Lemma 2

Let $G$ be a Hausdorff topological group and let $H$ be a compact normal subgroup of $G$, and $\varphi$ the canonical mapping $G \to G/H$. If $C$ is the identity component of $G$, then $\varphi(C)$ is the identity component of $G' = G/H$.

Once this lemma is established, we shall have $f_\alpha(C) = C_\alpha$ for each $\alpha \in I$, and since $C$ is a closed subgroup of $G$ (§ 2, no. 2, Proposition 7) it suffices to apply Proposition 3 a).

To prove the lemma, notice first that if $C'$ is the component of the identity element $e'$ in $G'$, then $\varphi(C) \subset C'$ since $\varphi(C)$ is connected. Suppose that $\varphi(C) \neq C'$. Since $C$ is a closed normal subgroup of $G$ (§ 2, no. 2, Proposition 7), $\varphi(C)$ is a normal subgroup of $G'$; if $\psi$ is the canonical mapping $G' \to G'/\varphi(C)$, then $\psi(C')$ is connected and does not consist of the identity element alone; hence the identity component of $G'/\varphi(C)$ does not consist of the identity element alone. But $G'/\varphi(C)$ is isomorphic to $(G/H)/(HC/H)$, hence to $G/HC$, and consequently also to $(G/C)/(HC/C)$ (§ 2, no. 7, Corollary to Proposition 22, and Proposition 20). Now, $G/C$ is Hausdorff and totally disconnected (Chapter I, § 11, no. 5, Proposition 9), and $HC/C$, being the canonical image of the compact normal subgroup $H$ of $G$, is a compact subgroup of $G/C$. It is therefore sufficient to prove Lemma 2 under the additional hypothesis that $G$ is totally disconnected, i.e. $C = \{e\}$.

Suppose then that $C' \neq \{e'\}$; replacing $G$ by its subgroup $\overline{f'}(C')$, which is totally disconnected and contains $H$, we may suppose that $G'$ is connected and does not consist of a single point.

Let $\mathfrak{M}$ be the set of closed subgroups $L$ of $G$ such that $LH = G$. We shall show that the set $\mathfrak{M}$, ordered by the relation $\supset$, is inductive. Indeed, if $\mathfrak{T}$ is a linearly ordered subset of $\mathfrak{M}$, then for each $x \in G$ the set of sets $xH \cap L$ with $L \in \mathfrak{T}$ is a filter base formed of closed sets in the compact space $xH$; hence the intersection of these sets is not empty, which shows that the intersection of the subgroups $L \in \mathfrak{T}$ belongs to $\mathfrak{M}$. Applying Zorn's lemma, we see therefore that $\mathfrak{M}$ has a minimal element $L_0$. Since $H$ is compact, $G/H = L_0H/H$ is isomorphic to $L_0/(L_0 \cap H)$ (§ 4, no. 1, Proposition 1, Corollary 3); since $L_0$ is totally disconnected and $L_0 \cap H$ is compact, we see that we may replace $G$ by $L_0$; in other words, we may suppose in addition that there is no closed subgroup $L \neq G$ such that $LH = G$.

Now let $F$ be the intersection of the neighbourhoods of the identity element which are both open and closed in $G$, and let us show that $F$ is a closed subgroup of $G$. Clearly $F$ is closed in $G$; hence it is enough to show that $F^{-1}F \subset F$. But if $x \in F$ and if $V$ is an open and closed neighbourhood of $e$ in $G$, then so is $xV$; for otherwise $e$ would belong to the complement $W$ of $xV$ in $G$, which is again open and closed, and we should have $x \notin W$; hence $x \notin F$, contrary to hypothesis. It follows that $xF$, the intersection of the sets $xV$ as $V$ runs through the open and closed neighbourhoods of $e$ in $G$, contains $F$; in other words, $x^{-1}F \subset F$, which proves our assertion. Since $G$ is totally disconnected and $G \neq \{e\}$, we have $F \neq G$. But if $V$ is an open and closed neighbourhood of $e$ in $G$, then $VH$ is also open and closed in $G$ (§ 4, no. 1, Corollary 1 to Proposition 1), hence $\varphi(V)$ is both open and closed in $G/H$, and this implies that $\varphi(V) = G/H$ by virtue of the hypothesis. We shall conclude from this that $FH = G$, which will give us the desired contradiction and complete the proof of the lemma. Indeed, for each $x \in G$, $xH$ meets every neighbourhood $V$ of $e$ which is both open and closed, hence also meets the intersection $F$ of these neighbourhoods, since the sets $Vx \cap H$ form a filter base consisting of closed sets in the compact space $xH$.

Q.E.D.