Torsion in free centre-by-nilpotent-by-abelian Lie rings of rank 2

TL;DR

This paper characterizes the torsion subgroup in the additive group of a specific class of free Lie rings constructed from two generators, focusing on cases where the nilpotency class is a prime number.

Contribution

It provides a complete description of torsion in free centre-by-nilpotent-by-abelian Lie rings of rank 2 for prime nilpotency classes.

Findings

Identifies the torsion subgroup structure for the case |X|=2 and c prime.

Provides explicit descriptions of torsion elements.

Advances understanding of torsion phenomena in specific Lie ring constructions.

Abstract

For $c\geq 2$, the free centre-by-(nilpotent-of-class-c-1)-by abelian Lie ring on a set $X$ is the quotient $L/[(L')^c,L]$ where $L$ is the free Lie ring on $X$, and $(L')^c$ denotes the $c$th term of the lower central series of the derived ideal $L'=L^2$ of $L$. In this paper we give a complete description of the torsion subgroup of its additive group in the case where $|X|=2$ and $c$ is a prime number.

Equations

$$L/[(L^{\prime }{)}^c,L]$$

$$L/([(L^{\prime }{)}^c,L]+L^{\prime \prime \prime })$$

$$L^c(A)\stackrel{{\nu }_c}{\to }{T}^c(A)\stackrel{{\rho }_c}{\to }{L}^c(A),{\nu }_c{\rho }_c=c.$$

$$0\to B^c(A)\to L^c(A)\stackrel{{\eta }_c}{\to }{M}^c(A)\mapsto 0$$

$$0\to M^c(A)\stackrel{{\mu }_c}{\to }A\otimes A^{c-1}\stackrel{{\kappa }_c}{\to }{A}^c\to 0$$

$$[a_1,a_2,\dots ,a_c]\mapsto a_1\otimes (a_2\circ \cdots \circ a_c)-a_2\otimes (a_1\circ \cdots \circ a_c)$$

$$a_1\otimes (a_2\circ \cdots \circ a_c)\mapsto [a_1,a_2,a_3,\dots ,a_c]+[a_1,a_3,a_2,\dots ,a_c]+\cdots +[a_1,a_c,a_2,\dots ,a_{c-1}],$$

$$M^c(A)\stackrel{{\mu }_c}{\to }A\otimes A^{c-1}\stackrel{{\lambda }_c}{\to }{M}^c(A),{\mu }_c{\lambda }_c=c$$

$$[a_1,\dots ,a_c]\mapsto \frac{1}{c}\left(\sum _{\sigma }\left[a_1,a_{\pi \left(2\right)},...,a_{\pi \left(c\right)}\right]-\sum _{\tau }\left[a_2,a_{\tau \left(1\right)},...,a_{\tau \left(c\right)}\right]\right),$$

$$M^c(A)\stackrel{{\theta }_c}{\to }{L}^c(A)\stackrel{{\eta }_c}{\to }{M}^c(A),{\theta }_c{\eta }_c=(c-2)!$$

$$[a_1,a_2,\dots ,a_c]x=\sum _{i=1}^c[a_1,\dots ,a_{i}x,\dots ,a_c],$$

$$cu=u{H}_k({\nu }_c{\rho }_c)=(u{H}_k({\nu }_c))H_k({\rho }_c)=0H_k({\rho }_c)=0.$$

$$\cdots \to H_1(M^p(A))\to B^p(A){\otimes }_U\mathbb{Z}\to L^p(A){\otimes }_U\mathbb{Z}\stackrel{{\eta }_p\otimes 1}{\to }{M}^p(A){\otimes }_U\mathbb{Z}\mapsto 0.$$

$$0\to B^p(A)\stackrel{p}{\to }{B}^p(A)\to B^p(A)\otimes {\mathbb{Z}}_p\to 0$$

$$\cdots \to H_1(B^p(A)\otimes {\mathbb{Z}}_p)\to B^p(A){\otimes }_U\mathbb{Z}\stackrel{p}{\to }{B}^p(A){\otimes }_U\mathbb{Z}\to B^p(A){\otimes }_U{\mathbb{Z}}_p\to 0.$$

$$M^p(A){\otimes }_U\mathbb{Z}\stackrel{{\theta }_p\otimes 1}{\to }{L}^p(A){\otimes }_U\mathbb{Z}\stackrel{{\eta }_p\otimes 1}{\to }{M}^p(A){\otimes }_U\mathbb{Z},{\theta }_k{\eta }_k\otimes 1=(p-2)!$$

$$c=b_1^{(1)}\otimes \cdots \otimes b_{k_1}^{(1)}\otimes b_1^{(2)}\otimes \cdots \otimes b_{k_2}^{(2)}\otimes \cdots \otimes b_1^{(p)}\otimes \cdots \otimes b_{k_p}^{(p)},$$

$$T^p(V)=X_1>X_2>\cdots >X_m>X_{m+1}=0.$$

$$b_{{\pi }_1(1)}^{(1)}\otimes \cdots \otimes b_{{\pi }_1(k_1)}^{(1)}\otimes \cdots \otimes b_{{\pi }_p(1)}^{(p)}\otimes \cdots \otimes b_{{\pi }_p(k_p)}^{(p)}+X_{i+1}=c+X_{i+1}.$$

$$\overline{c}=(b_1^{(1)}\circ \cdots \circ b_{k_1}^{(1)})\otimes (b_1^{(2)}\circ \cdots \circ b_{k_2}^{(2)})\otimes \cdots \otimes (b_1^{(p)}\circ \cdots \circ b_{k_p}^{(p)}).$$

$$X_i/X_{i+1}\cong S^{k_1}(L^1(V))\otimes \cdots \otimes S^{k_p}(L^p(V)).$$

$${\sigma }_i(\overline{c})=\frac{1}{k_1!\cdots k_p!}\sum _{\genfrac{}{}{0ex}{}{{\pi }_1\in \mathrm{Sym}(k_1),\dots ,}{{\pi }_p\in \mathrm{Sym}(k_p)}}{b}_{{\pi }_1(1)}^{(1)}\otimes \cdots \otimes b_{{\pi }_1(k_1)}^{(1)}\otimes \cdots \otimes b_{{\pi }_p(1)}^{(p)}\otimes \cdots \otimes b_{{\pi }_p(k_p)}^{(p)},$$

$$X_2=\mathrm{Im}({\sigma }_2)\oplus \cdots \oplus \mathrm{Im}({\sigma }_{m-1})\oplus L^p(V).$$

$$W=\mathrm{Im}({\sigma }_2)\oplus \cdots \oplus \mathrm{Im}({\sigma }_{m-1})\oplus (L(V{)}^{\prime \prime }\cap L^p(V)).$$

$$\alpha :T^p(V)\to V\otimes V^{p-1},\beta :V\otimes V^{p-1}\to T^p(V)$$

$$a_1\otimes a_2\otimes \cdots \otimes a_p\mapsto a_1\otimes (a_2\circ \cdots \circ a_p)$$

$$a_1\otimes (a_2\circ \cdots \circ a_p)\mapsto \frac{1}{(p-1)!}\sum _{\pi }{a}_1\otimes a_{\pi (2)}\otimes \cdots \otimes a_{\pi (p)},$$

$$T^p(V)=\mathrm{Ker}(\alpha )\oplus \mathrm{Im}(\beta )\cong \mathrm{Ker}(\alpha )\oplus (V\otimes V^{p-1}).$$

$$b_1^{(1)}\otimes b_2^{(1)}\otimes \cdots \otimes b_p^{(1)}\text{with}{b}_1^{(1)},\dots ,b_p^{(1)}\in {\mathcal{B}}^{(1)}\text{and}{b}_1^{(1)}\le \cdots \le b_p^{(1)}$$

$$[b_1^{(1)},b_2^{(1)},\dots ,b_p^{(1)}]\text{with}{b}_1^{(1)},\dots ,b_p^{(1)}\in {\mathcal{B}}^{(1)}\text{and}{b}_1^{(1)}>b_2^{(1)}\le \cdots \le b_p^{(1)}$$

Full text

1. Introduction

For an integer $c\geq 2$, the free centre-by-(nilpotent-of-class $c-1$)-by-abelian Lie ring on a set $X$ is the quotient

[TABLE]

where $L$ is an (absolutely) free Lie ring on $X$ and $(L^{\prime})^{c}$ is the $c$th term of the lower central series of the derived ideal $L^{\prime}=L^{2}$ of $L$. In this note we determine the torsion subgroup of the additive group of (1.1) in the case where $L$ has rank $2$, that is $X$ is a set of two elements, and $c$ is a prime number. It is easily observed that for any $c\geq 2$ and any $X$ with $|X|\geq 2$ the torsion subgroup of (1.1) is contained in the central ideal $(L^{\prime})^{c}/[(L^{\prime})^{c},L]$. If $c=p$, where $p$ is a prime, then the ideal $(L^{\prime})^{p}/[(L^{\prime})^{p},L]$ is a direct sum of a free abelian group and an infinite elementary abelian $p$-group. We exhibit an explicit $\mathbb{Z}_{p}$-basis of the torsion part if $|X|=2$. This is the main result of this paper, see Theorem 5.1 in Section 5. In the earlier paper [1] a complete description of the torsion subgroup of the additive group of the more reduced quotient

[TABLE]

was obtained, again in the case of rank $2$, but for arbitrary $c\geq 2$. Note that (1.2) coincides with (1.1) for $c=2$ and $c=3$, but not for $c\geq 4$. We make essential use of the results from [1].

The interest in torsion in relatively free Lie rings of the form (1.1) as well as their group-theoretic counterparts, the relatively free groups $F/[\gamma_{c}(F^{\prime}),F]$,where $F$ is a free group and $\gamma_{c}(F^{\prime})$ is the $c$th term of the lower central series of the derived subgroup $F^{\prime}$, has a long history. If $c=2$, these turn into the free centre-by-metabelian Lie rings and the free centre-by-metabelian groups, respectively, and it was the latter in which Kanta Gupta [8] first discovered torsion elements, a major surprise at the time. Later torsion was detected in $F/[\gamma_{c}(F^{\prime}),F]$ if $c$ is a prime [15] and if $c=4$ [16]. Quite surprising, though, it turned out that $F/[\gamma_{c}(F^{\prime}),F]$ is torsion free if $c$ is divisible two distint primes [10]. As to Lie rings, early work in [12] on the free centre-by-metabelian Lie rings $L/[L^{\prime\prime},L]$, and in particular on torsion in the additive group, turned out in need of some rectification, and this was eventually accomplished in [14] and [11]. For larger values of $c$, Drensky [7] proved that for any prime $p$ the free Lie ring $L/[(L^{\prime})^{p},L]$ contains non-trivial multilinear elements of degree $2p+1$ which have order $p$.

The paper is organized as follows. In Section 2 we introduce notation and assemble a number of preliminary results on the homogeneous components of free Lie rings. These are further examined in Section 3 where we derive our main result on Lie powers of free modules in prime degree, postponing, however, one key ingredient on Lie powers of prime degree $p$ over fields of characteristic $p$ to Section 4. In the final Section 5 we exploit the results of the previous sections to derive our main result.

2. Notation and some preliminaries

We write maps on the right and use left-normed notation for Lie products. Let $A$ be a free abelian group. By $L(A)$ we denote the free Lie ring on $A$. For a positive integer $c$, we let $L^{c}(A)$ denote the degree $c$ homogeneous component of $L(A)$, that is the span of all left normed simple Lie products $[a_{1},a_{2},\dots,a_{c}]$ with $a_{i}\in A$. In particular, $L^{1}(A)=A$. The universal envelope of $L(A)$ can be identified with the tensor ring $T(A)=\bigoplus_{c\geq 0}T^{i}(A)$ where $T^{0}(A)=\mathbb{Z}$ and, for $c>0$, $T^{c}(A)=\underbrace{A\otimes\dots\otimes A}_{c}$, the $c$th tensor power of $A$. By $A^{c}$ we denote the $c$th symmetric power of $A$. The free metabelian Lie ring on $A$ is the quotient of $L(A)$ by its second derived ideal: $M(A)=L(A)/L(A)^{\prime\prime}$. This too is a graded Lie ring and we let $M^{c}(A)$ denote its $c$th homogeneous component, that is $M^{c}(A)=L^{c}(A)/(L^{c}(A)\cap L(A)^{\prime\prime})$. We call $L^{c}(A)$ and $M^{c}(A)$ the $c$th free Lie power and the $c$th free metabelian Lie power of $A$, respectively. It is well-known that if $\mathcal{A}$ is an ordered $\mathbb{Z}$-basis of $A$, then the left normed simple Lie products $[b_{1},b_{2},\dots,b_{c}]$ with $b_{i}\in\mathcal{A}$ and $b_{1}>b_{2}\leq\dots\leq b_{c}$ form a $\mathbb{Z}$-basis of $M^{c}(A)$ (see [2, Section 4.2.2]).

The canonical embedding of $L(A)$ into its universal envelope $T(A)$ induces in each degree $c$ an embedding $\nu_{c}:L^{c}(A)\rightarrow T^{c}(A)$. By a well-known theorem of Wever (see [13, Chapter 5, Theorem 5.16]), the composite of this embedding with the natural projection $\rho_{c}:T^{c}(A)\rightarrow L^{c}(A)$ defined by $a_{1}\otimes\dots\otimes a_{c}\mapsto[a_{1},\dots,a_{c}]$ amounts to multiplication by $c$ on $L^{c}(A)$:

[TABLE]

The definition of $M^{c}(A)$ gives rise to a short exact sequence

[TABLE]

where $B^{c}(A)=L^{c}(A)\cap L(A)^{\prime\prime}$ and $\eta_{c}$ is the natural projection map. Moreover, for $c\geqslant 2$ the metabelian Lie power $M^{c}(A)$ fits into a short exact sequence

[TABLE]

where the maps $\mu_{c}$ and $\kappa_{c}$ are given by

[TABLE]

and $a_{1}\otimes(a_{2}\circ\dots\circ a_{c})\mapsto a_{1}\circ a_{2}\circ\dots\circ a_{c}$, respectively (see [9, Corollary 3.2]). Moreover, there is a map $\lambda_{c}:A\otimes A^{c-1}\rightarrow M^{c}(A)$, given by

[TABLE]

such that the composite of $\mu_{c}$ and $\lambda_{c}$ amounts to multiplication by $c$ on $M^{c}(A)$:

[TABLE]

see [1, Section 3]. Finally, for $c\geq 2$ there is a map $\theta_{c}:M^{c}(A)\rightarrow L^{c}(A)$ given by

[TABLE]

where the sums run over all permutations $\sigma$ of $\{2,3,...,n\}$ and all permutations $\tau$ of $\{1,3,...,n\},$ respectively. Although we work over $\mathbb{Z}$, the factor $1/c$ in (2.5) makes sense as the expression on the right hand side can be written as a Lie polynomial with integer coefficients(see [4, Section 2]). This Lie polynomial has been calculated in [5, Proposition 7.3], but since it is rather involved, we prefer to use the compact form (2.5) in what follows. The composite of $\theta_{c}$ and the natural projection $\eta_{c}$ as in (2.2) amounts to multiplication by $(c-2)!$ on $M^{c}(A)$:

[TABLE]

(see [4, Section 2]).

Now suppose that $A$ carries the structure of a module for the polynomial ring $U=\mathbb{Z}[X]$ where $X$ is a finite set of variables. Then all the objects introduced in this section such as Lie powers, symmetric powers etc. will be regarded as $U$-modules under the derivation action. For example, for $x\in X$, $a_{i}\in A$,

[TABLE]

Note that all the maps introduced in this section are compatible with the derivation action, that is, all these maps are, in fact, $\mathbb{Z}[X]$-module homomorphisms. This will be used in what follows without further reference being given. The ring of integers $\mathbb{Z}$ will be regarded as a trivial $U$-module.

3. Lie powers of free modules

In this section we retain the notation introduced in Section 2, but now we assume throughout that $A$ is a free $U$-module for the polynomial ring $U=\mathbb{Z}(X)$. All homology groups in this section will be over the ground ring $U$. For brevity, if $W$ is a $U$-module, the homology groups $H_{k}(U,W)={\rm Tor}^{U}_{k}(W,\mathbb{Z})$, $k\geq 0,$ will be written as $H_{k}(W)$.

It is well known (see, for example, [14, Lemma 5.2]) that if $A$ is a free $U$-module then both $T^{c}(A)$ and $A\otimes A^{c-1}$ with $c\geq 2$ are also free $U$-modules under the derivation action.

Lemma 3.1**.**

Let $A$ be a free $U$-module, $c\geq 2$. Then

1. (i)

the tensor products $L_{c}(A)\otimes_{U}\mathbb{Z}$ and $M_{c}(A)\otimes_{U}\mathbb{Z}$ are direct sums of a free abelian group and a torsion group of exponent dividing $c$, 2. (ii)

for $k\geq 1$ the homology groups $H_{k}(L_{c}(A))$ and $H_{k}(M_{c}(A))$ are torsion groups of exponent dividing $c$.

Proof.

By applying the homology functor to the maps in (2.1) we get that $H_{k}(\nu_{c}\rho_{c})$ is multiplication by $c$ on $H_{k}(L_{c}(A))$ for all $k\geq 0$. Since $A$ is a free $U$-module, $H_{0}(T_{c}(A))$ is free abelian and $H_{k}(T_{c}(A))=0$ for $k\geq 1$. Then, for any $u\in H_{k}(L^{c}A)$ with $k>0$,

[TABLE]

The same holds if $k=0$ and $u\in{\rm Ker}(H_{0}(\nu_{c}))$. Hence multiplication by $c$ annihilates both the homology groups $H_{k}(L_{c}(A))$ for $k\geq 1$ and the kernel of the homomorphism $H_{0}(\nu_{c})$. The image of this homomorphism is contained in the free abelian group $H_{0}(T_{c}(A))$, and hence itself free abelian. The results (i) and (ii) for $L_{c}(A)$ follow. The proof of (i) and (ii) for $M_{c}(A)$ are obtained by a similar argument using the maps in (2.4) instead of those in (2.1). ∎

Now we consider Lie powers of prime degree. Let $p$ be a prime. By applying the homology functor to the short exact sequence (2.2) we obtain the exact sequence

[TABLE]

By Lemma 3.1., $L^{p}(A)\otimes_{U}\mathbb{Z}$ is a direct sum of a free abelian group and an elementary abelian $p$-group, and $H_{1}(M^{p}(A))$ is an elementary abelian $p$-group. Note that this does not exclude the possibility that these torsion subgroups are trivial. Now the exact sequence (3.1) yields that $B^{p}(A)\otimes_{U}\mathbb{Z}$ is a direct sum of a free abelian group and a (possibly trivial) $p$-group of exponent dividing $p^{2}$. In fact, we will show that this group has actually no torsion, in other words, we will prove the following result.

Lemma 3.2**.**

Let $A$ be a free $U$-module and $p$ a prime. Then the tensor product $B^{p}(A)\otimes_{U}\mathbb{Z}$ is a free abelian group.

Proof.

Since we already know that $B^{p}(A)\otimes_{U}\mathbb{Z}$ is a direct sum of a free abelian group and a $p$-group of finite exponent, it is sufficient to show that no non-zero element in $B^{p}(A)\otimes_{U}\mathbb{Z}$ is annihilated by $p$. We use reduction modulo $p$, that is the short exact sequence

[TABLE]

which, in its turn, gives rise to the exact sequence

[TABLE]

The Lemma will be proved once we show that the homology group on the left is zero, and this will certainly follow if we can verify that $B^{p}(A)\otimes\mathbb{Z}_{p}$, regarded as a module for the polynomial ring $\mathbb{Z}_{p}[X]$, is projective. The proof of this fact will be given in the next section (see Corollary 4.2). This will then complete the proof of Lemma 3.2. ∎

Now we have all the ingredients in place to prove the main result of this section. Recall the homomorphism $\theta_{c}:M^{c}(A)\rightarrow L^{c}(A)$ defined by (2.5).

Proposition 3.3**.**

Let $A$ be a free $U$-module and $p$ a prime. Then the torsion subgroups of $L^{p}(A)\otimes_{U}\mathbb{Z}$ and $M^{p}(A)\otimes_{U}\mathbb{Z}$ are isomorphic, and the homomorphism $\theta_{p}\otimes 1$ maps the latter isomorphically onto the former.

Proof.

By Lemma 3.1(i) both $L^{p}(A)\otimes_{U}\mathbb{Z}$ and $M^{p}(A)\otimes_{U}\mathbb{Z}$ are direct sums of a free abelian group and an elementary abelian $p$-group. Consider the maps in (2.6). By trivializing the $U$-action we obtain homomorphisms

[TABLE]

So the restriction of the composite $\theta_{k}\eta_{k}\otimes 1$ to the torsion subgroup of the tensor product $M^{p}(A)\otimes_{U}\mathbb{Z}$, an elementary abelian $p$-group, is multiplication by $(p-2)!$, that is, it is an isomorphism. It follows that the homomorphism $\theta_{p}\otimes 1$ maps the torsion subgroup of $M^{p}(A)\otimes_{U}\mathbb{Z}$ isomorphically into the torsion subgroup of $L^{p}(A)\otimes_{U}\mathbb{Z}$. To prove the proposition, we need to verify that this map is also surjective, that is, the homomorphism $\theta_{k}\otimes 1$ maps the torsion subgroup of $M^{p}(A)\otimes_{U}\mathbb{Z}$ isomorphically onto the torsion subgroup of $L^{p}(A)\otimes_{U}\mathbb{Z}$. But this is true since otherwise the restriction of $\eta_{p}\otimes 1$ to the torsion subgroup of $L^{p}(A)\otimes_{U}\mathbb{Z}$ would have a non-trivial kernel. This, however, is not the case, as follows from the exactness of (3.1). Since $B^{p}(A)\otimes_{U}\mathbb{Z}$ is free abelian by Lemma 3.2, and $H_{1}(M^{p}(A))$ is torsion by Lemma 3.1(ii), there cannot be any torsion elements in the kernel of $\eta_{p}\otimes 1$. This proves the proposition. ∎

In the next section we fill in the gap left in the proof of Lemma 3.2.

4. The degree $p$ Lie power in characteristic $p$

In this section $V$ denotes a vector space over a field $K$ of prime characteristic $p$. Moreover, we will assume that $V$ is a module for the polynomial ring $K[X]$ where $X$ is a finite set of indeterminates. Otherwise we will use all the notation introduced in Section 2, in particular, $L^{p}(V)$, $M^{p}(V)$ and $T^{p}(V)$ are the $p$th Lie, metabelian Lie, and tensor powers of $V$, respectively, $B^{p}(V)=L^{p}(V)\cap L(V)^{\prime\prime}$ is the kernel of the natural projection $L^{p}(V)\rightarrow M^{p}(V)$, and all of these will be regarded as $K[X]$-modules under the derivation action. Recall that $L^{p}(V)$ may be regarded as a submodule of the tensor power $T^{p}(V)$, and hence $B^{p}(V)$ is also a submodule of $T^{p}(V)$.

Lemma 4.1**.**

The submodule $B^{p}(V)$ is a direct summand of the $K[X]$-module $T^{p}(V)$.

This result is essentially proved as Theorem 3.1 in [6], except that there the module $V$ is assumed to be finite-dimensional. In what follows we reproduce the proof from [6] with some minor amendments necessary to accommodate infinite dimensional modules.

Proof.

For each $r\geq 1$ choose a basis ${\mathcal{B}}^{(r)}$ of $L^{r}(V)$ and let ${\mathcal{B}}=\bigcup_{r}{\mathcal{B}}^{(r)}$. Thus $\mathcal{B}$ is a basis of $L(V)$. For $b\in{\mathcal{B}}$, let ${\rm deg}(b)$ denote the degree of $b$, that is, ${\rm deg}(b)=r$ for $b\in{\mathcal{B}}^{(r)}$. Order $\mathcal{B}$ in any way subject to $b<b^{\prime}$ whenever ${\rm deg}(b)<{\rm deg}(b^{\prime})$. By the Poincaré–Birkhoff–Witt Theorem, $T^{p}(V)$ has a basis ${\mathcal{C}}$ consisting of all products of the form $b_{1}\otimes b_{2}\otimes\cdots\otimes b_{k}$ with $b_{1},\dots,b_{k}\in{\mathcal{B}}$, $b_{1}\leq b_{2}\leq\cdots\leq b_{k}$ and ${\rm deg}(b_{1})+\cdots+{\rm deg}(b_{k})=p$. More specifically, any basis element $c\in{\mathcal{C}}$ has the form

[TABLE]

where $k_{1},\dots,k_{p}$ are non-negative integers such that $k_{1}+2k_{2}+\cdots+pk_{p}=p$ and where, for $i=1,\dots,p$, we have $b_{1}^{(i)},\dots,b_{k_{i}}^{(i)}\in{\mathcal{B}}^{(i)}$ and $b_{1}^{(i)}\leq\cdots\leq b_{k_{i}}^{(i)}$. We call the $p$-tuple $(k_{1},\dots,k_{p})$ the type of $c$ and denote it by ${\rm type}(c)$. Let $\Omega$ denote the set of all such types. We order $\Omega$ (lexicographically) by $(k_{1},\dots,k_{p})>(k_{1}^{\prime},\dots,k_{p}^{\prime})$ if for some $j\in\{1,\dots,p\}$ we have $k_{i}=k_{i}^{\prime}$ for all $i<j$ but $k_{j}>k_{j}^{\prime}$. Using this ordering, write $\Omega=\{\omega_{1},\omega_{2},\dots,\omega_{m}\}$ where $\omega_{1}>\omega_{2}>\cdots>\omega_{m}$. Thus $\omega_{1}=(p,0,\dots,0)$ and $\omega_{m}=(0,\dots,0,1)$.

For $i=1,\dots,m$, define ${\mathcal{C}}_{i}=\{c\in{\mathcal{C}}:{\rm type}(c)=\omega_{i}\}$ and let $X_{i}$ denote the subspace of $T^{p}(V)$ spanned by ${\mathcal{C}}_{i}\cup{\mathcal{C}}_{i+1}\cup\cdots\cup{\mathcal{C}}_{m}$. Also, write $X_{m+1}=0$. Thus

[TABLE]

Note that $X_{i}/X_{i+1}$ has basis ${\mathcal{C}}_{i}$ modulo $X_{i+1}$. Furthermore, ${\mathcal{C}}_{1}$ consists of all products $b_{1}^{(1)}\otimes b_{2}^{(1)}\otimes\cdots\otimes b_{p}^{(1)}$ with $b_{1}^{(1)},\dots,b_{p}^{(1)}\in{\mathcal{B}}^{(1)}$ and $b_{1}^{(1)}\leq\cdots\leq b_{p}^{(1)}$. Also, ${\mathcal{C}}_{m}={\mathcal{B}}^{(p)}$ and $X_{m}=L^{p}(V)$.

Let $i\in\{1,\dots,m\}$ where $\omega_{i}=(k_{1},\dots,k_{p})$. For $c\in{\mathcal{C}}_{i}$ written as in (4.1) it is well known and easy to verify that the value of $c$ modulo $X_{i+1}$ is unchanged by any permutation of the factors $b_{1}^{(1)},\dots,b_{k_{p}}^{(p)}$. In particular, for all $\pi_{1}\in{\rm Sym}(k_{1})$, …, $\pi_{p}\in{\rm Sym}(k_{p})$, we have

[TABLE]

It follows easily that $X_{i}$ is a $K[X]$-submodule of $T^{p}(V)$. For $c$ written as before let $\overline{c}\in S^{k_{1}}(L^{1}(V))\otimes\cdots\otimes S^{k_{p}}(L^{p}(V))$ be defined by

[TABLE]

Clearly $\{\overline{c}:c\in{\mathcal{C}}_{i}\}$ is a basis of $S^{k_{1}}(L^{1}(V))\otimes\cdots\otimes S^{k_{p}}(L^{p}(V))$. Furthermore, it follows easily from (4.2) that the linear map given by $c+X_{i+1}\mapsto\overline{c}$ is a $K[X]$-module isomorphism from $X_{i}/X_{i+1}$ to $S^{k_{1}}(L^{1}(V))\otimes\cdots\otimes S^{k_{p}}(L^{p}(V))$. Thus

[TABLE]

Suppose that $i\in\{2,\dots,m-1\}$ where $\omega_{i}=(k_{1},\dots,k_{p})$. Thus $k_{1},\dots,k_{p}<p$ (and, in fact, $k_{p}=0$). Let $\sigma_{i}:S^{k_{1}}(L^{1}(V))\otimes\cdots\otimes S^{k_{p}}(L^{p}(V))\to T^{p}(V)$ be the linear map defined on the basis $\{\overline{c}:c\in{\mathcal{C}}_{i}\}$ by

[TABLE]

where $c$ is written as in (4.1) It is straightforward to verify that $\sigma_{i}$ is a $K[X]$-module homomorphism. It follows easily from (4.2) that $\sigma_{i}(\overline{c})\in X_{i}$ and, indeed, $\sigma_{i}(\overline{c})+X_{i+1}=c+X_{i+1}$ for all $c\in{\mathcal{C}}_{i}$. Thus the map $\sigma_{i}$ is injective and we have $X_{i}={\rm Im}(\sigma_{i})\oplus X_{i+1}$. Since $X_{m}=L^{p}(V)$ we therefore have

[TABLE]

Let $W$ be the submodule of $X_{2}$ given by

[TABLE]

Consider the maps

[TABLE]

defined by

[TABLE]

and

[TABLE]

where $a_{i}\in V$ and $\pi$ runs over all permutations of $\{2,3,\dots,c\}$. Then $\alpha$ is surjective and the composite $\beta\alpha$ is the identity map on $V\otimes V^{p-1}$. Hence we have a direct decomposition

[TABLE]

We claim that ${\rm Ker}(\alpha)=W$. It is easily seen that $W$ is contained in $\rm Ker(\alpha)$. To verify that we have actually equality, note that the elements

[TABLE]

form a basis of $T^{p}(V)$ modulo $X_{2}$. Furthermore, the Lie products

[TABLE]

form a basis of $L^{p}(V)$ modulo $L^{p}(V)\cap L(V)^{\prime\prime}$ (see Section 2). It follows that the elements (4.5) together with the elements (4.6) form a basis of $T^{c}(V)$ modulo $W$. Moreover, the images of these elements under the map $\alpha$ form a basis of $V\otimes V^{p-1}$. Indeed, we have

[TABLE]

and

[TABLE]

This can easily be seen from the short exact sequence (2.3) (with $V$ instead of $A$). Indeed, the elements (4.7) are mapped by $\kappa_{p}$ one-to-one onto the canonical basis of $V^{p}$, and the elements (4.8) are precisely the images of the canonical basis elements of $M^{p}(V)$ under the map $\mu_{p}$. Consequently, we have the desired equality ${\rm Ker(\alpha)}=W$, and so $T^{p}(V)=W\oplus{\rm Im}(\beta)$. By (4.4), $L^{p}(V)\cap L(V)^{\prime\prime}$ is a direct summand of $W$. Thus $L^{p}(V)\cap L(V)^{\prime\prime}$ is a direct summand of $T^{p}(V)$ and we have Lemma 4.1. ∎

Now the result we need to complete the proof of Proposition 3.3 follows easily.

Corollary 4.2**.**

If $V$ is a free $K[X]$-module, then $B^{p}(V)$ is a projective $K[X]$-module.

Proof.

If $V$ is a free $K[X]$-module, then the tensor power $T^{p}(V)$ is also a free $K[X]$-module, see [14, Lemma 5.2]). Since $B^{p}(V)$ is a direct summand of $T^{p}(V)$, it is projective. ∎

5. The main result

In this Section $L$ denotes a free Lie ring of finite rank with free generating set $X$. Our aim is to determine the torsion subgroup of the additive group of the quotient (1.1). In view of the short exact sequence

[TABLE]

this is a free central extension of the free (nilpotent-of-class-c-1)-by-abelian Lie ring $L/(L^{\prime})^{c}$. The additive structure of the latter is well-understood. Its underlying abelian group is free abelian [3]. Consequently, any torsion elements must be contained in the central quotient $(L^{\prime})^{c}/[(L^{\prime})^{c},L]$, and it is this quotient we will focus on from now on. By the Shirshov-Witt Theorem, the derived ideal $L^{\prime}$ is itself a free Lie ring, namely, the free Lie ring on $L^{\prime}/L^{\prime\prime}$: $L^{\prime}=L(L^{\prime}/L^{\prime\prime})$. This is a graded Lie ring and its degree $c$ homogeneous component $L^{c}(L^{\prime}/L^{\prime\prime})$ is isomorphic to the lower central quotient

[TABLE]

The adjoint representation induces on these lower central quotients the structure of an $L/L^{\prime}$-module, and hence of a module for its universal envelope $U=U(L/L^{\prime})$. The latter may be identified with the polynomial ring on $X$: $U=\mathbb{Z}[X]$. Thus (5.1) is actually a $U$-module isomorphism. In view of the canonical isomorphism

[TABLE]

trivializing the $U$-action on both sides of (4.1) gives an isomorphism

[TABLE]

We will exploit this isomorphism to investigate the additive structure of the quotient on the left hand side by examining the tensor product on the right hand side.

Suppose that $L$ has rank $2$ and, say, $X=\{x,y\}$. Then $L^{\prime}/L^{\prime\prime}$ is a free cyclic module over the polynomial ring $U=\mathbb{Z}[x,y]$ with free generator $[y,x]$, see [1, Proof of Theorem 6.1]. If $c$ is a prime, say $c=p$, then Proposition 3.3 applies to the tensor product on the right hand side of (5.2). Hence this tensor product is a direct sum of a free abelian group and an elementary abelian $p$-group. Moreover, the torsion subgroup is the image in $L^{p}(L^{\prime}/L^{\prime\prime})\otimes_{U}\mathbb{Z}$ of the torsion subgroup of $M^{p}(L^{\prime}/L^{\prime\prime})\otimes_{U}\mathbb{Z}$ under the map $\theta_{p}\otimes 1$. A complete description of the latter is given in [1, Corollary 6.2]. The elements

[TABLE]

where $u=[y,x,\underbrace{x,\dots,x}_{s},\underbrace{y,\dots,y}_{t}]$ with $s,t\geqslant 0$ form a basis of this torsion subgroup as a $\mathbb{Z}_{p}$-module. Applying $\theta_{p}\otimes 1$ to such a basis element gives

[TABLE]

Since this is an element of order $p$ we can drop the factor of $(p-2)!$ in the statement of our main result, which summarizes the discussion in this final section.

Theorem 5.1**.**

Let $L$ be a free Lie ring of rank $2$ with free generators $x$ and $y$, and let $p$ be a prime. Then the quotient $(L^{\prime})^{p}/[(L^{\prime})^{p},L]$ is a direct sum of a free abelian group and an elementary abelian $p$-group. Modulo $[(L^{\prime})^{p},L]$ the elements

[TABLE]

where $u=[y,x,\underbrace{x,\dots,x}_{s},\underbrace{y,\dots,y}_{t}]$ with $s,t\geqslant 0$ form a basis of this torsion subgroup as a $\mathbb{Z}_{p}$-module. ∎

The legality of the factor $1/p$ in the statement of the theorem is explained in Section 2.