Proposition 2

The set $\Gamma$ is a subgroup of $A^*$ and $(\Gamma_\alpha)$ is a central filtration on $\Gamma$.

$\Gamma = \bigcup_{\alpha > 0} \Gamma_\alpha$ by construction and the relation $\Gamma_\alpha = \bigcap_{\beta < \alpha} \Gamma_\beta$ follows from $A_\alpha = \bigcap_{\beta < \alpha} A_\beta$.

We show that $\Gamma_\alpha$ is a subgroup of $A^*$. Now $1 \in \Gamma_\alpha$; let $x, y$ be in $\Gamma_\alpha$, whence $x - 1 \in A_\alpha, y - 1 \in A_\alpha$. As $A_\alpha$ is a two-sided ideal of $A$, the formulae

imply $xy - 1 \in A_\alpha$ and $x^{-1} - 1 \in A_\alpha$, whence $xy \in \Gamma_\alpha$ and $x^{-1} \in \Gamma_\alpha$.

As $\Gamma = \bigcup_{\alpha > 0} \Gamma_\alpha$, this is a subgroup of $A^*$.

Finally let $\alpha > 0, \beta > 0, x \in \Gamma_\alpha$ and $y \in \Gamma_\beta$. Let $x - 1 = \xi$ and $y - 1 = \eta$. Then

by hypothesis, $\xi \in A_\alpha$ and $\eta \in A_\beta$, whence $\xi \eta - \eta \xi \in A_{\alpha + \beta}$. As $A_{\alpha + \beta}$ is a two-sided ideal of $A$, $(x, y) - 1 \in A_{\alpha + \beta}$, whence $(x, y) \in \Gamma_{\alpha + \beta}$.