J Y is the ideal of L ( Y ) generatedby the set R ,L = { [ y a , y b ] : ( a, b ) ∈ θ n,δ } . By (4 . J Y is a direct summand of L ( Y ). The set P Y = { [ s ; u ] : u ∈ M < Y } is a free generating set for L ( { s } ≀ Y ). Define J P Y = J N ∩ L ( P Y ) andnote that J P Y is an ideal of gr( F n +1 ). By Lemma 1 and Lemma 4, we have J N = J Y ⊕ J P Y .Write Ω = { s, [ s, y i ] : i ∈ [ n ] } and E = ( { s } ≀ Y ) \ Ω. By Theorem 2, L ( N t ) = L ( { s } ≀ Y ) = L ( P Y ) = L (Ω) ⊕ L ( E ≀ Ω) . (6 . J P Y = ( J N ∩ L (Ω)) ⊕ L ( E ≀ Ω) (see (4 . J N = J Y ⊕ ( J N ∩ L (Ω)) ⊕ L ( E ≀ Ω).By (6 . 1) and (6 . F n +1 ) = L ( Y ) ⊕ L (Ω) ⊕ L ( E ≀ Ω) . (6 . L ( E ≀ Ω) is an ideal in gr( F n +1 ) and J N ∩ L (Ω) is the ideal of L (Ω) generatedby the set R ,L = { [[ s, y a ] , [ s, y b ]] : ( a, b ) ∈ θ n,δ } . Write I Ω = J N ∩ L (Ω). Thus, J N = J Y ⊕ I Ω ⊕ L ( E ≀ Ω) . (6 . . 3) and (6 . F n +1 ) /J N ∼ = ( L ( Y ) /J Y ) ⊕ ( L (Ω) /I Ω ) (6 . Z -modules in a natural way. As shown in Section 2 (for N = R F n +1 ), L ( N ) = L d ≥ I d ( N ),where I d ( N ) = ( N d γ d +1 ( F n +1 )) /γ d +1 ( F n +1 ) and N d = N ∩ γ d ( F n +1 ), is an ideal of gr( F n +1 ).Since R L ⊂ I ( N ) ⊕ I ( N ) ⊕ I ( N ), we get J N ⊆ L ( N ). We set R = { [ x a , x b ] : ( a, b ) ∈ θ n,δ } , R = { [ t, x i , x j ] : i, j ∈ [ n ] } and R = { [[ t, x a ] , [ t, x b ]] : ( a, b ) ∈ θ n,δ } . Thus, R = R ∪ R ∪ R . We denote by F Ω the subgroup of F n +1 generated by the set { t, [ t, x ] , . . . , [ t, x n ] } . It is clear enough that F Ω is a free group of finite rank n + 1. Inparticular, F n +1 is naturally isomorphic to F Ω . J Y in gr( F n +1 ) Let N R = R F n be the normal closure of R in the free group F n , with a free generatingset { x , . . . , x n } , and X R be the subgroup of F n generated by the set R . By Lemma 5 (2)(for G = F n , S = R and N S = N R ), N R = X R [ X R , F n ]. Thus, N R ⊆ F ′ n +1 and, byProposition 4 (1), N R ∩ N t = { } . In the next few lines, we write Y x = { y x, , . . . , y x,n } ,where y x,j = x j F ′ n , j ∈ [ n ]. So, gr( F n ) is a free Lie algebra on Y x . We denote by J Y x the idealof gr( F n ) generated the set R x,L, = { [ y x,a , y x,b ] : ( a, b ) ∈ θ n,δ } . Since J Y x is a homogeneousideal, we have J Y x = L d ≥ J d Y x , where J d Y x = J Y x ∩ gr d ( F n ). Let L Y x ( N R ) = L d ≥ I x,d ( N R ),where I x,d ( N R ) = ( N R ,x,d γ d +1 ( F n )) /γ d +1 ( F n ) and, for any positive integer d ≥ N R ,x,d = N R ∩ γ d ( F n ). Since N R is normal in F n , we have L Y x ( N R ) is an ideal in gr( F n ). Since R x,L, ⊂ I x, ( N R ) and L Y x ( N R ) is an ideal of gr( F n ), we have J Y x ⊆ L Y x ( N R ). Moreover,for any positive integer d , with d ≥ J d Y x ⊆ I x,d ( N R ).By Corollary 1, J Y x is a direct summand of gr( F n ) and so, gr( F n ) /J Y x is a free Z -module.By Theorem 4, the Z -module J Y x / [ J Y x , J Y x ] is a free (right) U (gr( F n ) /J Y x )-module with afree generating set R x,L, + [ J Y x , J Y x ]. By a result of Labute [12, Theorem 1], gr( F n /N R ) ∼ =gr( F n ) /J Y x as Lie algebras. (Note that in the proof of Theorem 1 in [12], the left action ofthe universal enveloping algebra U ( L/I ) on the Z -module I/ [ I, I ] it is used.) In the proof ofTheorem 1 in [12], since J Y x / [ J Y x , J Y x ] is a free (right) U (gr( F n ) /J Y x )-module, the inducedhomomorphism ξ (Lemma 6 for N = N R and I = J Y x ) is onto and so, L Y x ( N R ) = J Y x .Let L Y ( N R ) = L d ≥ I d ( N R ), where I d ( N R ) = ( N R ,d γ d +1 ( F n +1 )) /γ d +1 ( F n +1 ) and, forany positive integer d ≥ N R ,d = N R ∩ γ d ( F n +1 ). Since N R ∩ N t = { } , we have N R ∩ γ d ( F n +1 ) = N R ∩ γ d ( F n ). Hence, for any positive integer d , with d ≥ N R ,x,d = N R ,d . It is28lear enough that L Y ( N R ) is an ideal in L ( F n ) = L ( Y ). The Lie algebra gr( F n ) is isomorphicas a Lie algebra to L ( F n ) = L ( Y ) by a Lie algebra isomorphism χ defined as χ ( y x,j ) = y j forall j ∈ [ n ]. Clearly, χ ( L Y x ( N R )) = L Y ( N R ) and χ ( J Y x ) = J Y . Since L Y x ( N R ) = J Y x and χ is a Lie algebra isomorphism, we obtain L Y ( N R ) = J Y . So, for any d ≥ I d ( N R ) = J d Y . For d = 2, I ( N R ) = J Y = hR ,L i . Thus, we assume that d ≥ 3. By (4 . P ′Y κ , , ≀ P ′′Y κ , Lemma 5 (1) and the Jacobi identity in the form [ x, [ y, z ]] = [ x, y, z ] − [ x, z, y ](where it is needed), we have I d ( N R )(= J d Y ) ⊆ ([ F n ] X R ,d − γ d +1 ( F n +1 )) /γ d +1 ( F n +1 ). Since[ F n ] X R ,d − ⊆ N R ,d = N R ∩ γ d ( F n ), we get ([ F n ] X R ,d − γ d +1 ( F n +1 )) /γ d +1 ( F n +1 ) ⊆ I d ( N R )and so, we have ([ F n ] X R ,d − γ d +1 ( F n +1 )) /γ d +1 ( F n +1 ) = I d ( N R ) = J d Y . We summarize theabove observations in the following result. Proposition 5 With the previous notations, gr( F n /N R ) ∼ = L ( F n ) / L Y ( N R ) as Lie algebrasin a natural way, with L ( F n ) = L ( Y ) and L Y ( N R ) = J Y . Moreover, for any d ≥ , ([ F n ] X R ,d − γ d +1 ( F n +1 )) /γ d +1 ( F n +1 ) = I d ( N R ) = J d Y . Remark 1 The group F n /N R is the raag H . A description of L ( F n ) / L Y ( N R ) has beenalso given in [7] (see, also, [8]). We point out that in [8], it has been shown that F n /N R is aMagnus group. L ( E ≀ Ω) in gr( F n +1 ) In this section, we describe the ideal L ( E ≀ Ω) in gr( F n +1 ). Proposition 6 With the previous notations,1. L ( E ≀ Ω) is the ideal of gr( F n +1 ) generated by the set R ,L = { [ s, y i , y j ] : i, j ∈ [ n ] } .2. Let N R = R F n +1 be the normal closure of R in F n +1 . Then, gr( F n +1 /N R ) ∼ =gr( F n +1 ) / L ( N R ) as Lie algebras in a natural way, with L ( N R ) = L ( E ≀ Ω) . For any d ≥ , L d grad ( E ≀ Ω) = I d ( N R ) = ([ F n +1 ] X R ,d − γ d +1 ( F n +1 )) /γ d +1 ( F n +1 ) , where X R isthe subgroup of F n +1 generated by the set R .Proof. 1. Write I R ,L for the ideal of gr( F n +1 ) generated by the set R ,L . Since E ≀ Ω ⊂ I R ,L , wehave L ( E ≀ Ω) ⊆ I R ,L . On the other hand, since R ,L ⊂ L ( E ≀ Ω) and, by Proposition3 (3), L ( E ≀ Ω) is an ideal of gr( F n +1 ), we have I R ,L ⊆ L ( E ≀ Ω) and so, we get therequired result.2. Recall that L ( N R ) = L d ≥ I d ( N R ), where I d ( N R ) = ( N R ,d γ d +1 ( F n +1 )) /γ d +1 ( F n +1 )and N R ,d = N R ∩ γ d ( F n +1 ). Since R ,L ⊂ I ( N R ) = L ( E ≀ Ω) and L ( N R ) isan ideal of gr( F n +1 ), we have, by Proposition 6 (1), L ( E ≀ Ω) ⊆ L ( N R ). For anypositive integer d , with d ≥ L d grad ( E ≀ Ω) ⊆ I d ( N R ). By (6 . L ( E ≀ Ω) is a directsummand of gr( F n +1 ) and so, gr( F n +1 ) /L ( E ≀ Ω) is a free Z -module. By Theorem 6,the Z -module L ( E ≀ Ω) / [ L ( E ≀ Ω) , L ( E ≀ Ω)] is a free (right) U (gr( F n +1 ) /L ( E ≀ Ω))-modulewith a free generating set R ,L + [ L ( E ≀ Ω) , L ( E ≀ Ω)]. Using similar arguments as before,29e have L ( N R ) = L ( E ≀ Ω). Hence, for any d ≥ L d grad ( E ≀ Ω) = I d ( N R ). For non-negative integers κ and λ , with κ ≥ 2, let V κ + λ be the free Z -module spanned by theset { [ s, y i , . . . , y i κ , z j , . . . , z j λ ] : i , . . . , i κ ∈ [ n ]; z j , . . . , z j λ ∈ Ω } . Clearly, L ( E ≀ Ω) = L M κ ≥ ,λ ≥ κ + λ ≥ V κ + λ . Then, for all d ≥ L d grad ( E ≀ Ω) is the Z -module of L ( E ≀ Ω) consisting of all Liecommutators of the form [ v , . . . , v ν ] where v i ∈ V κ i + λ i , i ∈ [ ν ] and ( κ + λ ) + · · · +( κ ν + λ ν ) = d − ν . For d ≥ 3, write I d ( N R ) = [ F n +1 ] X R ,d − γ d +1 ( F n +1 ) /γ d +1 ( F n +1 ).Since [ F n +1 ] X R ,d − ⊆ N R ,d = N R ∩ γ d ( F n +1 ) for all d ≥ 3, we have I d ( N R ) ⊆ I d ( N R ) = L d grad ( E ≀ Ω). On the other hand, since any Lie commutator [ v , . . . , v ν ] ofthe above form belongs to I d ( N R ), we get L d grad ( E ≀ Ω) ⊆ I d ( N R ) and so, we obtainthe required result. (cid:3) Remark 2 The group F n +1 /N R is a Formanek-Procesi group F P ( F n ) with F n a free groupof rank n ≥ 2. That is, F P ( F n ) = h t, g , . . . , g n : [ t, g i , g j ] = 1 , i, j ∈ [ n ] i . A presentation ofgr( F n +1 /N R ) has been also given in [16, Theorem 1 (1)]. Furthermore, it has been shownthat F P ( F n ) is a Magnus group [16, Theorem 1 (3)]. I Ω in gr( F n +1 ) For i ∈ [ n ], we write q i = [ t, x i ], ω i = q i γ ( F n +1 ) and Ω = { ω , . . . , ω n } . Let F Ω be the freegroup on the set { q , . . . , q n } . Note that F Ω ⊆ F ′ n +1 and F Ω γ ( F n +1 ) /γ ( F n +1 ) = [ X, F n ] γ ( F n +1 ) /γ ( F n +1 ) , where X is the subgroup of F n +1 generated by t . For a positive integer κ , let L od ,κ ( F Ω ) = γ κ ( F Ω ) γ κ +1 ( F n +1 ) /γ κ +1 ( F n +1 ) and form the (restricted) direct sum L od ( F Ω ) of the abeliangroups L od ,k ( F Ω ). Since [ γ κ ( F Ω ) , γ λ ( F Ω )] ⊆ γ κ + λ ( F Ω ) for all κ, λ ∈ N , we have L od ( F Ω )is a Lie subalgebra of gr( F n +1 ). In fact, L od ( F Ω ) = L (Ω ) = L ( I ( N t )). By (6 . L ( N t ) = L (Ω) ⊕ L ( E ≀ Ω). We point out that L (Ω) = L ( I ( N t ) ⊕ I ( N t )). For a positive integer d , L d grad (Ω) is the Z -submodule of L (Ω) spanned by all Lie commutators [ v , . . . , v µ ], µ ≥ v j ∈ I i ( j ) ( N t ), i ( j ) ∈ [2], j ∈ [ µ ] and i (1) + · · · + i ( µ ) = d . Thus, by (6 . 2) and Proposition6 (2), for d ≥ I d ( N t ) = L d grad (Ω) ⊕ I d ( N R ) . (6 . I ( N t ) = L (Ω) = h s i and I ( N t ) = L (Ω) = h [ s, y j ] : j ∈ [ n ] i . For d ≥ 3, by Propo-sition 4, it is easily verified that I d ( N t ) = ([ F n +1 ] F Ω0 ,d − ) γ d +1 ( F n +1 ) /γ d +1 ( F n +1 ). By Propo-sition 6, I d ( N R ) = ([ F n +1 ] [ F Ω0 ,F n ] ,d − ) γ d +1 ( F n +1 ) /γ d +1 ( F n +1 ). We denote by L d grad , (Ω) the Z -submodule of L d grad (Ω) spanned by all Lie commutators [ v , . . . , v µ ], with v j ∈ I ( N t ), j ∈ [ µ ] and 2 µ = d . Thus, L d grad , (Ω) = { } if d is odd, and L µ grad , (Ω) = L od ,µ ( F Ω ). Wewrite L d grad , (Ω) for the Z -submodule of L d grad (Ω) spanned by all Lie commutators [ v , . . . , v µ ],with v j ∈ I i ( j ) ( N t ), j ∈ [ µ ], i (1) , . . . , i ( µ ) ∈ [2], at least one of i (1) , . . . , i ( µ ) is equal to 1and i (1) + · · · + i ( µ ) = d . By Theorem 2, L (Ω) = L ( I ( N t )) ⊕ L ( I ( N t ) ≀ I ( N t )). Thus, for30ll d ∈ N , we have L d grad (Ω) = L d grad ( I ( N t )) ⊕ L d grad ( I ( N t ) ≀ I ( N t )). By the definitions of L d grad , (Ω) and L d grad , (Ω), we have, for all d ∈ N , L d grad (Ω) = L d grad , (Ω) ⊕ L d grad , (Ω) (6 . . d ∈ N , L d grad ( I ( N t ) ≀ I ( N t )) = L d grad , (Ω) . (6 . µ ≥ j (1) , . . . , j ( µ ), with j (1) + · · · + j ( µ ) = d and d ≥ X ( j (1) ,...,j ( µ )) = [ X j (1) , . . . , X j ( µ ) ], where X j ( i ) ∈ { X, F Ω } , i ∈ [ µ ], j (1) , . . . , j ( µ ) ∈ [2],and at least one of X j (1) , . . . , X j ( µ ) equals X . Let X grad ,d = Q j (1)+ ··· + j ( µ )= d X ( j (1) ,...,j ( µ )) and M ,d = X grad ,d γ d +1 ( F n +1 ) /γ d +1 ( F n +1 ). By the definitions of M ,d and L d grad , (Ω), we have M ,d = L d grad , (Ω) for all d ≥ 1. Form the (restricted) direct sum L M = L d ≥ M ,d of theabelian groups M ,d . By using the Jacobi identity in the form [ x, [ y, z ]] = [ x, y, z ]+ [ x, z, y ], wemay show that L M is a Lie subalgebra of gr( F n +1 ). By (6 . L M = L ( I ( N t ) ≀ I ( N t )).By (6 . L (Ω) = L od ( F Ω ) ⊕ L M . (6 . κ , let F Ω ,κ = F Ω ∩ γ κ ( F n +1 ). Form the (restricted) direct sum L ( F Ω )of the abelian groups I κ ( F Ω ) = F Ω ,κ γ κ +1 ( F n +1 ) /γ κ +1 ( F n +1 ). It is clear enough that L ( F Ω ) isa Lie subalgebra of gr( F n +1 ). Proposition 7 With the previous notations, L ( F Ω ) = L (Ω) . In particular, L ( F Ω ) is a freeLie algebra on I ( F Ω ) ⊕ I ( F Ω ) . For any positive integer d , I d ( F Ω ) = L d grad (Ω) .Proof. Since F Ω ⊆ F ′ n +1 , we have I ( F Ω ) = I ( N t ) and I ( F Ω ) = I ( N t ). Since L ( F Ω ) isa Lie subalgebra of gr( F n +1 ) and L (Ω) is free on I ( N t ) ⊕ I ( N t ), we have L (Ω) ⊆ L ( F Ω ).Clearly, L ( F Ω ) ⊆ L ( N t ). By (6 . 2) and the modular law, we get L ( F Ω ) = L (Ω) ⊕ ( L ( E ≀ Ω) ∩ L ( F Ω )) . (6 . L ( E ≀ Ω) ∩ L ( F Ω ) = { } . Let u ∈ L ( E ≀ Ω) ∩ L ( F Ω ). By Proposition 6 (2), L ( N R ) = L ( E ≀ Ω). Since both L ( N R ) and L ( F Ω ) are graded Lie algebras, we may assumethat u = uγ c +1 ( F n +1 ) ∈ I c ( N R ) ∩ I c ( F Ω ) with u ∈ γ c ( F n +1 ) and some c ∈ N . It is clearenough that c ≥ 3. Then, u = u γ c +1 ( F n +1 ) = u γ c +1 ( F n +1 ), where u ∈ [ F n +1 ] X R ,c − and u ∈ F Ω ,c = F Ω ∩ γ c ( F n +1 ). To get a contradiction, we assume that u / ∈ γ c +1 ( F n +1 ). Hence, u , u ∈ γ c ( F n +1 ) \ γ c +1 ( F n +1 ). Since F Ω is a free group on the set F Ω = { t, q , . . . , q n } , u is (uniquely) written as u = z µ i · · · z µ κ i κ , with z i , . . . , z i κ ∈ F Ω , i λ = i λ +1 , λ ∈ [ κ − 1] and µ , . . . , µ κ ∈ Z \ { } . In the next few lines, we write t = tγ c +1 ( F n +1 ) , x i = x i γ c +1 ( F n +1 ), i ∈ [ n ] and F n +1 ,c = F n +1 /γ c +1 ( F n +1 ). Thus, F n +1 ,c is a free nilpotent group of rank n + 1and class c with a free generating set { t, x , . . . , x n } . Note that, for any j ∈ [ n ], q j = q j γ c +1 ( F n +1 ) = [ t, x j ] ∈ F ′ n +1 ,c . Using the identity ab = ba [ a, b ] as many times as needed inthe above expression of u and working in F n +1 ,c , the element u = u γ c +1 ( F n +1 ) is writtenin F n +1 ,c as u = t α q α · · · q α n n w ( t, q , . . . , q n ) · · · w c ( t, q , . . . , q n ) , (6 . α, α , . . . , α n ∈ Z , w ( t, q , . . . , q n ) , . . . , w c ( t, q , . . . , q n ) are products of group commu-tators in t, q , . . . , q n and have weights 3 , , . . . , c in F n +1 , respectively. Thus, for j ∈ [ c ] \{ , } , w j ( t, q , . . . , q n ) ∈ F Ω ,j = F Ω ∩ γ j ( F n +1 ). 31orm the (restricted) direct sum gr( F n +1 ,c ) of the abelian groups γ r ( F n +1 ,c ) /γ r +1 ( F n +1 ,c ).It is well known that gr( F n +1 ,c ) is a free nilpotent Lie algebra of rank n + 1 and class c witha free generating set { tF ′ n +1 ,c , x F ′ n +1 ,c , . . . , x n F ′ n +1 ,c } . By (6 . u is written in gr( F n +1 ,c )as u = t α F ′ n +1 ,c + q α · · · q α n n γ ( F n +1 ,c ) + w ( t, q , . . . , q n ) γ ( F n +1 ,c ) + · · · + w c − ( t, q , . . . , q n ) γ c ( F n +1 ,c ) + w c ( t, q , . . . , q n ) . Since u ∈ γ c ( F n +1 ) \ γ c +1 ( F n +1 ), that is, u ∈ γ c ( F n +1 ,c ) (a non-trivial element in F n +1 ,c ),we get u = w c ( t, q , . . . , q n ). Therefore, u γ c +1 ( F n +1 ) ∈ M ,c . By (6 . . 8) and (6 . u γ c +1 ( F n +1 ) ∈ L c grad , (Ω). On the other hand, u γ c +1 ( F n +1 ) ∈ I c ( N R ). By (6 . 7) and(6 . L ( E ≀ Ω) ∩ L ( F Ω ) = { } and so, by(6 . L ( F Ω ) = L (Ω). Since L (Ω) is free on Ω and h Ω i = I ( F Ω ) ⊕ I ( F Ω ), we have L ( F Ω ) isa free Lie algebra on I ( F Ω ) ⊕ I ( F Ω ). Since both L ( F Ω ) and L (Ω) are graded Lie algebras,we have I d ( F Ω ) = L d grad (Ω) for all d . (cid:3) Let N R = R F Ω be the normal closure of R in F Ω and X R be the subgroup of F Ω generated by the set R . In fact, X R ⊆ F ′ Ω . By Lemma 5 (2), N R = X R [ X R , F Ω ]. Thus, N R ⊆ F ′ Ω ∩ γ ( F n +1 ). For a positive integer c , let N R ,c = N R ∩ γ c ( F n +1 ). For c ∈ [4], N R ,c = N R . For c ≥ 4, we write I c ( N R ) = ( N R ,c γ c +1 ( F n +1 )) /γ c +1 ( F n +1 ). Form the(restricted) direct sum L ( N R ) of the abelian groups I c ( N R ). Since N R ,c ⊆ F Ω ,c for all c ,we have L ( N R ) is a Lie subalgebra of L ( F Ω ). Since N R is a normal subgroup of F Ω , weobtain L ( N R ) is an ideal of L ( F Ω ). Let I Ω be the ideal of L ( F Ω ) = L (Ω) generated by theset R ,L = { [ ω a , ω b ] : ( a, b ) ∈ θ n,δ } . For µ ≥ j (1) , . . . , j ( µ ), with j (1) + · · · + j ( µ ) = c ≥ 1, let Y R , ( j (1) ,...,j ( µ )) = [ X R , Y j (1) , . . . , Y j ( µ ) ], with Y j ( i ) ∈ { X, F Ω } , i ∈ [ µ ], j (1) , . . . , j ( µ ) ∈ [2], and at least one of Y j (1) , . . . , Y j ( µ ) equals X . Moreover, let Y R , grad ,c = Q j (1)+ ··· + j ( µ )= c Y R , ( j (1) ,...,j ( µ )) . We point out that F Ω /N R is raag. Proposition 8 With the previous notations, gr( F Ω /N R ) ∼ = L ( F Ω ) / L ( N R ) as Lie algebrasin a natural way with L ( N R ) = I Ω . For all non-negative integers c , I c +4 ( N R ) = I c +4Ω . If c = 2 ν + 1 , then I ν +5 ( N R ) = Y R , grad , ν +1 γ ν +6 ( F n +1 ) /γ ν +6 ( F n +1 ) and, if c = 2 ν , then I ν +4 ( N R ) = ([ F Ω ] X R ,ν Y R , grad , ν ) γ ν +5 ( F n +1 ) /γ ν +5 ( F n +1 ) .Proof. Since R ,L ⊂ I ( N R ) and L ( N R ) is an ideal of L ( F Ω ), we have I Ω ⊆ L ( N R ).Thus, for any positive integer d , with d ≥ I d Ω ⊆ I d ( N R ). As in Section 4.2 (taking I Ω for J ∩ L (Ω) and having in mind Proposition 3 (4)), we may show that I Ω is a direct summandof L ( F Ω ) = L (Ω) (by using (4 . . . 16) and (4 . L ( F Ω ) /I Ω is a free Z -module. By Theorem 4 (for I m = I Ω and Y m = { s } ∪ Ω ), the Z -module I Ω / [ I Ω , I Ω ] is afree (right) U ( L ( F Ω ) /I Ω )-module with a free generating set R ,L + [ I Ω , I Ω ]. By a result ofLabute [12, Theorem 1], gr( F Ω /N R ) ∼ = L ( F Ω ) /I Ω as Lie algebras. In the proof of Theorem1 in [12], since I Ω / [ I Ω , I Ω ] is a free (right) U ( L ( F Ω ) /I Ω )-module, the induced homomorphism ξ (Lemma 6 for N = N R and I = I Ω ) is onto and so, L ( N R ) = I Ω . Hence, for all d ≥ I d ( N R ) = I d Ω . For a non-negative integer c , [ F Ω ] X R ,c ⊆ N R , c +4 = N R ∩ γ c +4 ( F n +1 ) andso, ([ F Ω ] X R ,c γ c +5 ( F n +1 )) /γ c +5 ( F n +1 ) ⊆ I c +4 ( N R ) = I c +4Ω . (6 . Y R , grad ,c ⊆ N R ,c +4 = N R ∩ γ c +4 ( F n +1 ) for all c ≥ 1, we have( Y R , grad ,c γ c +5 ( F n +1 )) /γ c +5 ( F n +1 ) ⊆ I c +4 ( N R ) = I c +4Ω . (6 . I Ω (by (4 . 17) and (4 . d ≥ I d Ω = L d (Ω (1)2 ) ⊕ L d grad ( P ′ Ω , , ≀ P ′′ Ω ) ⊕ · · · ⊕ L d grad ( P ′ Ω n − , , ≀ P ′′ Ω n − ) ⊕ L d grad ( P ′ Ω ≀ P ′′ Ω ) , (6 . (1)2 = { [ ω , ω , a ω , b ω ] : a + b ≥ } . That is, Ω (1)2 is the standard free generatingset for γ ( L (Ω )). It follows, by (6 . 12) (for c = 0) and (6 . 14) (for d = 4), that I ( N R ) = I = hR ,L i . Thus, we may assume that d ≥ 5. Write d = c + 4, with c ≥ 1. Let c = 2 ν + 1.By (6 . P ′′ Ω ,ω , . . . , P ′′ Ω n − ,ω n , P ′ Ω , , ≀ P ′′ Ω , . . . , P ′ Ω n − , , ≀ P ′′ Ω n − , P ′ Ω ≀ P ′′ Ω , Lemma 5 (1) and the Jacobi identity in the form [ x, [ y, z ]] = [ x, y, z ] − [ x, z, y ](where it is needed), we have I ν +5 ( N R ) ⊆ Y R , grad , ν +1 γ ν +6 ( F n +1 ) /γ ν +6 ( F n +1 ). By (6 . c = 2 ν + 1), we get I ν +5 ( N R ) = Y R , grad , ν +1 γ ν +6 ( F n +1 ) /γ ν +6 ( F n +1 ). Let c = 2 ν .By using similar arguments as above and by (6 . 12) and (6 . I ν +4 ( N R ) =([ F Ω ] X R ,ν Y R , grad , ν ) γ ν +5 ( F n +1 ) /γ ν +5 ( F n +1 ). (cid:3) gr( F P ( H ))For j ∈ [4] \ { } , let H j be the subgroup of F n +1 generated by the set R j and H be thesubgroup of F n +1 generated by the set R . It follows, by Lemma 5 (for G = F n +1 , G = F = { x , . . . , x n , t } , S = R and N S = N ), that N = H [ H, F n +1 ]. For j ∈ [4] \ { } , let H j,n +1 bethe set { [ r, z ε , . . . , z ε κ κ ] : r ∈ R j ; κ ≥ ε i = ± i ∈ [ κ ]; z , . . . , z κ ∈ F } . By Lemma 5 (2),the set R j ∪ H j,n +1 is a generating set for H j [ H j , F n +1 ]. In other words, H j [ H j , F n +1 ] is thenormal closure of R j in F n +1 . Lemma 7 1. Let N = R F n +1 be the normal closure of R in F n +1 . Then, N = H [ H , F n +1 ][ H , F n +1 ][ H , F n +1 ] . 2. For v = [ r, z ε , . . . , z ε κ κ ] with r ∈ R , κ ≥ , ε i = ± , i ∈ [ κ ] and at least one of z , . . . , z κ is t , we have vγ κ +3 ( F n +1 ) ∈ I κ +2 ( N R ) . Otherwise, vγ κ +3 ( F n +1 ) ∈ I κ +2 ( N R ) .3. For v = [ r, z ε , . . . , z ε κ κ ] with r ∈ R , κ ≥ , ε i = ± , i ∈ [ κ ] and z , . . . , z κ ∈ F , wehave vγ κ +5 ( F n +1 ) ∈ I κ +4 ( N R ) ⊕ I κ +4 ( N R ) .Proof. 1. It is enough to show that [ H, F n +1 ] ⊆ [ H , F n +1 ][ H , F n +1 ][ H , F n +1 ]. By using theidentities [ ab, c ] = [ a, c ][ a, c, b ][ b, c ], [ a, bc ] = [ a, c ][ a, b ][ a, b, c ] and Lemma 5 (1), we have N = H [ H , F n +1 ][ H , F n +1 ][ H , F n +1 ].2. Clearly, vγ κ +3 ( F n +1 ) = [ r, z , . . . , z κ ] ε γ κ +3 ( F n +1 ), with ε = ± 1. Without loss the gener-ality, we may assume v = [ r, z , . . . , z κ ], with r ∈ R and z , . . . , z κ ∈ F . In the next fewlines, we consider the word z · · · z κ as the κ -tuple ( z , . . . , z κ ). Let 1 ≤ i < · · · < i ν ≤ κ be the ν different positions of t in the κ -tuple ( z , . . . , z κ ). By our hypothesis, ν ≥ v (0) = [ r, z , . . . , z i − ] with z , . . . , z i − ∈ X . Note that, for i = 1, v (0) = r . ByLemma 5 (1)( e ) and, working modulo γ κ +3 ( F n +1 ), v = − [[ t, v (0)] , z i +1 , . . . , z κ ]= P i − λ =0 ( P σ ∈ S i − ,λ ( − λ +1 [ t, z σ (1) , . . . , z σ ( i − , z i +1 , . . . , z κ ]) . 33e point out that r occurs in the word z σ (1) · · · , z σ ( i − . By Lemma 5 (1)( d ) (Jacobiidentity), each of the above group commutators belongs to [ F n +1 ] X R ,κ − . By Propo-sition 7, we have vγ κ +3 ( F n +1 ) ∈ I κ +2 ( N R ). Otherwise, by Proposition 5, we have vγ κ +3 ( F n +1 ) ∈ I κ +2 ( N R ).3. We proceed as above. Without loss of generality, we may assume v = [ r, z , . . . , z κ ], with r ∈ R and z , . . . , z κ ∈ F . We separate two cases. First, we assume that z , . . . , z κ ∈ X .Since [[ t, x a ] , z ] , [[ t, x b ] , z ] ∈ R for all r = [[ t, x a ] , [ t, x b ]], with ( a, b ) ∈ θ n,δ , and z ∈ X ,we have vγ κ +5 ( F n +1 ) ∈ I κ +4 ( N R ). Thus, we assume that at least one of z , . . . , z κ is t .Let 1 ≤ i < · · · < i ν ≤ κ be the ν different positions of t in the κ -tuple ( z , . . . , z κ ). Byour hypothesis, ν ≥ 1. If i ≥ 2, then, z ∈ X and so, by using the above argument, weget vγ κ +5 ( F n +1 ) ∈ I κ +4 ( N R ). Therefore, z = t . By working modulo I κ +4 ( N R ), usingthe Jacobi identity in the form [ x, y, z ] = [ x, z, y ] + [ x, [ y, z ]] as many times as neededand by Proposition 8, we obtain vγ κ +5 ( F n +1 ) ∈ I κ +4 ( N R ) ⊕ I κ +4 ( N R ). Thus, in anycase, vγ κ +5 ( F n +1 ) ∈ I κ +4 ( N R ) ⊕ I κ +4 ( N R ). Proposition 9 With the previous notations,1. L ( N ) = J N .2. gr(FP(H)) = gr(F n+1 / N) ∼ = gr(F n+1 ) / J N as Lie algebras.Proof. 1. Write L = L Y ( N R ) + L ( N R ) + L ( N R ). By Propositions 5, 8, 6 (2) and by (6 . L is the additive direct sum of the Lie subalgebras L Y ( N R ), L ( N R ) and L ( N R ) of gr( F n +1 ), and L = J N . Hence, L is the ideal of gr( F n +1 ) generated bythe set R L . Since R L ⊂ I ( N ) ⊕ I ( N ) ⊕ I ( N ) and L ( N ) is an ideal of gr( F n +1 ),we have J N ⊆ L ( N ). Suppose that there exists d ∈ N such that J dN " I d ( N ) and J sN = I s ( N ) for all s ≤ d − 1. It is clear enough that d ≥ 5. Thus, there exists u ∈ N d = N ∩ γ d ( F n +1 ) such that uγ d +1 ( F n +1 ) / ∈ J dN and so, u / ∈ γ d +1 ( F n +1 ). It iseasily seen that [ H j , F n +1 ] ⊆ γ j +1 ( F n +1 ) for j ∈ [4] \ { } . By Lemma 7 (1), everyelement u ∈ N \ { } is written as u = hh h h with h ∈ H and h j ∈ [ H j , F n +1 ], j ∈ [4] \ { } . Using the identity ab = ba [ a, b ] as many times as needed in the aboveexpression of u and working in F n +1 ,d , the element u = uγ d +1 ( F n +1 ) is written in F n +1 ,d as a product w w · · · w ν , where each w j is of weight j in F n +1 and it is a product ofgroup commutators in elements of R ∪ ( S j ∈ [4] \{ } H j,n +1 ). Thus, for example, w isa product of [ x a , x b ], ( a, b ) ∈ θ n,δ . As in the proof of Proposition 7, u is written ingr( F n +1 ,d ) as u = w γ ( F n +1 ,d ) + w γ ( F n +1 ,d ) + · · · + w d − γ d ( F n +1 ,d ) + w d . Since u ∈ γ d ( F n +1 ) \ γ d +1 ( F n +1 ), that is, u ∈ γ d ( F n +1 ,d ) (a non-trivial element in F n +1 ,d ), weget u = w d . Therefore, by working modulo γ d +1 ( F n +1 ), u = Q u µ is a product of groupcommutators of weight d in F n +1 where in each u µ at least one element of R occurs.We claim that u = u + u + u , where u j ∈ I d ( N R j ), j ∈ [4] \ { } . We point out thatif r ∈ R ∪ H ,n +1 occurs in u µ for some µ , then, by Proposition 6, u µ ∈ I d ( N R ). Forsome µ , let u µ = w ( a , . . . , a λ ) ∈ gr d ( F n +1 ) with a , . . . , a λ ∈ R ∪ R ∪ H ,n +1 ∪ H ,n +1 .By using the Jacobi identity as many as needed and by Lemma 7 (2)-(3), we get u µ ∈ I d ( N R ) ⊕ I d ( N R ). Therefore, u = u + u + u , where u j ∈ I d ( N R j ), j [4] \ { } . By(6 . u ∈ J dN which is the required contradiction.Thus, for all d ≥ J dN = I d ( N ) and so, L ( N ) = J N .34. Write G = F n +1 /N . Since N ⊆ F ′ n +1 , we have G/G ′ ∼ = F n +1 /F ′ n +1 and gr( G ) isgenerated as a Lie algebra by the set G ′ = { α i , b : i ∈ [ n ] } where α i = x i G ′ , i ∈ [ n ],and b = tG ′ . Since gr( F n +1 ) is free on the set A = { y , . . . , y n , s } , where y i = x i F ′ n +1 , i ∈ [ n ] and s = tF ′ n +1 , the map ζ from gr( F n +1 ) into gr( G ) satisfying the conditions ζ ( y i ) = α i , i ∈ [ n ], and ζ ( s ) = b , extends uniquely to a Lie algebra homomorphism.Since gr( G ) is generated as a Lie algebra by the set G ′ , we have ζ is onto. Hence,gr( F n +1 ) / Ker ζ ∼ = gr( G ) as Lie algebras. By definition, J N ⊆ Ker ζ and so, ζ induces aLie algebra epimorphism ζ from gr( F n +1 ) /J N onto gr( G ). In particular, ζ induces ζ d ,say, a Z -linear mapping from (gr d ( F n +1 ) + J N ) /J N onto gr d ( G ) = γ d ( G ) /γ d +1 ( G ). For d ≥ 2, gr d ( G ) ∼ = ( γ d ( F n +1 ) γ d +1 ( F n +1 ) N ) / ( γ d +1 ( F n +1 ) N ) ∼ = γ d ( F n +1 ) / ( γ d ( F n +1 ) ∩ ( γ d +1 ( F n +1 ) N )) . Since γ d +1 ( F n +1 ) ⊆ γ d ( F n +1 ), we have, by the modular law, γ d ( F n +1 ) / ( γ d ( F n +1 ) ∩ ( γ d +1 ( F n +1 ) N )) = γ d ( F n +1 ) / ( γ d +1 ( F n +1 ) N d ) ∼ = gr d ( F n +1 ) /I d ( N d ) . For d ≥ 2, and in the view of Proposition 9 (1), we have I d ( N ) = J dN and so, we getrank(gr d ( G )) = rank(gr d ( F n +1 ) /J dN ). Since J N = L d ≥ J dN , we have(gr d ( F n +1 ) + J N ) /J N ∼ = gr d ( F n +1 ) / (gr d ( F n +1 ) ∩ J N )= gr d ( F n +1 ) /J dN . By Theorem 5, we obtain Ker ζ d is torsion-free. Since rank(gr d ( G )) = rank(gr d ( F n +1 ) /J dN ),we have Ker ζ d = { } and so, ζ d is an isomorphism. Since ζ is an epimorphism andeach ζ d is isomorphism, we have ζ is a Lie algebra isomorphism. Hence, we have therequired result. (cid:3) Corollary 3 With the previous notations, gr(FP(H)) = gr(F n+1 / N) ∼ = gr(F n / N R ) ⊕ gr(F Ω / N R ) as Z -modules.Proof. By Proposition 5, Proposition 8 and by (6 . (cid:3) F n +1 /N is residually nilpotent Let χ n be the natural epimorphism from F n +1 onto F n satisfying the conditions χ n ( x i ) = x i , i ∈ [ n ], and χ n ( t ) = 1. Note that Ker χ n = N t , the normal closure of { t } in F n +1 .By Proposition 4 (1), F n +1 is the semidirect product of N t by F n . Let π F be the naturalepimorphism from F n onto F n /N R . By the definitions of Ker( π F χ n ) and N t N R , we haveKer( π F χ n ) = N t N R . Hence, F n +1 /N t N R ∼ = F n /N R . By a result of Duchamp and Krob[8], F n /N R is residually nilpotent. For j ∈ [4] \ { } , let N R j ,θ be the normal closure of R j in F n +1 . By Lemma 7 (1) and Lemma 5 (1)( a ), we get N = N R ,θ N R ,θ N R ,θ . We claimthat N ⊆ N t N R . Indeed, it is enough to show that N R ,θ ⊆ N t N R . Furthermore, it isenough to show that w − [ x a , x b ] w ∈ N t N R for all w ∈ F n +1 and ( a, b ) ∈ θ n,δ . By writing35 − [ x a , x b ] w = [ x a , x b ][[ x a , x b ] , w ] and the definitions of N t and N R , we obtain the requiredresult.In the next few lines, let G = F n +1 /N and H = N t N R . Since N ⊆ F ′ n +1 , we have G/G ′ ∼ = F n +1 /F ′ n +1 . Let h = wN ∈ G with w / ∈ N . If w / ∈ F ′ n +1 and since G/G ′ ∼ = F n +1 /F ′ n +1 ∼ = Z n +1 in a natural way, we get h / ∈ G ′ and G/G ′ is nilpotent. Thus, we assume that w ∈ F ′ n +1 andso, h ∈ G ′ . The element w is written as a product of group commutators w ( x , . . . , x n , t ). Weseparate two cases.1. Let w / ∈ H . Since N t is normal in F n +1 , we have H contains some group commutator w ( x , . . . , x n , t ), occurring in the product of w , having the letter t . Hence, w modulo H is a product of group commutators w ( x , . . . , x n ). Set w = Q w ( x , . . . , x n ) and so, w = w h t h with h t ∈ N t and h ∈ N R . Note that wH = w H and w / ∈ H .Now, π F χ n ( w ) = π F ( w h ) = w N R . Since F n /N R is residually nilpotent and w / ∈ N R (otherwise, w ∈ H which is a contradiction), there exists a normal subgroupΛ w /N R , with Λ w normal in F n , such that w / ∈ Λ w and F n / Λ w is nilpotent. Let K w =( π F χ n ) − (Λ w /N R ). Since π F χ n is epimorphism, we have K w is a normal subgroup of F n +1 and w / ∈ K w (otherwise, w N R ∈ Λ w /N R . Hence, there exists w ′ ∈ Λ w suchthat w N R = w ′ N R . Since N R ⊆ Λ w , we get a contradiction). Since H ⊆ K w ,we get F n +1 /K w ∼ = F n / Λ w . Since F n / Λ w is nilpotent, we obtain ( F n +1 /H ) / ( K w /H ) isnilpotent.2. Let w ∈ H . Thus, h = wN = w ′ N for some w ′ ∈ N t . So, without loss of generality, wemay regard w ∈ N t . Since N t,d = N t ∩ γ d ( F n +1 ) ⊆ γ d ( F n +1 ) for all d ∈ N and F n +1 isresidually nilpotent, we get T d ≥ N t,d = { } . Therefore, there exists d ∈ N such that w ∈ N t,d \ N t,d +1 . Clearly, we may assume that d ≥ 2. Let d = 2. By Proposition 4 (2), wγ ( F n +1 ) is a non-trivial element of I ( N t ) = F Ω γ ( F n +1 ) /γ ( F n +1 ). By Propositions6 (2) and 8, there exists w ∈ F Ω \ F ′ Ω and w γ ( F n +1 ) / ∈ I ( N ) (that is, w γ ( F n +1 ) = v γ ( F n +1 ) for all v ∈ N ) such that wγ ( F n +1 ) = w γ ( F n +1 ). Suppose that wN ∈ γ ( G ). Since γ ( G ) ∼ = γ ( F n +1 ) /N in a natural way, we have wN = hN for some h ∈ γ ( F n +1 ). Hence, w ∈ γ ( F n +1 ), which is a contradiction. Thus, wN / ∈ γ ( G ) and G/γ ( G ) is nilpotent. So, we assume that d ≥ 3. Clearly, wγ d +1 ( F n +1 ) is a non-trivialelement of I d ( N t ). By (6 . I d ( N t ) = L d grad (Ω) ⊕ I d ( N R ). By (6 . I d ( N R ) ⊆ I d ( N ). We point out that N R = N R ,θ . By theanalysis of I d ( N t ) given in Section 6.3, Proposition 7, Proposition 8 and Proposition6 (2), there exists w ∈ F Ω ,d and w γ d +1 ( F n +1 ) / ∈ I d ( N ) (that is, w γ d +1 ( F n +1 ) = vγ d +1 ( F n +1 ) for all v ∈ N d ) such that wγ d +1 ( F n +1 ) = w γ d +1 ( F n +1 ). As before, let wN ∈ γ d +1 ( G ). Since γ d +1 ( G ) ∼ = γ d +1 ( F n +1 ) /N d +1 in a natural way, we have wN d +1 = h ′ N d +1 for some h ′ ∈ γ d +1 ( F n +1 ), which is a contradiction. Thus, wN ∈ γ d ( G ) \ γ d +1 ( G ).Since G/γ d +1 ( G ) is nilpotent, we get the required result. Therefore, F n +1 /N is residuallynilpotent. (cid:3) References [1] S. Andreadakis, E. Raptis and D. Varsos, A characterization of residually finite HNN-extensionsof finitely generated abelian groups. Arch. Math. (Basel) (1988), no. 6, 495–501.[2] Yu.A. Bahturin, Identical Relations in Lie Algebras, Nauka, Moscow, 1985 (in Russian). Englishtranslation: VNU Science Press, Utrecht, 1987. 3] N. Bourbaki, Lie Groups and Lie Algebras Part I, Hermann, Paris, 1987 (Chapters 1-3).[4] R.M. Bryant and R. St¨ohr, Fixed points of automorphisms of free Lie algebras, Arch. Math., (1996), 281–289.[5] R.M. Bryant, L.G. Kov´acs and R. St¨ohr, Invariant bases for free Lie rings, Q. J. Math. (2002),no. 1, 1–17.[6] R. Charney, An introduction to right angled Artin groups, Geom. Dedicata (2007), 141–158.[7] G. Duchamp and D. Krob, The free partially commutative Lie algebra: bases and ranks, Adv.Math. (1992), 92–126.[8] G. Duchamp and D. Krob, The lower central series of the free partially commutative group, Semigroup Forum (1992), 385–394.[9] E. Formanek and C. Procesi, The automorphism group of a free group is not linear, J. Algebra (1992), 494–499.[10] M. Hall, A basis for free Lie rings and higher commutators in free groups, Proc. Amer. Math.Soc. , (1950), 575–581.[11] J.P. Labute, On the descending central series of groups with a single defining relation, J. Algebra , (1970), 16–23.[12] J.P. Labute, The determination of the Lie algebra associated to the lower central series of a group, Trans. Amer. Math. Soc. , (1985), no. 1, 51–57.[13] M. Lazard, Sur les groupes nilpotents et les anneaux de Lie, Ann. Sci. Ecole Norm. Sup. , (3)(1954), 101–190.[14] W. Magnus, A. Karrass and D. Solitar, Combinatorial Group Theory: presentations of groups interms of generators and relations, Dover Publications, 1978.[15] V. Metaftsis, E. Raptis and D. Varsos, On the linearity of HNN-extensions with abelian base. J.Pure Appl. Algebra, (2012), no. 5, 997–1003.[16] I. Sevaslidou, The associated Lie algebra of an HNN-extension of a group, J. Gen. Lie TheoryAppl. , accepted.V. Metaftsis, Department of Mathematics, University of the Aegean, Karlovassi, 832 00 Samos, Greece. e-mail: [email protected]. Papistas, Department of Mathematics, Aristotle University of Thessaloniki, 541 24 Thessaloniki,Greece. e-mail: [email protected]@math.auth.gr