TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 352, Number 2, Pages 901–934 S 0002-9947(99)02369-7 Article electronically published on October 6, 1999

ON THE MODULE STRUCTURE OF FREE LIE ALGEBRAS

R. M. BRYANT AND RALPH STOHR¨

Abstract. We study the free Lie algebra L over a field of non-zero charac- teristic p as a module for the cyclic group of order p acting on L by cyclically permuting the elements of a free generating set. Our main result is a complete decomposition of L as a direct sum of indecomposable modules.

1. Introduction and statement of results For a group G,afieldK(or, more generally, a commutative ring with identity), and a KG-module V ,letL(V) denote the free Lie algebra on V .Foreachpositive integer n,letLn(V) be the homogeneous component of L(V ) of degree n.Then L(V)andLn(V) are themselves KG-modules, and these modules have been studied intensively over the past 50 years. We mention the pioneering work of Thrall [12], Brandt [2] and Wever [14], and for an account of some of the later work we refer to [9]. Most of the results on the module structure of L(V ) are concerned with the case where K is a field of characteristic zero, and very little seems to be known for prime characteristic. In fact, many of the methods used in characteristic zero fail in the modular case. In a recent paper, Donkin and Erdmann [5] studied L(V )asa module for GL(V ) over an infinite field of prime characteristic p. They obtained a formula for the multiplicities of the indecomposable components of Ln(V )forvalues of n not divisible by p. Their results also confirm that considerable difficulties arise when p divides n. In this paper we study the case where G is the cyclic group of order p, K is an arbitrary field of characteristic p and V is a free KG-module. Let IG denote the augmentation ideal of KG.NotethatIG is an indecomposable KG-module of dimension p 1. For a given positive integer m,andforn 1, let − ≥ 1 (1.1) a(n)= µ(d)mn/d, n − d p d n X| | where the sum ranges over all divisors d of n which are multiples of p,andwhere µis the M¨obius function. Our main result reads as follows. Theorem 1. Let p be a prime, G the cyclic group of order p, K a field of charac- teristic p and V afreeKG-module of finite dimension m.ThenLn(V)is a direct sumofafreeKG-module and a(n) isomorphic copies of IG.

Thus we obtain a complete decomposition of Ln(V ) into indecomposable KG- modules. The non-trivial part of Theorem 1 is in the case where p divides n.If (n, p)=1,wehavea(n) = 0 and thus Ln(V )isafreeKG-module. This is, in fact,

Received by the editors August 20, 1997. 1991 Mathematics Subject Classification. Primary 17B01, Secondary 20C20.

c 1999 American Mathematical Society

901

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not hard to show, and will be established in Section 2 (Corollary 2.2). For p =2, Theorem 1 was proved in our earlier paper [4]. An easy consequence of the theorem is the following result (derived in Section 6). G Corollary 1. The dimension of the fixed point subspace Ln(V ) is given by 1 dim L (V )G = µ(d)mn/d, n pn d n (d,pX|)=1 where the sum runs over all divisors d of n which are coprime to p. In the case where m = p, the dimension formula in Corollary 1 was conjectured by Short [10]. Another consequence of Theorem 1 concerns the module structure of the free Lie ring.

Corollary 2. Let V be a free ZG-module of finite rank. Then the free Lie ring L(V ), regarded as a ZG-module, has no non-zero direct summand upon which G acts trivially, and G p 1 L(V ) = L(V )(1 + g + ...+g − ), where g is a generator of G. Theorem 1 will be deduced from Theorem 2 below, which gives additional insight into the structure of the free Lie algebra on an arbitrary (not necessarily finite dimensional) free KG-module V . To state it, we need to introduce some more notation. Let A(p) denote the free associative ring on free generators x1,x2,... ,xp and take the usual Lie bracket operation defined by [a, b]=ab ba for all a, b A(p). Define a word w by − ∈ 1 w = w(x ,... ,x )= x x [x ,x ,... ,x ] , 1 p p π(1) π(p) 1 κ(2) κ(p) π ··· − κ ! X X where π runs over all permutations of 1, 2,... ,p ,κ runs over all permutations of 2,... ,p and the terms in the second{ sum are left-normed} Lie products. It turns out{ that the} expression within parentheses belongs to pA(p); hence w A(p). For a KG-module V ,letA(V) denote the free associative algebra on V .Thus∈ A(V)isthe universal enveloping algebra of L(V ) and, with A(V ) regarded as a KG-module, L(V ) is a submodule. For arbitrary elements u1,... ,up of A(V ) we may take the corresponding value w(u1,... ,up) of the word w. Theorem 2. Let G be the cyclic group of order p, g a generator of G, K a field of characteristic p and V afreeKG-module. Then there exists a set of homogeneous 2 p 1 L elements of L(V ) such that the elements u, ug, ug , ... , ug − , for u , together with the elements ∈L pα pα pα p 1 j (1.2) w(u , (u )g,...,(u )g − )(1 g ), − for u ,α 0,1 j p 1, form a basis of L(V ). ∈L ≥ ≤ ≤ − The basis elements (1.2) are, as written, elements of the enveloping algebra A(V ), but they turn out to be elements of L(V ) itself. In Section 7 we obtain an explicit expression for such a basis element as an element of L(V ), that is, as a p 1 linear combination of Lie products of the elements u,ug,...,ug − . Our approach to Theorem 2 is based on the following strategy. Write A(V )=

n 0An(V), where An(V )isthen-th homogeneous component of A(V ). We ≥ L

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introduce a Lie subalgebra L∗ of A(V ) which is, in many ways, close to L(V ): for example, L∗ An(V )=L(V) An(V) for all n except n =1andn=p.A ∩ ∩ crucial point about L∗ is that it is a free Lie algebra with a free generating set which consists of homogeneous elements of degree at least 2 (as elements of W A(V )). Furthermore G acts freely on this set :thusL∗is the free Lie algebra W on the free KG-module with basis . The free generating set provides L∗ with a grading of its own, and, by construction,W elements of degreeWn with respect to have degree at least 2n with respect to the original free generating set of A(V ). ThisW enables us to obtain information about the homogeneous components of L(V ) from homogeneous components of L∗ which are of smaller degree (with respect to ). In other words, it opens the way for an inductive proof. W In addition to the free Lie and associative algebras, some other free algebras play an important role in this paper. These are the free restricted Lie algebra, the free metabelian Lie algebra and the symmetric algebra (that is, the free commutative associative algebra). In Section 2 we assemble some facts about these free algebras which will be assumed later in the paper. Most of this material is well known and we record it without giving a particular reference (useful general references, however, are [1], [6] and [9]). Specific references are given for a few facts thought to be less familiar. In Section 3 we examine symmetric powers of the regular module KG (in other words, the homogeneous components of the symmetric algebra on KG) and prove that certain submodules of these symmetric powers are free. In Section 4 the free metabelian Lie algebra M(V ) is used to obtain the structure of Ln(V )inthe special but seminal case where n = p.SinceM(V) is isomorphic to a quotient algebra of L(V ), there is a module homomorphism ν from Lp(V )ontoMp(V), the homogeneous component of degree p in M(V ). We use the word w, introduced above, to construct a right inverse of ν. Then we are able to use a direct sum decomposition of Mp(V ) to obtain a direct sum decomposition of Lp(V ). Free restricted Lie algebras and symmetric powers are crucial components in the implementation of our proof strategy. They play a major role in Section 5, which is entirely devoted to the proof of the key technical result of this paper, Proposition 5.2. In particular, the results of Section 3 are used in this proof. In Section 6 we exploit Proposition 5.2 to carry out the proof strategy outlined above. This culminates in Theorem 6.4, from which the main results are easily deduced. With a view to further applications ([7] in particular) we also include a modification of Theorem 6.4 (namely Proposition 6.5) which refers to the Lie powers of projective modules for the normalizer of a p-cycle in the symmetric group of degree p. We are deeply grateful to L. G. Kov´acs for many helpful discussions about this paper. We are also indebted to G. E. Wall for pointing out the relevance of the word w, which had an invaluable impact on our progress, and to M. R. Vaughan-Lee for helpful comments regarding the calculation in Section 7. Part of this work was carried out while the second author was visiting the Australian National University, Canberra. Financial support for this visit from the Royal Society and from the ANU is gratefully acknowledged.

2. Notation and preliminaries Throughout this paper p will be a prime number and G will be a cyclic group of order p with generator g.Also,Kwill be a field of characteristic p. We shall require

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some notation and simple facts about Lie algebras and other algebras over both K and Z. In order to cover both cases we temporarily work with a coefficient ring J which is an arbitrary commutative ring with identity. We shall use vector space terminology to describe J-modules, in conformity with the special case J = K. Let JG be the group algebra of G over J and let V be a (right) JG-module which is free as a J-module. We write L(V ) for the free Lie algebra on V :thus L(V) is the free Lie algebra over J with the property that L(V ) contains V as a subspace (that is, J-submodule) and every basis of V (that is, free generating set for V as J-module) is a free generating set of L(V ). The action of G on V extends uniquely to L(V ) subject to L(V ) becoming a JG-module on which the elements of G act as algebra automorphisms. For n 1, the n-th homogeneous ≥ component Ln(V )ofL(V) is called the n-th Lie power of V : it is the subspace of L(V ) spanned by all left-normed Lie products [v1,v2,... ,vn] with v1,... ,vn V, and it is a JG-submodule of L(V ). Thus L(V ) is a graded module with a direct∈ decomposition

L(V )= Ln(V). n 1 M≥

If V has finite dimension (that is, J-rank) m, the dimension of Ln(V )isgivenby Witt’s formula: 1 dim L (V )= µ(d)mn/d, n n dn X| where d ranges over all positive divisors of n and µ is the M¨obius function. In an analogous way we obtain the free associative algebra A(V ), the symmetric algebra (or free commutative associative algebra) S(V ) and the free metabelian Lie algebra M(V ), where M(V ) is identified with the quotient of L(V )byits second derived subalgebra L(V )00. All these algebras will be regarded as graded JG-modules in the obvious way, and their homogeneous components An(V )(for n 0), Sn(V )(forn 0) and Mn(V )(forn 1) will be called, respectively, the n-th≥ free associative power,≥ n-th symmetric power≥ and n-th free metabelian Lie power of V . We regard A(V ) as a Lie algebra under the Lie bracket operation given by [a, b]=ab ba for all a, b A(V ). If u ,... ,un are (not necessarily distinct) − ∈ 1 elements of A(V ), then by a Lie product of u1,... ,un we mean u1 itself in the case n = 1 and, inductively, for n>1, any Lie product [u, v], where u is a Lie product of u1,... ,uk and v is a Lie product of uk+1,... ,un for some k with 1 kb2 b3 ... bn, form a basis of Mn(V ) (see [1], Section 4.7, for the∈V last result). ≤ ≤ ≤ We note that, in the free metabelian Lie algebra M(V ), left-normed Lie products [u ,u ,u ,... ,uk] with u ,... ,uk M(V) are symmetric with respect to the 1 2 3 1 ∈ entries u3,... ,uk. This follows immediately from the Jacobi identity and the fact that M(V )00 =0.

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An easy induction shows that, for u, u ,u ,... ,uk A(V), 1 2 ∈ (2.1) u u uku = uu u uk 1 2 ··· 1 2 ··· k 1 − u u [u, u ,... ,u ], − ω(1) ··· ω(i) ω(i+1) ω(k) i=0 ω X X where the inner sum ranges over all permutations ω of 1, 2,... ,k such that ω(1) <...<ω(i)andω(i+1)<...<ω(k). { } We now take J to be our field K of characteristic p and we consider restricted Lie algebras over K (see [1], [6], [9]). We write R(V ) for the free restricted Lie algebra on V , regarded as a graded KG-module. Its homogeneous component Rn(V )is called the n-th restricted Lie power of V .ItiswellknownthatA(V) becomes a restricted Lie algebra under the Lie bracket operation together with the unary operation given by a[p] = ap for all a A(V ). Then R(V ) can be identified with the restricted Lie subalgebra of A(V ) generated∈ by V ,andL(V) can be identified with the Lie subalgebra generated by V . Making these identifications, we have L(V ) ⊆ R(V ) A(V )andL(V)=R(V)=A(V)=V.Also,Ln(V)=An(V) L(V) ⊆ 1 1 1 ∩ and Rn(V )=An(V) R(V) for all n 1. For all a, b R(V )wehave∩ ≥ ∈ (2.2) [a, b] L(V ), ∈ and therefore [R(V ),R(V)] L(V ). There is an element v(x ,x )ofthep-th ⊆ 1 2 homogeneous component of the free Lie algebra on x1 and x2 such that, for all a, b A(V ), ∈ (2.3) (a + b)p = ap + bp + v(a, b). By (2.2), if a, b R(V )thenv(a, b) L(V ). If is a subset∈ of A(V ) then it is easily∈ verified that the restricted Lie subalgebra of A(VU) generated by is spanned by all elements which have the form vpα ,where vis a Lie product of elementsU from and α is a non-negative integer. If is a basis of L(V ), then theU elements bpα ,whereb and α 0, form a basisB [p] of R(V ), and, if is an ordered basis of R(V ), then∈B the elements≥ B R (2.4) bα1 bα2 bαk, 1 2 ··· k where b1,... ,bk ,b1

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form part of a basis of A(V ): thus φ is an embedding. The image of φ is clearly the associative subalgebra of A(V ) generated by R∗, and, since φ is an embedding, this subalgebra is freely generated by . Now we return to the case of a generalU coefficient ring J. For a subset of a G-module, we write G for the set defined by U U G = ugi : u ,0 i p 1 . U { ∈U ≤ ≤ − } Note that if V is a free JG-module, and if is a free generating set of V ,then G is a G-free basis of V , that is, a basis of V Xon which G acts freely. Conversely,X if is a G-free basis of V ,thenanyG-transversal of is a free generating set of VV as a JG-module, and = G. Clearly, a JG-moduleX V is free if and only if it has a G-free basis. V X Let V be a JG-module which is free as a J-module. Let be a (fixed) basis of V , and hence a free generating set of the Lie algebra L(V ). ByV a monomial in L(V ) we mean a non-zero Lie product of elements of . Every monomial of L(V )hasa multidegree with respect to : this is the sequenceV which lists, for each element b of , the multiplicity with whichV b occurs in the monomial. The same applies to theV algebras A(V ), S(V )andM(V). It will be convenient to label multidegrees by elements of the standard basis of S(V ). Let < be an arbitrary ordering of V and denote by n the basis of Sn(V ) consisting of the monomials b b bn with C 1 2 ··· b ,... ,bn and b b ... bn. Clearly, the multidegrees with total degree 1 ∈V 1 ≤ 2 ≤ ≤ n are in one-to-one correspondence with the elements of n.ForP A, L, S, M C ∈{ } and c n we let Pn,c( ) denote the span in Pn(V ) of all monomials of multidegree ∈C V c. Then it is easily verified that Pn(V ) is the direct sum of its subspaces Pn,c( ). Now assume that is G-free. We call a monomial of P (V ) balanced (with respectV to ) if all the elementsV within each G-orbit of occur with the same multiplicity; V V b otherwise the monomial is said to be unbalanced. We write n for the set of all u C balanced elements of n,and n for the set of all unbalanced elements of n.Then we define C C C b u P ( )= Pn,c( ),P()= Pn,c( ). n V V nV V c b c u M∈Cn M∈Cn In the case where J = K we define Rn,c( )=An,c( ) R(V )foreachc n.We also define Rb ( )=Ab( ) R(V)andRVu( )=AuV( ∩) R(V). ∈C n V n V ∩ n V n V ∩ Lemma 2.1. Let V be a free JG-module and a G-free basis of V .LetP A, L, S, M, R ,whereP =Rapplies only inV the case J = K. Then, for all∈ { b } u b u n 1, Pn( ) and Pn ( ) are submodules of Pn(V ),andPn(V)=Pn( ) Pn( ). Moreover,≥ PVu( ) is a freeV JG-module, and P b( ) has a module direct decompositionV ⊕ V n V n V b P ( )= Pn,c( ). n V V c b M∈Cn

Proof. First we consider the case P = S,usingthebasis n of Sn(V ). It is easily u C b verified that G acts freely on n and trivially on each element of n. The result in this case follows immediately.C Now let P A, L, M, R .ForeachC ctake a basis ∈{ } i n,c of Pn,c( ). Note that for i =0,1,... ,p 1wehavePn,c( )g = Pn,cgi ( ). It B V − b V u V follows that Pn(V ) is the direct sum of its submodules Pn( )andPn( ). Moreover, i i V V since g acts as an automorphism, n,cg is a basis of Pn,cgi ( ). Consequently, if In B u V u is a G-transversal of the G-free set n,then c I n,cG is a G-free basis of Pn ( ). C ∈ n B V S

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u b Thus Pn ( )isafreeJG-module. Finally, it is clear that Pn( ) is the direct sum V b V of its submodules Pn,c( ) with c . V ∈Cn Corollary 2.2. Let V be a free JG-module and let P A, L, S, M, R ,where ∈{ } P=Rapplies only in the case J = K.ThenPn(V)is a free JG-module for all n such that (n, p)=1. Proof. Let be a G-free basis of V . Since the degree of any balanced monomial is V b u divisible by p,if(n, p)=1thenP ( ) = 0 and hence Pn(V )=P ( ). n V n V Remark. In the case where P = A we can do better: the free associative powers An(V )ofafreeJG-module V are free JG-modules for all n 1. Indeed, one can easily check that G acts freely on the set of all monomials of≥ degree n. For a free JG-module V , the module structure of the free metabelian Lie powers Mn(V ) is straightforward. Let be an ordered free generating set for V , and write = G.Thus is a G-free basisX of V . We extend the order of to by setting V X Vp 1 p 1 X V xisomorphism J(G 1) Mpq,c( ) given by 1 g [x1g ,x1] cˆ(1,j) for j =1,2,... ,p 1is,infact,a− → JG-moduleV isomorphism,− 7→ and that ◦G acts −

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a freely modulo Mpq,c( ) on the remaining basis elements (2.6). The lemma now follows. V

b p 1 Remark. In the case n = p,theset p consists of the elements x(xg) (xg − ) a C b ··· where x . Hence Mp,c( )=Mp,c( ) for all c.ThusMp( ) is the direct sum of the modules∈XM a ( ) and hasV a basisV consisting of the elementsV p,c V j j 1 j+1 p 1 (2.7) [xg ,x,xg,...,xg − ,xg ,... ,xg − ],

f for x ,j=1,2,... ,p 1. Furthermore, the complement Mp ( )maybetaken to be ∈XM u( ). − V p V p 1 The elements 1,g,...,g − of the cyclic group G form a G-free basis of the regular module KG. When working in the symmetric algebra S(KG)wewritegi instead of gi,fori=0,1,... ,p 1. Then we identify S(KG) with the polyno- − mial algebra K[g0,g1,... ,gp 1] and write monomials in this algebra in the form α0 α1 αp 1 − g0 g1 gp −1 (where α0,... ,αp 1 0), thus avoiding confusion with the group ··· − − ≥ α0 α1 (p 1)αp 1 element 1 g g − − . (Note, in particular, that g0 is not the identity el- ement of S(KG···).) Similarly, when working in S(IG), where IG is the augmen- j tation ideal of KG,wewriteaj instead of 1 g ,forj=1,... ,p 1. The original multiplication of G should be forgotten− in S(KG)andS(IG−). All we should remember is the action of G on these algebras: namely, g acts on S(KG)by g0 g1 ... gp 1 g0 and g acts on S(IG)byaj aj a1,forj

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p 1 c =1+g+...+g − ,wehave p 1 n p+1 p 1 n p+1 (x− x − )c =(x − )c x − 1 2 1 · 2 p 1 p 1 p 1 p 1 n p+1 =(x − +(x +x ) − +(x +2x ) − +...+(x +(p 1)x ) − )x − . 1 1 2 1 2 1 − 2 2 n The coefficient of x2 in this expression (over our field of characteristic p)isgiven by p 1 p 1 p 1 1 − +2 − +...+(p 1) − =1+1+...+1=p 1. − − p 1 n p+1 Thus (x − x − )c = 0, which gives the required result. 1 2 6 (iii) Let θ : V (n 1,r) V(n, r) be the map defined by θ(v)=vxr for all v V (n 1,r), and− let φ :→V (n, r) V (n, r 1) be the map induced from the ∈ − → − homomorphism K[x1,... ,xr] K[x1,... ,xr 1]inwhichxi xi for i

Proposition 3.2. V (n, r) is a free KG-module whenever n + r p +1.Inpartic- ular, V (n, p) is free for all n 1. ≥ ≥ Proof. Define sets of ordered pairs of positive integers by Γ= (n, r):1 n p 1,2 r p, n + r p +1 { ≤ ≤ − ≤ ≤ ≥ } and ∆ = (n, r):n p 1,2 r p. Thus Γ can be thought of as the set of integral{ points of a≥ triangular− ≤ region≤ in} the plane, and ∆ as the set of integral points of a semi-infinite horizontal strip. We first claim that V (n, r) is free for all (n, r) Γ. If not, then choose (n, r) Γ with r maximal so that V (n, r) is not free. By Lemma∈ 3.1(i), r2. By∈ the result for Γ, n>p 1. Thus (n 1,r)and(n, r 1) belong to ∆. Thus, by minimality, V (n 1,r−)andV(n, r −1) are free. Thus,− by Lemma 3.1(iii), V (n, r) is free—a contradiction.− − The result follows since Γ ∆= (n, r):n 1,2 r p, n + r p +1 . ∪ { ≥ ≤ ≤ ≥ } We now interpret this result in S(KG). As noted above, Vp is a regular module. Indeed, there is a KG-module isomorphism from Vp to KG given by x1 1, p 1 p 1 7→ x2 g 1, ... , xp (g 1) − . (Note that g 1, ... ,(g 1) − span the 7→ − 7→ − − 2− p 1 augmentation ideal IG, which is also spanned by 1 g,1 g,... ,1 g− .) − − − We use the induced isomorphism from S(Vp)toS(KG), and in S(KG)wewritegi i j instead of g and aj instead of 1 g , as explained in Section 2. It is easy to check − that, under this induced isomorphism, V (n, p) is mapped to the submodule S˜n(KG) α0 α1 αp 1 of Sn(KG) which is spanned over K by all elements of the form g0 a1 ap −1 , ··· − where 0 α0 p 1andα0+α1+...+αp 1 =n. From≤ Proposition≤ − 3.2 we therefore obtain the− following result.

Corollary 3.3. S˜n(KG) is a free KG-module for all n 1. ≥

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We shall also need to use some other submodules of Sn(KG). For n satisfy- ing 0 n p(p 1), let Sˆn(KG) denote the span in Sn(KG) of all monomials α0 α1≤ ≤αp 1 − g0 g1 gp −1 where α0+α1+...+αp 1 =nand 0 αi p 1fori=0,... ,p 1. ··· − − ≤ ≤ − − ˆ ˆu Clearly these monomials form a basis for Sn(KG). Let Sn(KG) denote the span of all unbalanced monomials of this kind; that is, monomials in which α0,α1,... ,αp 1 ˆb − are not all equal. If p n let Sn(KG) denote the subspace spanned by the element α α α | g0 g1 gp 1,whereα=n/p. By arguments similar to those in Section 2 we obtain the following··· − result. Proposition 3.4. Suppose 0 n p(p 1). ≤ ≤ ˆ− ˆu ˆ (i) If n is not divisible by p,thenSn(KG)=Sn(KG),andSn(KG) is a free KG-module. ˆ ˆu ˆb ˆu (ii) If p n,thenSn(KG)=Sn(KG) Sn(KG),whereSn(KG) is a free KG- | ˆb ⊕ module and Sn(KG) is a one-dimensional trivial KG-module.

4. The p-th Lie power of a free G-module In this section we shall initially work with integer coefficients. Let L(p), A(p) and M(p) denote, respectively, the free Lie ring, the free associative ring and the free metabelian Lie ring on free generators x1,x2,... ,xp,andidentifyL(p)withthe Lie subring of A(p) generated by x1,... ,xp. Wedenotebyνthe natural surjection ν : L(p) M(p). Consider→ the element

xπ(1) xπ(p) [x1,xκ(2),... ,xκ(p)] π ··· − κ X X of A(p), in which π and κ range over all permutations of 1, 2,... ,p and 2,... ,p , respectively. It follows from [13] (Lemma 1 on page 677){ that this} element{ belongs} to pA(p). Thus the element defined by

w = w(x1,... ,xp) (4.1) 1 = x x [x ,x ,... ,x ] p π(1) π(p) 1 κ(2) κ(p) π ··· − κ ! X X is an element of A(p). Note that w is symmetric in the variables x2,... ,xp.Let

wˆ=ˆw(x,... ,xp)=w(x ,x ,x ,... ,xp) w(x ,x ,x ,... ,xp) 1 2 1 3 − 1 2 3 1 = [x ,x ,... ,x ] [x ,x ,... ,x ] , p 1 κ(2) κ(p) − 2 λ(1) λ(p) κ λ ! X X where the range of κ is as before and λ ranges over all permutations of 1, 3,... ,p . Since pwˆ L(p)andL(p) is a direct summand of A(p), it follows that{ w ˆ L(p}); that is,w ˆ∈is a Lie element. However w is not a Lie element because, using∈ the natural surjection δ : A(p) Z[x1,... ,xp], we have → (4.2) δ(w)=(p 1)! x x xp, − 1 2 ··· and this is a non-zero element of Z[x1,... ,xp]. We shall now calculate the image ν(ˆw) in the free metabelian Lie ring M(p). Recall that a left-normed Lie product [u1,u2,u3,... ,up]inM(p) is symmetric

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with respect to the entries u3,... ,up.Thus

1 ν(ˆw)= [x,x ,... ,x ] [x ,x ,... ,x ] p 1 κ(2) κ(p) − 2 λ(1) λ(p) κ λ ! X Xp (p 2)! = − [x ,x ,x ,... ,x ]+ [x ,x ,x ,...] p 1 2 3 p 1 i 2 i=3 X p

[x ,x ,x ,... ,xp] [x ,xi,x ,...] . − 2 1 3 − 2 1 i=3 ! X But, by the Jacobi identity, p p p

[x ,xi,x ,...] [x ,xi,x ,...]= [x ,x ,xi,...] 1 2 − 2 1 1 2 i=3 i=3 i=3 X X X =(p 2)[x ,x ,... ,xp]. − 1 2 Hence

(4.3) ν(ˆw)=(p 2)![x ,x ,... ,xp]. − 1 2 Now suppose that the cyclic group G acts on the free generators x1,... ,xp via xig = xi+1 (subscripts modulo p). Then A(p), L(p)andM(p) become ZG-modules. Observe that w(g 1) =w ˆ.Thuswhas the property that, although it is not itself a Lie element, it becomes− one when it is acted on by any element of the augmentation ideal of ZG. For any associative algebra A on which G acts by algebra automorphisms, and for any element u of A,wewritew(u) for the value of w at the elements of the G-orbit of u in A;thatis,

p 1 w(u)=w(u,ug,...,ug − ). It follows that

p 1 (4.4) w(u)(g 1) =w ˆ(u,ug,...,ug − ). − Now let V be a ZG-module which is free as a Z-module. We consider the Z- algebras L(V ), A(V ), S(V )andM(V). It is easy to verify that there is a ZG-module map Mp(V ) V Sp 1(V ) such that → ⊗ − [v ,v ,... ,vp] v (v v vp) v (v v vp) 1 2 7→ 1 ⊗ 2 3 ··· − 2 ⊗ 1 3 ··· for all v1,... ,vp V. Furthermore, there is a ZG-module map ∈ V Sp 1(V ) Ap(V ) ⊗ − → such that, for all v ,... ,vp V, 1 ∈

v1 (v2v3 vp) v1 vκ(2)vκ(3) vκ(p) , ⊗ ··· 7→ κ ··· ! X where κ ranges over all permutations of 2, 3,... ,p .Also,thereisaZG-module { } map Ap(V ) Lp(V ) such that, for all v ,... ,vp V, → 1 ∈ v v vp [v ,v ,... ,vp]. 1 2 ··· 7→ 1 2

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The composite of these three maps is a ZG-module map Mp(V ) Lp(V )such → that, for all v ,... ,vp V, 1 ∈ [v ,v ,... ,vp] pwˆ(v ,v ,... ,vp). 1 2 7→ 1 2 Since Lp(V ) is torsion-free as a Z-module, we finally obtain a ZG-module homo- morphism ψ : Mp(V ) Lp(V ) such that → ψ([v1,v2,... ,vp]) =w ˆ(v1,v2,... ,vp)

for all v ,... ,vp V. 1 ∈ Remark. The definition of the homomorphism ψ does not depend upon G being cyclic of prime order. In fact, it can be defined for Z-free modules for an arbitrary group.

Suppose that V is a free ZG-module. Then it follows from [11] (Lemma 3.4 com- bined with Theorem 3.11) that Lp(V ) L(V )00 has a finite filtration with quotients ∩ which are free ZG-modules. Hence Lp(V ) L(V )00 is itself a free ZG-module. We now consider algebras over K. Suppose∩ that V is a free KG-module with a free generating set , and take the basis defined by = G. The natural X V V X surjection ν : Lp(V ) Mp(V )isaKG-module homomorphism with kernel Lp(V ) → ∩ L(V )00.SinceVcan be obtained from a free ZG-module by tensoring with K, while L(V ) is obtained in the same way from the corresponding Lie algebra over Z,the result stated above in the integral case implies that Lp(V ) L(V )00 is a free KG- module. ∩ By Lemma 2.3 (with n = p) and the remark after it, Mp(V ) has a direct sum decomposition u b Mp(V )=M ( ) M ( ), p V ⊕ p V b where Mp ( ) is a direct sum of isomorphic copies of IG (one for each x ). Let V b ∈X ε : Mp(V ) Mp ( ) be the projection map. Thus the kernel of the composite map → b V εν : Lp(V ) Mp ( ) is an extension of Lp(V ) L(V )00 by a module isomorphic to u → V ∩ f Mp ( ), and hence is a free KG-module. We denote this module by Lp (V ). Clearly V u it is spanned by Lp(V ) L(V )00 and L ( ). ∩ p V Since L(V )andM(V) can be obtained from the corresponding algebras over Z by tensoring with K,themapψdefined above in the integral case yields a KG-module homomorphism ψ : Mp(V ) Lp(V ) such that → ψ([v1,v2,... ,vp]) =w ˆ(v1,v2,... ,vp)

for all v ,... ,vp V. In view of (4.3) and the fact that (p 2)!=1inK,the 1 ∈ − composite map νψ : Mp(V ) Mp(V ) is the identity map. Hence the restriction b → a b ψ : Mp ( ) L p ( V ) is a right inverse of εν.WriteLp(V)=ψ(Mp( )). Then V −→ a f V it follows that Lp(V ) is the direct sum of Lp(V ) and the kernel Lp (V )ofεν.By Lemma 2.3 and the remark after it, the balanced Lie products (2.7) form a basis of M b( ). By the definitions of ψ, w andw ˆ, and exploiting the symmetry of w in p V x2,... ,xp,wehave j p 1 j p 1 ψ([xg ,x,...,xg − ]) =w ˆ(xg ,x,...,xg − ) = w(x,...) w(xgj ,...) − = w(x) w(xgj ) − = w(x)(1 gj). −

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Since ψ is an embedding, the elements w(x)(1 gj)(forx ,1 j p 1) are a − ∈X ≤ ≤ − distinct and form a basis of Lp(V ). Thus we have proved the following result. Proposition 4.1. Let V be a free KG-module with free generating set and basis ,where = G.Then,withLf(V)and La(V ) as defined above, X V V X p p f a Lp(V )=L (V) L (V), p ⊕ p f a Lp(V)is a free KG-module, and Lp(V ) is a direct sum of isomorphic copies of IG, f one for each element of . Moreover, Lp (V ) is spanned by Lp(V ) L(V )00 and u X j ∩ Lp ( ). Furthermore, the elements w(x)(1 g ),wherex and 1 j p 1, V a − ∈X ≤ ≤ − are distinct and form a basis of Lp(V ).

5. The associative algebra As before, let V be a free KG-module with free generating set and basis , X V where = G.WewriteL=L(V), R = R(V ), A = A(V ), Ln = Ln(V ), and so V X f f a a on. Thus, in relation to Proposition 4.1, we write Lp = Lp (V )andLp =Lp(V). Note that is a free generating set of each of the algebras L, R and A. For a subset of a K-space,V we write for the subspace spanned by . U hUi U Let be a basis of L which has the form = i 1 i,where i Li for each i B f a f f a Ba ≥ B B ⊆ and p = with L and L . Furthermore, we take = and B Bp ∪Bp Bp ⊆ p Bp ⊆ p S B1 V a = w(x)(1 gj):x ,j=1,... ,p 1 . Bp { − ∈X − } i j For each x ,writegi(x)=xg for i =0,1,... ,p 1andaj(x)=w(x)(1 g ) ∈X − − for j =1,... ,p 1. These elements gi(x)andaj(x) will be called x-factors:the − gi(x) will be called free x-factors and the aj(x) will be called augmentation x- factors.Byan -factor we mean an x-factor for some x .Thus 1consists of X a ∈X B all free -factors and p consists of all augmentation -factors. X B α X Let = [p],where [p] = bp :b ,α 0 .Thus is a basis of R.We R B] B ] { ∈Ba [p] ≥ } R write = ∗,where = ( ) and R R ∪R R B1∪ Bp ] [p] [p] [p] f [p] [p] ∗ = =( 1 1) 2 ... p 1 ( p) p+1 ... . R R\R B \B ∪B ∪ ∪B − ∪ B ∪B ∪ Thus ] consists of all free -factors and all p-power powers of augmentation -factors.R X X p f p Note that x : x and Lp span their direct sum in Rp.LetRp∗= x :x f h ∈Vi h ∈Vi L and let R∗ be the restricted Lie subalgebra of R which is generated by ⊕ p L2,... ,Lp 1,Rp∗,Lp+1,... . − Lemma 5.1. The restricted Lie algebra R∗ has basis ∗. R Proof. Clearly ∗ generates R∗, and the elements of ∗ are linearly independent R R because ∗ . Thus it suffices to verify that the subspace ∗ is a restricted Lie algebra.R ⊆R hR i For all d, e ∗ we have [d, e] L by (2.2), and so [d, e] Ln for some n 2. ∈R ∈ ∈ ≥ If n = p then [d, e] n ∗ .Butifn=pthen both d and e must belong to 6 ∈hB i⊆hR i f f 2 ... p 1,andso[d, e] Lp L00 Lp = p ∗ .Thus ∗ is a Lie Balgebra.∪ ∪B − ∈ ∩ ⊆ hB i⊆hR i hR i p By its definition, the set ∗ is closed under the unary operation u u .In R 7→ view of (2.3) and the fact that ∗ is a Lie algebra, it follows that ∗ is closed hR i hR i under this unary operation. Hence ∗ is a restricted Lie algebra. hR i

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Since R∗ is a subalgebra of R, it is a free restricted Lie algebra, by Witt’s Theorem. Let A∗ be the associative subalgebra of A generated by R∗;thatis,A∗ is the associative subalgebra generated by L2,... ,Lp 1,Rp∗,Lp+1,... .Asshown − in Section 2 (see the paragraph after (2.4)), A∗ is a free associative algebra, freely generated as associative algebra by any free generating set of R∗.Foreachk 0 ≥ write Ak∗ = A∗ Ak. Then it is easily verified that A∗ = A0∗ A1∗ ...,andA∗ is a KG-submodule∩ of A. The main result of this section is as⊕ follows.⊕

Proposition 5.2. For each k 1, A∗ is a free KG-module. ≥ k A product e e em of -factors e ,... ,em will be called an -product if, for 1 2 ··· X 1 X every free -factor b, the number of values of i for which ei = b does not exceed X p 1. The free degree of an -product e em is the number of values of i for − X 1 ··· which ei is a free -factor. X Let x .An -product e em in which each ei is an x-factor is called an ∈X X 1 ··· x-product.Anx-block is an x-product e1 em such that e1 e2 ... em in the ordering of x-factors given by ··· ≤ ≤ ≤

g0(x) <...

α0 αp 1 β1 βp 1 (5.1) u(x)=g0(x) gp 1(x) − a1(x) ap 1(x) − , ··· − ··· − where α0,... ,αp 1 0,1,... ,p 1 and β1,... ,βp 1 0. Note that α0 + ...+ − ∈{ − } − ≥ αp 1 is the free degree of u(x), and this is bounded above by p(p 1). −We next define an ordering on the whole of . In this ordering− every element of ] R ] is taken to precede every element of ∗, while ∗ is ordered arbitrarily and Ris ordered as follows. For each x theR powersR of x-factors which lie in ] areR ordered by ∈X R p p g0(x) <...

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a basis of Ak∗. Recall from Section 2 the definition of a Lie product of elements u1,... ,un of A.

Lemma 5.3. Let v be a Lie product of -factors e ,... ,en,wheren 2.If X 1 ≥ v A∗ then n = p, e1,... ,ep = g0(x),... ,gp 1(x) for some x ,andv 6∈ { } { − } ∈X may be written as a linear combination of a1(x),... ,ap 1(x) together with elements − of Ap∗ which are Lie products of elements in g0(x),... ,gp 1(x) . { − } Proof. Suppose v A∗. Clearly v Lm for some m 2. But Lm A∗ for all 6∈ ∈ ≥ ⊆ m 2 except m = p.Thusv Lp. Hence each ei is a free -factor, and n = p.All ≥ ∈ f X f unbalanced monomials of degree p belong to L , by Proposition 4.1, and L A∗. p p ⊆ Thus v is balanced. Hence e1,... ,ep = g0(x),... ,gp 1(x) for some x . { } { − } ∈X By Proposition 4.1 applied to the subalgebra Lx of L generated by e ,... ,ep we { 1 } see that v is a linear combination of a1(x),... ,ap 1(x) together with elements of f − L Lx. This gives the result. p ∩ For l =0,1,... ,k,wherek 1, let Ak,l denote the span in Ak of all basis elements (5.3) of degree k such that≥ deg(z) l. Clearly ≥ Ak∗ = Ak,k Ak,k 1 ... Ak,1 Ak,0 = Ak. ≤ − ≤ ≤ ≤ We shall now prove that each Ak,l is a submodule of Ak (by submodule we always mean KG-submodule).

Lemma 5.4. (i) Let v be an -product and let z be an element of (A∗) such that X B deg(vz)=kand deg(z) l.Thenvz Ak,l. ≥ ∈ (ii) Ak,l is a submodule of Ak.

Proof. (i) Write v = e1 em,whereeachei is an -factor. There is a permuta- tion π of 1,... ,m such··· that e e z hasX the form (5.3) and so belongs { } π(1) ··· π(m) to Ak,l. Thus we may assume that m 2 and, by a suitable induction, it suf- ≥ fices to prove that the element e1 es 1[es,es+1]es+2 emz belongs to Ak,l when 1 s

(u(x )g) (u(xn)g)(zg) Ak,l. 1 ··· ∈ Note that the action of g permutes the free xi-factors for each i,andthatifb is an augmentation xi-factor then bg is a linear combination of augmentation xi- factors. It follows easily that (u(x )g) (u(xn)g) is a linear combination of - 1 ··· X products of the same degree as u(x ) u(xn). Also, zg is a linear combination 1 ··· of elements of (A∗) of the same degree as z.Thus(u(x)g) (u(xn)g)(zg)isa B 1 ··· linear combination of elements which belong to Ak,l by (i). The result follows. For x we write P (x) for the subspace of A spanned by all x-products ∈X (including the empty product 1). For each k 1wewritePk(x) for the subspace ≥ spanned by all x-products of degree k,andBk(x) for the subspace spanned by all

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x-blocks of degree k.WewriteA∗(x) for the subspace of A∗ spanned by all elements of A∗ which are Lie products of -factors at least one of which is an x-factor. Note X that A∗(x) A∗ A∗ ... . ⊆ 1 ⊕ 2 ⊕ Lemma 5.5. (i) For k 1 and x , ≥ ∈X Pk(x) Bk(x)+P(x)A∗(x). ⊆ (ii) For x, y with x = y, ∈X 6 A∗(x)P (y) P (y)A∗(x). ⊆ Proof. (i) Let e em be an x-product of degree k (where each ei is an x-factor). 1 ··· There is a permutation π of 1,... ,m such that eπ(1) eπ(m) Bk(x). Thus we may assume that m 2 and,{ by a suitable} induction, it··· suffices∈ to prove that ≥ e1 es 1[es,es+1]es+2 em Bk(x)+P(x)A∗(x) ··· − ··· ∈ when 1 s

It is clear that, for l

Each V (l)k1,... ,kn is a submodule of A /A and, as KG-modules, Lemma 5.6. x1,... ,xn k,l k,l+1 k ,... ,k k k V (l) 1 n = V (0) 1 V(0) n A∗. x1,... ,xn ∼ x1 ⊗···⊗ xn ⊗ l Proof. As stated above, V (l)k1,... ,kn has a basis consisting of the elements x1,... ,xn u(x1) u(xn)z+Ak,l+1 with deg(u(xi)) = ki and deg(z)=l. The tensor product in the··· statement of the lemma has a basis consisting of corresponding elements

(u(x )+Ak , ) (u(xn)+Ak , ) z. 1 1 1 ⊗···⊗ n 1 ⊗ Let θ be the vector space isomorphism from V (l)k1,... ,kn to the tensor product x1,... ,xn which maps basis element to corresponding basis element.

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For i =1,... ,n, u(xi)g is a linear combination of xi-products of degree ki, and so, by Lemma 5.5(i), we can write u(xi)g = bi + ci where bi Bk (xi)and ∈ i ci P(xi)A∗(xi). Hence ∈ (u(x ) u(xn)z)g =b bn(zg)+v, 1 ··· 1 ··· where v is a linear combination of elements of the form d dn(zg), where, for 1 ··· each i, di Bk (xi)ordi P(xi)A∗(xi), and where di P (xi)A∗(xi) for at least ∈ i ∈ ∈ one value of i. Recall that A∗(xi) A1∗ A2∗ ... . By repeated use of Lemma 5.5(ii) we get ⊆ ⊕ ⊕

d dn(zg) P(x ) P(xn)(A∗ A∗ ...)(zg). 1 ··· ∈ 1 ··· 1 ⊕ 2 ⊕ Thus

v P (x ) P(xn)(A∗ A∗ ...). ∈ 1 ··· l+1 ⊕ l+2 ⊕ Since v has degree k it follows that v is a linear combination of elements of degree k of the form v0z0 where v0 is an -product, z0 (A∗) and deg(z0) l +1. Thus, X ∈B ≥ by Lemma 5.4, v Ak,l . Therefore ∈ +1 (u(x ) u(xn)z)g+Ak,l = b bn(zg)+Ak,l . 1 ··· +1 1 ··· +1 It follows that V (l)k1,... ,kn is a submodule of A /A . x1,... ,xn k,l k,l+1 A similar result holds for each V (0)ki , and we have xi

u(xi)g + Aki,1 = bi + Aki,1. Hence

θ((u(x ) u(xn)z+Ak,l )g)=(b +Ak , ) (bn +Ak , ) zg 1 ··· +1 1 1 1 ⊗···⊗ n 1 ⊗ =(u(x )+Ak , )g (u(xn)+Ak , )g zg 1 1 1 ⊗···⊗ n 1 ⊗ =(θ(u(x ) u(xn)z+Ak,l ))g. 1 ··· +1 Thus θ is a module isomorphism.

k Note that, for each x and k 1, V (0)x is a submodule of Ak/Ak,1.Our main task now will be the∈X proof of the≥ following result. Proposition 5.7. For x and k 1, V (0)k is a free KG-module. ∈X ≥ x Before starting on the proof we show that this result yields Proposition 5.2, the main result of this section. Suppose, then, that Proposition 5.7 has been proved. By Lemma 5.6 it follows that each module V (l)k1,... ,kn is free. Since A /A is a direct sum of modules x1,... ,xn k,l k,l+1 of this sort, it also is free. This holds for l =0,... ,k 1, and so Ak, /Ak,k is free. − 0 But Ak, = Ak and Ak,k = A∗.SinceAkis free for all k 1 (see the remark after 0 k ≥ Corollary 2.2) it follows that Ak∗ is free for all k 1. This gives Proposition 5.2. The rest of this section will be devoted to the proof≥ of Proposition 5.7. We make ˜ ˆ ˆu use of the modules Sn(KG), Sn(KG)andSn(KG) considered in Section 3. k We fix x and k 1, and write V (0) = X/Ak, ,whereAk, X Ak. ∈X ≥ x 1 1 ≤ ≤ Thus X/Ak,1 has a basis consisting of all elements u + Ak,1,whereuis an x-block of degree k. Write k = qp+r,where0 r

α0 αp 1 β1 βp 1 (5.4) u = g0(x) gp 1(x) − a1(x) ap 1(x) − , ··· − ··· −

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where α0,... ,αp 1 0,1,... ,p 1 and − ∈{ − } α0 + ...+αp 1 +p(β1 +...+βp 1)=k=qp + r. − −

Recall that α0 + ...+αp 1 is the free degree of u. This free degree cannot ex- ceed p(p 1). Therefore it− takes one of the values r, p + r,...,lp+r,wherel= min(q, [p − 1 (r/p)]). Every −x-product− of degree k is obtained by permuting the factors of an x-block of degree k, so its free degree also takes one of the values jp + r for j =0,1,... ,l. For j =0,1,... ,l,letXj be the subspace of X spanned by Ak,1 together with all x-blocks of degree k and free degree not exceeding jp+r. Also, put X 1 = Ak,1. Thus −

(5.5) Ak,1 = X 1 X0 ... Xl =X. − ≤ ≤ ≤ Note that, for j =0,... ,l,Xj/Xj 1 has a basis consisting of all elements u+Xj 1 where u is an x-block of degree k −and free degree jp + r. −

Lemma 5.8. Let j 0,1,... ,l . ∈{ } (i) The subspace Xj contains every x-product of degree k and free degree jp+ r. (ii) Let e em be an x-product of degree k and free degree jp + r (where each 1 ··· ei is an x-factor).Supposem 2and 1 s

e1 es 1eses+1es+2 em +Ak,1 = e1 es 1es+1eses+2 em +Ak,1. ··· − ··· ··· − ··· Proof. (i) Every x-product of degree k and free degree jp + r can be obtained by permuting the factors of an x-block of degree k and free degree jp + r. The latter belongs to Xj . Thus (i) follows from (ii). (ii) It suffices to show that the element e1 es 1[es,es+1]es+2 em belongs to ··· − ··· Xj 1,andthatifes or es+1 is an augmentation factor then this element belongs − to Ak,1. As in the proofs of Lemmas 5.4 and 5.5, e1 es 1[es,es+1]es+2 em is a ··· − ··· linear combination of elements of the form e1 es 1bs btc,where(bs,... ,bt)is ··· − ··· a subsequence of (es+2,... ,em)andcis a Lie product of es, es+1 and elements of es ,... ,em . By Lemma 5.3, c is a linear combination of augmentation x-factors { +2 } and elements of A∗(x). In the special case where es or es+1 is an augmentation factor, Lemma 5.3 shows that c A∗(x). In this special case we get ∈ e1 es 1[es,es+1]es+2 em Ak,1, ··· − ··· ∈ by Lemma 5.4(i). In the general case, e1 es 1[es,es+1]es+2 em is a linear ··· − ··· combination of elements of Ak,1 and x-products of degree k and free degree smaller than jp + r. The result therefore follows by induction on j.

Note that Lemma 5.8 shows, roughly speaking, that the x-factors commute in Xj/Xj 1. −

Lemma 5.9. For j =0,1,... ,l,Xj is a submodule of X and there is a KG-module isomorphism ˆ Xj /Xj 1 = Sjp+r(KG) Sq j(IG). − ∼ ⊗ −

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Proof. Since X 1 = Ak,1, we may use induction on j and assume that Xj 1 is a submodule of X−. − ˆ There is a vector space isomorphism χ from Xj/Xj 1 to Sjp+r(KG) Sq j(IG) − ⊗ − given on each basis element u + Xj 1,where − α0 αp 1 β1 βp 1 u=g0(x) gp 1(x) − a1(x) ap 1(x) − , ··· − ··· − by

α0 αp 1 β1 βp 1 χ(u + Xj 1)=g0 gp −1 a1 ap −1 . − ··· − ⊗ ··· − i Recall that when working in S(KG)andS(IG)wewritegi instead of g and ai i i instead of 1 g . Note that the map gi(x) g (for i =0,... ,p 1) extends − 7→ − to a KG-module isomorphism from g0(x),... ,gp 1(x) to KG, while the map i h − i ai(x) 1 g (for i =1,... ,p 1) extends to a KG-module isomorphism from 7→ − − a1(x),... ,ap 1(x) to IG. With u as above, we have h − i ˆ (χ(u + Xj 1))g Sjp+r(KG) Sq j(IG), − ∈ ⊗ − since the tensor product is a KG-module. Using Lemma 5.8 and the KG-module isomorphisms with KG and IG described above, we calculate that 1 (5.6) χ− ((χ(u + Xj 1))g)=ug + Xj 1. − − We deduce that ug Xj.SinceXj 1is a submodule of X it follows that Xj is a submodule of X.∈ Furthermore, equation− (5.6) shows that χ is a KG-module isomorphism. We can now prove Proposition 5.7 in the case where r =0. 6 k Lemma 5.10. Let k = qp + r,where0 r

Proof. Suppose that r = 0. Then the modules Sˆjp r(KG) are free modules, by 6 + Proposition 3.4. Hence, by Lemma 5.9, each module Xj/Xj 1 is free. Thus, in k − view of (5.5), X/Ak,1 is free; that is, V (0)x is free.

So, from now on, we assume that r = 0. Since the modules Sˆjp(KG) are not free—see Proposition 3.4—the method of Lemma 5.10 fails in this case. Because r =0wehavek=qp. The free degree of an x-block of degree k takes one of the values 0,p,...,lp,wherel= min(q, p 1). − Let be the set of all x-blocks of degree k.Thus u+Ak, : u is a basis E { 1 ∈E} of X/Ak, .Anelementuof will be said to be balanced if it has the form 1 E j j j β1 βp 1 (5.7) u = g0(x) g1(x) gp 1(x) a1(x) ap 1(x) − ··· − ··· − for some j 0,1,... ,l . Note that the element (5.7) belongs to Xj. ∈{ } Let 0 be the set of balanced elements of , and let 1 = 0.Foru 0 as in (5.7),E write E E E\E ∈E

j β1 βp 1 (5.8) u = w(x) a1(x) ap 1(x) − . ··· − From the definitions of w(x)andwit follows that w(x)j canbewrittenasalinear

combination of terms of the form gi1 (x) gijp (x) with each of g0(x),... ,gp 1(x) occurring exactly j times as a factor. But,··· by Lemma 5.8, −

β1 βp 1 gi (x) gi (x)a1(x) ap 1(x) − +Xj 1 =u+Xj 1. 1 ··· jp ··· − − −

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Thus u + Xj 1 is a scalar multiple of u + Xj 1. Calculating as in (4.2) and noting − − j that (p 1)! = 1inK,weseethatu+Xj 1 =( 1) u + Xj 1.Let 0= u:u . Then− it follows− by a simple induction− on j that− − E { ∈ E0} v + Ak, : v { 1 ∈ E0 ∪E1} is a basis of X/Ak, .LetXbe the subspace of X such that Ak, X X and 1 1 ≤ ≤ X/Ak, has basis u + Ak, : u . 1 { 1 ∈ E0} Lemma 5.11. Let e1,... ,em w(x),a1(x),... ,ap 1(x) ,wheree1 em has ∈{ − } ··· degree k and where there are at most l values of i for which ei = w(x). (i) e1 em X. (ii) If···m 2∈and 1 s

Note that Lemma 5.11 shows, roughly speaking, that the factors w(x), a1(x), ... , ap 1(x)commuteinX/Ak,1. − ˜ Lemma 5.12. X is a submodule of X,andX/Ak,1 ∼= Sq(KG).

Proof. Note that, if u is as in (5.8), then j, β1,... ,βp 1 satisfy the conditions − j + β1 + ...+ βp 1 = q and 0 j l = min(q, p 1). These conditions are − ≤ ≤ − equivalent to j + β1 + ...+βp 1 = q and 0 j p 1. Thus there is a vector − ˜ ≤ ≤ − space isomorphism ξ from X/Ak,1 to Sq(KG) given on each basis element u + Ak,1, where

j β1 βp 1 u = w(x) a1(x) ap 1(x) − , ··· − by

j β1 βp 1 ξ(u + Ak,1)=g0a1 ap −1 . ··· − i i Since ai(x)=w(x)(1 g ), the map w(x) 1, ai(x) 1 g for i =1,... ,p − 7→ 7→ − − 1, extends to a KG-module isomorphism from w(x),a1(x),... ,ap 1(x) to KG. ˜ h ˜ − i With u as above, (ξ(u + Ak,1))g belongs to Sq(KG) because Sq(KG)isaKG- module. Using Lemma 5.11 and the KG-module isomorphism with KG described above, we calculate that 1 ξ− ((ξ(u + Ak,1))g)=ug + Ak,1. Hence ug X, X is a submodule of X,andξis a KG-module isomorphism. ∈

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We can now prove Proposition 5.7 in the remaining case, where r =0.

k Lemma 5.13. Suppose that k is divisible by p and let k = qp.ThenV(0)x is a free KG-module. Proof. Consider the chain of submodules

Ak,1 X = X 1 + X X0 + X ... Xl +X =X. ≤ − ≤ ≤ ≤ We wish to show that X/Ak,1 is free. By Lemma 5.12 and Corollary 3.3, X/Ak,1 is free. Thus it suffices to show that (Xj + X)/(Xj 1 + X) is free for j =0,1,... ,l. − It is easily verified that (Xj + X)/(Xj 1 + X) has a basis consisting of all elements − of the form u +(Xj 1 +X), where u 1 and the free degree of u is jp.Bythe argument used in the− proof of Lemma∈E 5.9 we find that ˆu (Xj + X)/(Xj 1 + X) = Sjp(KG) Sq j(IG). − ∼ ⊗ − ˆu But Sjp(KG) is free, by Proposition 3.4. Therefore (Xj + X)/(Xj 1 + X) is free, as required. −

6. Main results We continue to use the notation of Section 5. In what follows we make use of various gradings of the free algebra A and its submodules. For most of the results it would be sufficient to use the natural grading by multidegree (with respect to a free generating set) as introduced in the paragraphs preceding Lemma 2.1 in Section 2. However, for Proposition 6.5 we require a different grading. For that reason we prove the results in this section for the more general notion of a ρ-grading as defined below. For each x and each monomial u of A,whereu=x1 xr with x1,... ,xr ,wewritedeg∈V(u) for the degree of u in x, that is, the··· number of values of∈i V x for which xi = x.Letρbe an equivalence relation on and suppose that ρ is G- V invariant;thatis,forallx1,x2 ,wehave(x1,x2) ρ if and only if (x1g,x2g) ρ. Let Φ be the set of all functions∈V φ : /ρ 0,∈1,2,... with finite support.∈ V →{ } For φ ΦletAφ denote the subspace of A spanned by all monomials u such that ∈ x E degx(u)=φ(E) for each equivalence class E. We say that an element of A is ∈ ρ-homogeneous if it belongs to Aφ for some φ. Clearly A = φ Φ Aφ. Furthermore P ∈ Aφ An,wheren= φ(E). Thus ρ-homogeneous elements are homogeneous. ⊆ E L A subspace B of A is said to be ρ-graded if B = φ Φ Bφ,whereBφ =B Aφ. P ∈ ∩ The subspaces Bφ are called the ρ-homogeneous components of B. (Note that, in the case where the equivalence relation ρ is simplyL equality, the ρ-homogeneous components are just the ordinary multihomogeneous components.) It is easily seen that a subspace B of A is ρ-graded if and only if it is spanned by a set of ρ- homogeneous elements. Furthermore, if B and C are ρ-graded subspaces, then so are B C and B + C. Clearly each An is ρ-graded. Since∩ G acts on and ρ is G-invariant, there is an induced action of G on /ρ. V V 1 Thus G also acts on Φ; the action of g is given by φ φg where (φg)(E)=φ(Eg− ) for each E /ρ. Clearly, if u is ρ-homogeneous,7→ then so is ug. In fact, for each ∈V φ Φ, (Aφ)g = Aφg.ThusGpermutes the ρ-homogeneous components Bφ of any ρ-graded∈ KG-submodule B of A. In this section we shall be concerned with direct decompositions of KG-modules which may have infinite dimension. It is a straightforward exercise to verify that

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every KG-module is a direct sum of finite-dimensional indecomposables. It follows that a direct summand of a free KG-module is free. Lemma 6.1. (i) Let B be a ρ-graded free KG-submodule of A.ThenBhas a G-free basis where each element of is ρ-homogeneous. (ii) Let B Sand C be ρ-graded KG-submodulesS of A such that C B and B/C is afreeKG-module. Then there exists a ρ-graded free KG-submodule≤ U of B such that B = C U. ⊕ Proof. Let Ω be the set of all G-orbits in the action of G on Φ.

(i) For I ΩletBI = φ IBφ. Thus we have a module decomposition B = ∈ ∈ I Ω BI . Hence each BI is free, and it suffices to consider the case where B = BI for∈ some I.IfIis a singleton,L I = φ , then the result is clear. If I contains p L p 1 { } elements, I = φ,φg,...,φg − , then we can take a basis φ of Bφ to obtain a { } S G-free basis of B,where = φG. S S S (ii) Clearly Cφ Bφ for each φ Φ. For I Ω define BI and CI as in (i). ⊆ ∈ ∈ Then B/C = I Ω BI/CI .ThuseachBI/CI is free. When I = φ let UI be ∼ ∈ { } any submodule of BI such that BI = CI UI. Clearly UI is free. When I = pL1 ⊕ φ,φg,...,φg − let uλ : λ Λ be a subset of Bφ such that uλ + Cφ : λ Λ { } { ∈ } { ∈ } is a basis of Bφ/Cφ, and let UI be the subspace of BI with basis uλ : λ Λ G. { ∈ } Then it is easily verified that UI is a ρ-graded free submodule of BI such that BI = CI UI . To complete the proof we define U = I Ω UI . ⊕ ∈ f It is easy to see that L, R and Lp are ρ-graded.L Recall from Section 5 that R∗ is the restricted Lie algebra generated by L2,... ,Lp 1,Rp∗,Lp+1,...,where p f − R∗= x :x L . Recall also that R∗ is free. Since R∗ is generated by (and p h ∈Vi⊕ p hence spanned by) ρ-homogeneous elements, it follows that R∗ is ρ-graded. If is a set of homogeneous elements of R,wesaythat is reduced if no element of belongsS to the restricted Lie algebra generated by theS other elements of .In theS proof of the next proposition we use the result that every reduced set freelyS generates the restricted Lie algebra that it generates: see Kukin [8]. Proposition 6.2. Let ρ be a G-invariant equivalence relation on . The free re- V stricted Lie algebra R∗ has a G-free free generating set consisting of ρ-homo- W geneous elements such that = n with n Ln for n = p,and p = W n 2 W W ⊆ 6 W f xp :x with f Lf . ≥ Wp ∪{ ∈V} Wp ⊆ p S Proof. For each n 2letR∗(n) denote the subalgebra of R∗ defined as follows: ≥ for 2 np,letR∗(n) be generated by L2,... ,Lp 1,Rp∗, − − Lp+1,... ,Ln. Note that each R∗(n)isρ-graded, since it is generated by (and hence spanned by) ρ-homogeneous elements. We shall define sets n inductively for n 2 W ≥ with the following properties: each n is G-free and consists of ρ-homogeneous W f p f f elements; n Ln for n = p; p = x :x with L ;andeach W ⊆ 6 W Wp ∪{ ∈V} Wp ⊆ p set ... n is reduced and generates R∗(n). These properties imply that, for W2 ∪ ∪W all n 2, the set 2 ... n is a free generating set for R∗(n), so that n 2 n ≥ W ∪ ∪W ≥ W is a free generating set for R∗. Thus we shall have proved the proposition. S We first find the sets n for 2 n2.) If W ≤ p>2 then, by Corollary 2.2, L2 is a free KG-module. Since L2 is ρ-graded, Lemma 6.1 shows that this module has a G-free basis which consists of ρ-homogeneous W2 elements. Since L generates R∗(2), so does . Clearly is reduced. 2 W2 W2

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Suppose next that 3 n

R∗(n 1) = (R∗(n 1) Ak). − − (i) ∩ i,k M

By consideration of the degrees of the elements of 2 ... n 1, W ∪ ∪W −

R∗(n 1) An =(R∗(n 1)(2) An) ... (R∗(n 1)(n 1) An). − ∩ − ∩ ⊕ ⊕ − − ∩

For i =1,...,p 1, R∗(n 1) is a free KG-module, by Corollary 2.2. Hence the − − (i) modules R∗(n 1)(2) An,... ,R∗(n 1)(n 1) An are free (since they are direct − ∩ − − ∩ summands of the R∗(n 1)(i)). Hence R∗(n 1) An is free. Since 2 ... n 1 − − ∩ W ∪ ∪W − ⊆ L,wehaveR∗(n 1) L for ifree module Un such − ∩ that Ln =(R∗(n 1) An) Un, together with a G-free basis n for Un which − ∩ ⊕ W consists of ρ-homogeneous elements. Since R∗(n) is generated by R∗(n 1) and − Ln, it is generated by ... n. It is easily verified that this set is reduced. W2 ∪ ∪W We now consider n = p. Suppose then that we have found 2,... , p 1.We W W − take the notational convention that 2 ... p 1 = and R∗(p 1) = 0 for p = 2. Just as in the previous case, W ∪ ∪W − ∅ −

R∗(p 1) Ap =(R∗(p 1)(2) Ap) ... (R∗(p 1)(p 1) Ap) − ∩ − ∩ ⊕ ⊕ − − ∩

and R∗(p 1) Ap is free. Also, 2 ... p 1 [L, L]. Hence R∗(p 1)(i) L00 for 2 i −p ∩1. Thus W ∪ ∪W − ⊆ − ⊆ ≤ ≤ − f R∗(p 1) Ap L00 Ap L . − ∩ ⊆ ∩ ⊆ p f Since Lp and R∗(p 1) Ap are ρ-graded and free, Lemma 6.1 yields a ρ-graded − ∩ f free module Up such that Lp =(R∗(p 1) Ap) Up, together with a G-free basis f − ∩ ⊕ f p of Up which consists of ρ-homogeneous elements. Let p = x : x . Wp W Wp ∪{ ∈V} Since R∗(p) is generated by R∗(p 1) and Rp∗, it is generated by 2 ... p.It is easily verified that this set is reduced.− W ∪ ∪W Finally, suppose that n>pand that we have found 2,... , n 1.LetA∗(n 1) W W − − be the associative subalgebra of A generated by R∗(n 1). Thus A∗(n 1) is ρ- − − graded. Moreover (see Section 2), A∗(n 1) is a free associative algebra with − 2 ... n 1 as a free generating set. Since this generating set is G-free, W ∪ ∪W − the homogeneous components A∗(n 1) are free KG-modules for i 1. Thus − (i) ≥ A∗(n 1) An is also a free module, since it is a direct sum of summands of the − ∩ A∗(n 1) . Recall from Section 5 that A∗ is the associative algebra generated by − (i) L2,... ,Lp 1,Rp∗,Lp+1,... . Hence A∗ is generated by R∗(n 1) Ln Ln+1 ..., − − ∪ ∪ ∪ and so by 2 ... n 1 Ln Ln+1 ... . Therefore A∗ An is spanned by all productsW ∪ of degree∪W −n formed∪ ∪ from this∪ last generating set.∩ It follows that A∗ An =(A∗(n 1) An)+Ln. By Proposition 5.2, A∗ An is free. Since ∩ − ∩ ∩ A∗(n 1) An is free, so also is − ∩ ((A∗(n 1) An)+Ln)/(A∗(n 1) An). − ∩ − ∩

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Thus Ln/(A∗(n 1) Ln) is free. Since Ln and A∗(n 1) Ln are ρ-graded, − ∩ − ∩ Lemma 6.1 yields a ρ-graded free module Un such that Ln =(A∗(n 1) Ln) Un, − ∩ ⊕ together with a G-free basis n of Un which consists of ρ-homogeneous elements. W Let L∗(n 1) be the Lie subalgebra of R∗(n 1) generated by 2 ... n 1. − − W ∪ ∪W − Thus L∗(n 1) is a free Lie algebra, with free generating set 2 ... n 1.Since − W ∪ ∪W − L∗(n 1) R∗ R,wehave[L∗(n 1),L∗(n 1)] L. It follows easily that − ⊆ ⊆ − − ⊆ p L∗(n 1) = x : x (L∗(n 1) L). − h ∈Vi⊕ − ∩ Let be any basis of L which includes both and a basis 0 of L∗(n 1) L. F p V F − ∩ Let 1 = x : x 0.Thus 1is a basis of L∗(n 1). Take an arbitrary F { [p] ∈V}∪F[p] pαF − [p] [p] ordering of ,where = b :b ,α 0 .Since 1 ,there F F [p] { ∈F ≥ } F ⊆F is an induced ordering of 1 .Let (A)and 1(A∗(n 1)) be the bases of A F F [p] F [p] − and A∗(n 1), respectively, obtained from and as in Section 2. Clearly − F F1 (A∗(n 1)) (A), (A), and (A∗(n 1)) = .Thusthe F1 − ⊆F F⊆F F1 − ∩F F0 intersection of the subspaces of A spanned by (A∗(n 1)) and is equal to the F1 − F subspace spanned by . In other words, A∗(n 1) L = L∗(n 1) L. F0 − ∩ − ∩ It follows that A∗(n 1) L = R∗(n 1) L, and so, by the choice of Un, − ∩ − ∩ Ln =(R∗(n 1) Ln) Un.SinceR∗(n) is generated by R∗(n 1) and Ln,it − ∩ ⊕ − is generated by 2 ... n. It is easily verified that this set is reduced. This completes the proof.W ∪ ∪W

Lemma 6.3. Let L∗ be the Lie subalgebra of R∗ generated by ,where is W W as in Proposition 6.2. Then L∗ is a free Lie algebra with free generating set . Furthermore, W

L∗ = L2 ... Lp 1 Rp∗ Lp+1 ... . ⊕ ⊕ − ⊕ ⊕ ⊕ For each i 1, let L∗ be the i-th homogeneous component of L∗ with respect to the ≥ i free generating set .ThenLi∗ An=Li∗ Lnfor all i, n 1 except for the case i =1,n=p. Also,W we have ∩ ∩ ≥

Ln =(L1∗ An) ... (Ln∗ 1 An) for n 2,n=p, ∩ ⊕ ⊕ − ∩ ≥ 6 and f Lp =(L1∗ Lp) ... (Lp∗ 1 Lp). ∩ ⊕ ⊕ − ∩ Proof. Clearly L∗ is a free Lie algebra with free generating set .SinceL∗ R p W ⊆ we have [L∗,L∗] L. It follows easily that L∗ = x : x (L∗ L). Recall ⊆ h ∈Vi⊕ ∩ that A∗ is the associative algebra generated by R∗. By the same argument as was used to prove that A∗(n 1) L = L∗(n 1) L in the proof of Proposition 6.2, − ∩ − ∩ we obtain A∗ L = L∗ L. Therefore ∩ ∩ f L2 ... Lp 1 Lp Lp+1 ... A∗ L L∗ L. ⊕ ⊕ − ⊕ ⊕ ⊕ ⊆ ∩ ⊆ ∩ But L∗ L L2 ... Lp 1 Lp Lp+1 ... .Thus ∩ ⊆ ⊕ ⊕ − ⊕ ⊕ ⊕ L∗ L=L2 ... Lp 1 (L∗ Lp) Lp+1 ... , ∩ ⊕ ⊕ − ⊕ ∩ ⊕ ⊕ f where L L∗ Lp. p ⊆ ∩ Each element u of [L∗,L∗] can be written as a linear combination of Lie mono- mials of length at least 2 in the elements of .Ifu [L∗,L∗] Ap,then W ∈ ∩ only elements of 2 ... p 1 are needed, and these belong to [L, L]. Thus W ∪ ∪W −

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f [L∗,L∗] Ap L00 Ap Lp . Since the elements of are homogeneous in ,we have ∩ ⊆ ∩ ⊆ W V f L∗ Ap =(L∗ Ap) ([L∗,L∗] Ap) (L∗ Ap)+L . ∩ 1∩ ⊕ ∩ ⊆ 1 ∩ p f f f Thus L∗ Lp =(L1∗ Lp)+Lp.ButL1∗ Lp Lp.ThusL∗ Lp=Lp. It follows that ∩ ∩ ∩ ⊆ ∩ f L∗ L = L2 ... Lp 1 Lp Lp+1 ... , ∩ ⊕ ⊕ − ⊕ ⊕ ⊕ and so

L∗ = L2 ... Lp 1 Rp∗ Lp+1 ... . ⊕ ⊕ − ⊕ ⊕ ⊕ Since [L∗,L∗] L,wehaveL∗ Ln =L∗ An for all n and all i 2. Also, for ⊆ i ∩ i ∩ ≥ n = p, L1∗ An L,soL1∗ Ln =L1∗ An. 6 Since each∩ element⊆ of ∩has degree∩ at least 2 in , W V L∗ An =(L1∗ An) ... (Ln∗ 1 An) ∩ ∩ ⊕ ⊕ − ∩ for all n 2. For n 2 with n = p we have L∗ An = Ln,andso ≥ ≥ 6 ∩ Ln=(L1∗ An) ... (Ln∗ 1 An). ∩ ⊕ ⊕ − ∩ f For n = p, since [L∗,L∗] Ap L L∗ Ap,wehave ∩ ⊆ p ⊆ ∩ f f Lp =(L1∗ Lp) (L2∗ Ap) ... (Lp∗ 1 Ap). ∩ ⊕ ∩ ⊕ ⊕ − ∩ f But L∗ L = L∗ Lp,andL∗ Ap=L∗ Lp for i 2. Hence 1 ∩ p 1 ∩ i ∩ i ∩ ≥ f Lp =(L1∗ Lp) ... (Lp∗ 1 Lp). ∩ ⊕ ⊕ − ∩

Our strategic aim, as outlined in the Introduction, has been achieved in the previous lemma. This result provides a direct decomposition of Ln into a sum of modules which can be found as direct summands of homogeneous components of L∗ of degree (in L∗) smaller than n.ThefactthatLnis a direct sum of isomorphic copies of KG and IG can now be established by a straightforward induction. We omit the details because we shall obtain the indecomposables of Ln together with their multiplicities (when is finite) as a consequence of the much stronger Theorem 2, which is proved later inV this section. In fact, Theorem 2 follows immediately from our next result. Theorem 6.4. Let L be the free Lie algebra on a G-free free generating set ,with Lregarded as a KG-module. Let ρ be any G-invariant equivalence relationV on . Then there exist subsets , ,... of L such that, for all n 1, the union of V L1 L2 ≥ i ug : u n,i=0,1,... ,p 1 { ∈L − } and

α 1 p − j w(u )(1 g ):u n/pα ,j=1,... ,p 1 α 1 { − ∈L − } pα≥n [|

is a basis of Ln. Furthermore, the elements of this basis are distinct, as written, and each set n consists of elements which are ρ-homogeneous. L

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Proof. If L is a free Lie algebra over K with a G-free free generating set ,where = G, with G-transversal ,andifρis a G-invariant equivalence relationV on VG,X then, for the purposes ofX this proof, let us say that (L, ,ρ)isaG-triple. X X For a G-triple (L, ,ρ)letR∗ be defined as in Section 5, and let be a free X W generating set for R∗ with the properties given in Proposition 6.2. Choose a G- transversal of such that xp : x .Thus = G.Eachelement of G is ρ-homogeneous.Y W Thus{ we can∈X}⊆Y define an equivalenceW relationY σ on G by Y Y taking (y1,y2) σ (for y1,y2 G) if and only if both y1 and y2 belong to the same ρ-homogeneous∈ component∈Y of A. Itiseasytoverifythatσis G-invariant. Let L∗ be the free Lie algebra with free generating set G as described in Lemma Y 6.3. It follows that (L∗, ,σ)isaG-triple. Define a function µ on G-triples by Y µ(L, ,ρ)=(L∗, ,σ). (Although there were choices made in the construction of fromX (L, ,ρ),Y we simply make arbitrary choices in order to define µ(L, ,ρ).) Y By an inductiveX definition we can define, for every G-triple (L, ,ρ),X subsets η (L, ,ρ),η (L, ,ρ),... of L satisfying η (L, ,ρ)= and, for allX n 2, 1 X 2 X 1 X X ≥ n 1 − ηn(L, ,ρ)= (ηi(µ(L, ,ρ)) Ln). X X ∩ i=1 [ ForafixedG-triple (L, ,ρ) we shall write, as before, A for the free associative al- X gebra on G and (L∗, ,σ)=µ(L, ,ρ). Furthermore, we write n = ηn(L, ,ρ) X Y X L X and n = ηn(L∗, ,σ) for all n 1. Thus = , = and M Y ≥ L1 X M1 Y n 1 − (6.1) n = ( i Ln) for all n 2. L M ∩ ≥ i=1 [ We now prove that the elements of n are ρ-homogeneous for all n 1. This is clear for n = 1. Arguing by inductionL on n for all G-triples, we may≥ assume that n>1 and that the elements of 1,... , n 1 are σ-homogeneous. But G consists of elements which are ρ-homogeneous.M M It− follows easily that the elementsY of 1,... , n 1 are ρ-homogeneous. Thus, by (6.1), the elements of n are ρ-homogeneous.M M − This completes the induction. L Note that the argument in the preceding paragraph also shows that the elements of each n are ρ-homogeneous. Thus, writing i,n = i An for all i, n 1, M M M ∩ ≥ we have i = i,n. By Lemma 6.3, i Ln = i,n except when i =1, M n 1 M M ∩ M n=p. Thus, by (6.1),≥ S n 1 − (6.2) n = i,n for all n 2 with n = p. L M ≥ 6 i=1 [ p Also, ( Lp) x :x = ,p. Thus, by (6.1), M1 ∩ ∪{ ∈X} M1 p 1 − p (6.3) p x :x = i,p. L ∪{ ∈X} M i=1 [ For any subset of A and α 0wewrite pα = upα :u and S ≥ S { ∈S} ζ( )= w(u)(1 gj ):u ,j=1,... ,p 1 . S { − ∈S − } It is easy to verify that, if consists of ρ-homogeneous elements of A,then pα and ζ( ) have the same property.S S S

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For each n 1 we define ≥ α 1 p − (6.4) (Ln)= nG ζ( n/pα ) B L ∪ α 1 L pα≥n [| and

α 1 p − (6.5) (Ln∗ )= nG ζ( n/pα ). B M ∪ α 1 M pα≥n [|

Clearly all elements of (Ln)and (L∗) are ρ-homogeneous. By (6.1), nG Ln B B n L ⊆ for all n.Thus n/pα Ln/pα , and it follows easily from (4.4) and (2.2) that α 1 L ⊆ p − ζ( α ) Ln. Hence (Ln) Ln. Similarly, (L∗ ) L∗ . Ln/p ⊆ B ⊆ B n ⊆ n We shall prove that (Ln)isabasisofLn, for all n 1. This is clear for n =1 because G = G. ArguingB by induction on n for all≥G-triples, we may assume L1 X that n>1andthat (Li∗)isabasisofLi∗ for i =1,... ,n 1. So, from now on, n is fixed with theseB properties. − By our assumptions, (L∗) An is a basis for L∗ An for i =1,... ,n 1(since B i ∩ i ∩ − elements of (L∗) are ρ-homogeneous). But, if n = p, B i 6

Ln =(L1∗ An) ... (Ln∗ 1 An) ∩ ⊕ ⊕ − ∩ by Lemma 6.3. Thus

n 1 − (6.6) ( (L∗) An) isabasisof Ln if n = p. B i ∩ 6 i=1 [

For i>1, L∗ Lp = L∗ Ap by Lemma 6.3. Thus, if n = p, (L∗) Lp is a basis i ∩ i ∩ B i ∩ for L∗ Lp for i>1. Also, since (L∗)= G, it is easy to check that (L∗) Lp i ∩ B 1 Y B 1 ∩ is a basis for L∗ Lp. By Lemma 6.3, 1 ∩ f Lp =(L1∗ Lp) ... (Lp∗ 1 Lp). ∩ ⊕ ⊕ − ∩ Thus

p 1 − f (6.7) ( (L∗) Lp) isabasisof L if n = p. B i ∩ p i=1 [

To prove that (Ln)isabasisofLn we first consider the case n = p. Then, by (6.4) and (6.1), B

(Lp)= pGζ( ) B L ∪L1 p1 − j = ( iG Lp) w(x)(1 g ):x ,j=1,... ,p 1 . M ∩ ∪{ − ∈X − } i=1 [ p 1 But, for i

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n 1 Now suppose that n = p. By (6.6), it suffices to prove that − ( (L∗) An)= 6 i=1 B i ∩ (Ln). By (6.5) and (6.2), B S n 1 n1 n 1 − − − α 1 p − ( (L∗) An)= ( iG An) (ζ( α ) An) B i ∩ M ∩ ∪ Mi/p ∩ i=1 i=1 i=1 α 1 pα≥ i [ [ [ [| n 1 n 1 − − α 1 p − = i,nG ζ(( α α ) ) M ∪ Mi/p ,n/p i=1 i=1 α 1 pα ≥(i,n) [ [ [| (n/pα) 1 − α 1 p − = nG ζ(( α ) ). L ∪ Mt,n/p α 1 t=1 pα≥n [| [ By comparison with (6.4) we see that it suffices to prove

(n/pα) 1 − α 1 α 1 p − p − (6.8) ( α ) = . Mt,n/p Ln/pα α 1 t=1 α 1 pα≥n pα≥n [| [ [| For α such that n/pα 1,p we have, by (6.2), 6∈ { } (n/pα) 1 − α = α . Mt,n/p Ln/p t=1 [ Thus, for these values of α, the terms on the two sides of (6.8) are equal. To prove (6.8) we need only compare the terms corresponding to values of α for which n/pα 1,p . These only occur when n is a power of p, n = pβ (where β 2since n 2and∈{ n}=p). In this case it remains to prove ≥ ≥ 6 (n/pα) 1 − α 1 α 1 p − p − (6.9) ( α ) = . Mt,n/p Ln/pα α β 1,β t=1 α β 1,β ∈{[− } [ ∈{[− } The term on the left hand side with α = β is empty. Evaluation of the other term on the left gives, by (6.3),

p 1 − β 2 β 2 β 2 β 2 β 1 p − p − p p − p − p − ( t,p) = x :x = , M Lp ∪{ ∈X} Lp ∪L1 t=1 [ which establishes equality with the right hand side of (6.9). We have now proved that (Ln)isabasisofLn for all n. To complete the proof of the theorem we must showB that the basis elements are distinct as written in the statement of the theorem.

Let = n 1 n and = n 1 n.Fori=0,1,... ,p 1 define κi : A A L ≥i L M ≥ M − → by κi(u)=ug for all u A.Forj=1,... ,p 1andα=1,2,3,... define S Sα 1 ∈ p − j − λj,α : A A by λj,α(u)=w(u )(1 g ). Let ∆ be the set of all the functions → − κi and λj,α. We must prove that the elements δ(u)foru ,δ ∆, are distinct. It is clearly sufficient to show that, for each n, the elements∈Lδ(u)∈ which belong to Ln are distinct. i The elements δ(u) which belong to L1 are the elements xg for x and i =0,1,... ,p 1. Since is a G-transversal of G, the result is true for∈Xn =1. − X X

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Working by induction on n over all G-triples, we may assume that n 2and ≥ that those elements δ(u), with u ,δ ∆, which belong to L1∗ ... Ln∗ 1 are distinct. ∈M ∈ ∪ ∪ − Suppose first that n = p. The elements δ(u) which belong to Lp are the elements i j ug for u p,i=0,1,... ,p 1, and the elements w(x)(1 g )forx , ∈L p −1 −i ∈X j=1,... ,p 1. Now p = i=1− ( i Lp), so the elements ug are distinct by − L M ∩ j a the inductive hypothesis. The elements w(x)(1 g ) are distinct and belong to Lp S i − f f a by Proposition 4.1. But the elements ug belong to Lp by (6.7), and Lp = Lp Lp by Proposition 4.1. Thus the result follows in this case. ⊕ Suppose now that n = p.Letδ(u) Lnwhere u ,δ ∆. If u i with 6 ∈ ∈L ∈ ∈L i>1, then u 1 ... i 1 by (6.1), and hence δ(u) L1∗ ... Ln∗ 1.But, ∈M ∪ ∪M− ∈ ∪ ∪ − if u 1,thenδ=λj,α for some j and α with α>1(sincen=p): in this case we ∈L p p p 6 have δ(u)=λj,α 1(u ), where u 1 and λj,α 1(u ) L1∗ ... Ln∗ 1. Hence the result follows− by the inductive∈M hypothesis. − ∈ ∪ ∪ −

With a view to future applications we now briefly describe a version of Theorem 6.4 in which the sets n may be chosen to have an additional property. Identify the group LG with a Sylow p-subgroup of the symmetric group of degree p, and let N be the normalizer of G in this symmetric group. Thus N has order p(p 1), and it is the semidirect product of G by a cyclic group H of order p 1. Let h be− a generator of H. Suppose that V is a KN-module such that the restriction− V G is a free KG-module. Note that V (IG)isaKN-submodule of V .SinceK contains↓ a primitive (p 1)-th root of unity, there is a KH-submodule X of V such that V = V (IG) X,and− Xhas a basis which consists of eigenvectors of h.It ⊕ X is not difficult to see that is a free generating set for V G.ThusVhas a G-free basis G with a G-transversalX which consists of eigenvectors↓ of h.NotethatH permutesX the set of all scalar multiplesX of elements of G. The free associative algebra A generated by V is naturallyX a KN-module, and the restriction from N to G gives, of course, the KG-module structure that we have already been using. Clearly, the algebras L and R are KN-submodules of A,and it is easily verified that the algebra R∗ is also a KN-submodule. Take any equivalence relation ρ on G such that (x, xg) ρ for all x G. X ∈ ∈X Thus ρ is G-invariant, and each ρ-homogeneous component Aφ is a KN-submodule of A. In order to obtain a version of Proposition 6.2 and hence the desired version of Theorem 6.4, we require a variant of Lemma 6.1. Suppose that B is any ρ-graded KN-submodule of A such that B G is free. Then each ρ-homogeneous component ↓ Bφ of B is a KN-submodule such that Bφ G is free. Hence (as already noted ↓ for V ) the module Bφ has a G-free basis φ with a G-transversal φ consisting of eigenvectors of h. By taking the union overSφ we obtain a G-free basisT of B which consists of ρ-homogeneous elements and which has a G-transversal Sconsisting of eigenvectors of h. This is the required variant of Lemma 6.1(i). SupposeT now that B and C are ρ-graded KN-submodules of A such that C B and (B/C) G is ≤ ↓ free. Then each (Bφ/Cφ) G is free, and so (since G is a Sylow p-subgroup of N) ↓ each Bφ/Cφ is projective. Hence we easily obtain a ρ-graded KN-submodule U of B such that U G is free and B = C U. This is the required variant of Lemma 6.1(ii). ↓ ⊕ It is easily checked that all the KG-modules considered in the proof of Proposi- tion 6.2 are KN-modules. Hence, by use of the above variant of Lemma 6.1, we can choose the sets n to have the extra property that each n has a G-transversal W W

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n consisting of eigenvectors of h.Thus has a G-transversal consisting of Yeigenvectors of h, and clearly we can take Wxp : x . Note alsoY that is a KN-module. { ∈X}⊆Y hWi Now let L∗ and σ be as defined in the proof of Theorem 6.4. Then it is easily checked that (y,yg) σ for all y = G. Thus in the proof of Theorem 6.4 we may restrict consideration∈ to G-triples∈W (L,Y ,ρ)where(x, xg) ρ for all x G X ∈ ∈X and where consists of eigenvectors of h.Thesets ndefined inductively by means of X= and equation (6.1) then consist of eigenvectorsL of h. L1 X Proposition 6.5. In the case described above, in which = G,where consists V X X of eigenvectors of h and (x, xg) ρ for all x ,thesets n of Theorem 6.4 may be chosen to have the additional∈ property that∈V they consistL of eigenvectors of h.

We can now prove our main results, as stated in Section 1.

Proof of Theorem 2. Theorem 2 follows immediately from Theorem 6.4 on putting

= n 1 n. L ≥ L S Remark. By choosing ρ to be the relation of equality, we see that the elements of in Theorem 2 may be taken to be multihomogeneous. L

Proof of Theorem 1. We derive Theorem 1 from Theorem 2, writing n = Ln L L∩ in accordance with Theorem 6.4. The fact that Ln is a direct sum of a free module and modules isomorphic to IG follows from Theorem 2, since for each u n the p 1 ∈L elements u,ug,...,ug − span a regular submodule of Ln,andforeachu n/pα α 1 ∈L with α 1andpα nthe elements w(up − )(1 gj ), for j =1,... ,p 1, span a ≥ | − − submodule of Ln that is isomorphic to IG. It remains to verify the formula for the number of copies of IG. We do this by induction on n.Letf(n)anda(n) denote, respectively, the multiplicity of the regular module and the multiplicity of IG when Ln is written as a direct sum of indecomposables. Note that, by Theorem 2, f(n) is equal to the number of elements of n. We need to verify that a(n) satisfies formula (1.1). Clearly, L

dim(Ln)=pf(n)+(p 1)a(n). − Hence 1 f(n)= (dim(Ln) (p 1)a(n)). p − − It follows from Theorem 2 that

a(n)=f(n/p)+f(n/p2)+f(n/p3)+... ,

where f(r) is interpreted as 0 when r is not an integer. If (n, p)=1thena(n)=0, which gives (1.1) in this case. In particular we have the basis of our induction. If p n,thenwehave | (6.10) a(n)=f(n/p)+a(n/p) 1 = (dim(L )+a(n/p)). p n/p

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Witt’s formula and the inductive hypothesis give

1 1 1 a(n)= µ(d)mn/pd µ(d)mn/pd pn/p − n/p  d (n/p) d | p d (n/p)  X | X|  1   (6.11) = µ(d)mn/pd n d (n/p) (d,p|X)=1 1 = µ(d)mn/d. n − d p d n X| | This completes the proof of Theorem 1.

G Proof of Corollary 1. Clearly dim (Ln) = f(n)+a(n). By (6.10), f(n)+a(n)= a(pn). Hence the result follows from (6.11) with pn in place of n.

Proof of Corollary 2. In the case where V is a regular ZG-module, Corollary 2 follows from Corollary 1 and the results of [3]. The general case is obtained in the same way, by means of a straightforward generalisation of the results of [3].

7. On the basis elements of L In this final section we obtain an identity which allows us to express basis ele- ments of the form (1.2) in Theorem 2 explicitly as elements of L(V ). We define numbers βn,k, for all non-negative integers n, k, by setting β0,0 =2, βn, =1foralln 1, and 0 ≥

k n n k 1 βn,k =( 1) − − − k k 1  −  for all n, k with k 1, where the binomial coefficient is taken to be 0 if n k 1 < 0, k 1 < 0orn k≥ 1

for all n 2andallk 1. Hence each βn,k is an integer. Using (7.1), we obtain ≥ ≥ n 2 − (7.2) βn,k = βi,k − +1 i=0 X for all n 2andallk 0. For elements≥ x and ≥y of any Lie ring, define the elements [x, yk] inductively, for 0 k k 1 each non-negative integer k,by[x, y ]=xand [x, y ]=[[x, y − ],y]fork>0. In the next two lemmas, x, y and z are arbitrary elements of a Lie ring.

Lemma 7.1. For all n 0, ≥ [n/2] n n k k n 2k [x, z ,y] [y,z ,x]= βn,k[[x, z ], [y,z ],z − ]. − k X=0

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Proof. The result is easy to verify for n =0andn= 1. We proceed by induction on n and assume n 2. Then ≥ [x, zn,y] [y,zn,x] −n 1 n 1 n 1 n 1 =[x, z − ,y,z] [y,z − ,x,z]+[x, z − , [z,y]] [y,z − ,[z,x]] n 1 − n 1 n 2 − n 2 =[[x, z − ,y] [y,z − ,x],z] [[x, z],z − ,[y,z]] [[y,z],z − ,[x, z]] . − − − Using the inductive hypothesis for n 1andn 2, the right hand side can
be rewritten as − − [(n 1)/2] − k k n 2k 1 βn 1,k[[x, z ], [y,z ],z − − ,z] − k X=0 [(n 2)/2] − k k n 2k 2 βn 2,k[[x, z, z ], [y,z,z ],z − − ] − − k X=0 [n/2] n k k n 2k = βn 1,0[x, y, z ]+ (βn 1,k βn 2,k 1)[[x, z ], [y,z ],z − ]. − − − − − k X=1 The lemma follows by (7.1) and the fact that βn 1,0 = βn,0. − Lemma 7.2. For all n 2, ≥ n 2 − i n i 2 i n i 2 ([x, z ,y,z − − ] [y,z ,x,z − − ]) − i=0 X [(n 2)/2] − k k n 2k 2 = βn,k [[x, z ], [y,z ],z − − ]. − +1 k X=0 Proof. For i =0,... ,n 2letui be the summand indexed by i on the left hand side. By Lemma 7.1, −

i i n i 2 ui =[[x, z ,y] [y,z ,x],z − − ] − [i/2] k k i 2k n i 2 = βi,k[[x, z ], [y,z ],z − ,z − − ] k X=0 [i/2] k k n 2k 2 = βi,k[[x, z ], [y,z ],z − − ]. k X=0 Summing from i =0toi=n 2, and then changing the order of summation, gives − n 2 [(n 2)/2] n 2 − − − k k n 2k 2 ui = βi,k [[x, z ], [y,z ],z − − ], i=0 k i=0 ! X X=0 X and, in view of (7.2), this is equal to the right hand side of the equation in the lemma.

Proposition 7.3. Let wˆ be as defined in Section 4. Then

wˆ = γp,k[[x1,xπ(3),... ,xπ(k+2)], [x2,xπ(k+3),... ,xπ(2k+2)],xπ(2k+3),... ,xπ(p)], k,π X

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p 2 where k runs from 0 to − , π ranges over all permutations of 3, 4,... ,p ,and 2 { } each γp,k is an integer given by   k 1 p k 2 γp,k =( 1) − − . − k +1 k   Proof. We apply the equation of Lemma 7.2, with n = p, to the free Lie ring on free generators x1,x2,... ,xp,asinSection4.Putx=x1,y=x2 and z = x3 +x4 +...+xp. We expand both sides of the equation and compare the multilinear parts (consisting of the terms with degree 1 in each generator x1,... ,xp). The multilinear part on the left hand side is [x ,x ,... ,x ] [x ,x ,... ,x ] 1 κ(2) κ(p) − 2 λ(1) λ(p) κ λ X X where κ and λ range over all permutations of 2, 3,... ,p and 1, 3,... ,p ,re- spectively. But this is equal to p wˆ. The multilinear{ part} on the{ right hand} side is

βp,k [[x ,x ,... ,x ], [x ,x ,... ,x ],x ,... ,x ], − +1 1 π(3) π(k+2) 2 π(k+3) π(2k+2) π(2k+3) π(p) k,π X with k and π as in the proposition. It is easily verified that βp,k+1 is divisible by p for all k 0. Hence the result follows on dividing through by p. ≥ Let be as in Theorem 2 and take u and α 0. Then, by (4.4) and α PropositionL 7.3, we may write w(up )(g 1)∈L as a linear combination≥ of Lie products pα pα pα −p 1 of elements from u , (u )g,...,(u )g − . These Lie products may be written { } p 1 pα i i pα as Lie products of elements from u,ug,...,ug − , because (u )g =(ug ) and, over a field of characteristic p,thevalueof[{ x, y}pα ]isgivenby

α α α [x, yp ]=xyp yp x =[x,y,y,...,y]. − pα

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[12] R. M. Thrall, ‘On symmetrized Kronecker powers and the structure of the free Lie ring’, Amer. J. Math., 64 (1942), 371–388. MR 3:262d [13] G. E. Wall, ‘On the Lie ring of a group of prime exponent’, in ‘Proc. Second Internat. Conf. Theory of Groups, Canberra, 1973’, Lecture Notes in Mathematics, 372, Springer, Berlin, etc., 1974, pp. 667–690. MR 50:10098 [14] F. Wever, ‘Uber¨ Invarianten von Lieschen Ringen’, Math. Annalen, 120 (1949), 563–580. MR 10:676e [15] E. Witt, ‘Die Unterringe der freien Lieschen Ringe’, Math. Z., 64 (1956), 195–216. MR 17:1050a

Department of Mathematics, UMIST, Manchester M60 1QD, United Kingdom E-mail address: [email protected]

E-mail address: [email protected]

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