Lemma 1

For an element a of $\hat{A}(X)$ to be invertible, it is necessary and sufficient that its constant term be invertible in K.

If $a$ is invertible in $\hat{\mathbf{A}}(\mathbf{X})$, $\varepsilon(a)$ is invertible in $K$. Conversely, if $\varepsilon(a)$ is invertible in $K$, there exists $u \in \hat{\mathbf{A}}_1(\mathbf{X})$ such that $a = \varepsilon(a)(1-u)$; we write $b = \left( \sum_{n \geq 0} u^n \right) \varepsilon(a)^{-1}$. Then $ab = ba = 1$ and $a$ is invertible.

The set of elements of $\hat{\mathbf{A}}(\mathbf{X})$ of constant term 1 is therefore a subgroup of the multiplicative monoid $\hat{\mathbf{A}}(\mathbf{X})$, called the Magnus group constructed on $\mathbf{X}$ (relative to $K$). In this chapter it will be denoted by $\Gamma(\mathbf{X})$ or simply $\Gamma$. For every integer $n \geq 1$, we denote by $\Gamma_n$ the set of $a \in \Gamma$ such that $\omega(a-1) \geq n$. By Proposition 2 of § 4, no. 5, the sequence $(\Gamma_n)_{n \geq 1}$ is an integral central filtration on $\Gamma$.