Remark

The above proof shows that $\exp(\hat{\mathbf{L}})$ is the subgroup $\Delta$ of $\Gamma$ consisting of the $b$ satisfying (4).

Hence the group law of $\Delta$ can be transported by the exponential mapping to $\hat{\mathbf{L}}$. In other words, $\hat{\mathbf{L}}$ is a complete topological group with the law of composition $(a, b) \mapsto a \mathbin{\mathbf{h}} b$ given by

The topological group thus obtained is called the Hausdorff group (derived from X relative to K).

Let $g$ be the homomorphism of the free group $F = F(X)$ into $\Gamma$ such that $g(x) = \exp x$ for $x \in X$. As $\exp x - 1 - x = \sum_{n \geq 2} x^n/n!$ is of order $\geq 2$, $g$ is injective by Theorem 1 of § 5, no. 3. Therefore the mapping $\log \circ g$ is an injective homomorphism of $F$ into the Hausdorff group which extends the canonical injection $X \to \hat{\mathbf{L}}$.

For every integer $m \geq 1$ we denote by $\hat{\mathbf{L}}_m$ the set of elements of order $\geq m$ in $\hat{\mathbf{L}}$ and by $\Gamma_m$ the set of $u \in \Gamma$ such that $u - 1$ is of order $\geq m$. Then $\hat{\mathbf{L}}_m = \exp(\Gamma_m)$ by Remark 2 of no. 1; as $(\Gamma_m)_{m \geq 1}$ is an integral central filtration on $\Gamma$ (§ 4, no. 5 Proposition 2), $(\hat{\mathbf{L}}_m)_{m \geq 1}$ is an integral central filtration on the group $\hat{\mathbf{L}}$.