Proposition 1

If $H$ is a stable subgroup of a topological group $G$ with operators, then the closure $\overline{H}$ of $H$ in $G$ is a stable subgroup of $G$.

We know already (§ 2, no. 1, Proposition 1) that $\overline{H}$ is a subgroup of $G$. Also if $\alpha$ is any operator on $G$, the image of $H$ under the continuous mapping $x \to x^\alpha$ is contained in $H$, and therefore the image of $\overline{H}$ is contained in $\overline{H}$ (Chapter I, § 2, no. 1, Theorem 1).

Let $H$ be a stable normal subgroup of a topological group $G$ with operators. Then for each operator $\alpha$ on $G$, the mapping of $G/H$ into itself induced by $x \to x^\alpha$ is continuous (§ 2, no. 8, Remark 3), and the structure of group with operators on $G/H$ is therefore compatible with the quotient topology on $G/H$.

Let $(G_i)_{i \in I}$ be a family of topological groups with operators, where each $G_i$ is assumed to have the same operator set $\Omega$. For each $\alpha \in \Omega$, the mapping $x \to ((\mathrm{pr}_i x)^\alpha)$ of $G = \prod_{i \in I} G_i$ into itself is continuous (Chapter I, § 4, no. 1, Proposition 1), and the structure of groups with operators on $G$ is therefore compatible with the product topology on $G$.

If $G$ is a Hausdorff topological group with operators and if $G$ has a completion $\hat{G}$ (§ 3, no. 4), then every endomorphism $x \to x^\alpha$ of $G$ defined by an operator on $G$ can be extended by continuity to an endomorphism of $\hat{G}$ (§ 3, no. 3, Proposition 5). Hence $\hat{G}$ has the structure of a topological group with operators, and the operator set is the same for $\hat{G}$ as for $G$.