Lemma 2

For every integer $p \geq 0$, the module $L_p$ admits the family $(\bar{w}_i)_{0 \leq i < p}$ as basis, the Lie algebra $g_p$ admits $(\bar{u})_{u \in P_p}$ as basic family and the module $L(X)$ is the direct sum of $L_p$ and $g_p$.

$L_0 = \{0\}$ and $g_0 = L(X)$ and the lemma is true for $p = 0$. We argue by induction on $p$. Suppose then that the lemma is true for some integer $p \geq 0$. Let $u_i, w = (\mathrm{ad}\ \bar{w}_p)^i.\bar{w} = \Psi(w_p^i w)$ for $i \geq 0, w \in P_p, w \neq w_p$. By the Corollary to Proposition 10 of no. 9, the free Lie algebra $g_p$ is the direct sum of the module $T_p$ of basis $\{\bar{w}_p\}$ and a Lie subalgebra $h_p$ admitting

as basic family. By Proposition 13 (d), the family $(\bar{u})_{u \in P_{p+1}}$ is equal to $\mathscr{F}$ and hence is a basic family of $h_p = g_{p+1}$. Hence $L(X) = L_p \oplus T_p \oplus g_{p+1}$ and, as $L_{p+1} = L_p + T_p$, $L(X) = L_{p+1} \oplus g_{p+1}$ and $(\bar{w}_0, \bar{w}_1, \ldots, \bar{w}_{p-1}, \bar{w}_p)$ is a basis of the module $L_{p+1}$.

Let $n$ be a positive integer. By Proposition 13 (c) there exists an integer $p(n)$ such that $P_p$ has only elements of length $> n$ for $p \geq p(n)$. For $p \geq p(n)$, the Lie subalgebra $g_p$ of $L(X)$ is generated by elements of degree $> n$ and hence $L^n(X) \cap g_p = \{0\}$. On the other hand, the elements $\bar{w}_i$ of $L(X)$ are homogeneous and the family $(w_i)_{0 \leq i < p}$ is a basis of a supplementary module of $g_p$. It follows immediately that the family of elements $\bar{w}_i$ of degree $n$ is a basis of the module $L^n(X)$ and that the sequence $(\bar{w}_i)_{i \geq 0}$ is a basis of the module $L(X)$.

(B) General case.

If $S$ is a subset of $X$, recall that $M(S)$ is identified with the submagna of $M(X)$ generated by $S$ and $L(S)$ is identified with the Lie subalgebra of $L(X)$ generated by $S$; we have seen that if $w \in M(S)$ is of length $\geq 2$ then $\alpha(w) \in M(S)$ and $\beta(w) \in M(S)$. It follows immediately that $H \cap M(S)$ is a Hall set relative to $S$.

For every finite subset $\Phi$ of $H$ there exists a finite subset $S$ of $X$ such that

Φ ⊂ M(S). Case (A) then shows that the elements $\bar{w}$ with $w \in \Phi$ are linearly independent in $L(S)$ and hence in $L(X)$. Therefore the family $(\bar{w})_{w \in H}$ is free.

For every element $a$ of $L(X)$ there exists a finite subset $S$ of $X$ such that $a \in L(S)$. By case (A), the subset $\Psi'(H \cap M(S))$ of $\Psi'(H)$ generates the module $L(S)$ and hence $a$ is a linear combination of elements of $\Psi'(H)$. Hence $\Psi'(H)$ generates the module $L(X)$, which completes the proof.