**Proof.**
We have the corresponding result for topological spaces, see Lemma 5.22.2. Combined with Lemma 5.30.3 we see that it suffices to prove that (1) implies (3).

We first prove that every neighbourhood $E$ of the neutral element $e$ contains an open subgroup. Namely, since $G$ is the cofiltered limit of finite discrete topological spaces (Lemma 5.22.2), we can choose a continuous map $f : G \to T$ to a finite discrete space $T$ such that $f^{-1}(f(\{ e\} )) \subset E$. Consider

\[ H = \{ g \in G \mid f(gg') = f(g')\text{ for all }g' \in G\} \]

This is a subgroup of $G$ and contained in $E$. Thus it suffices to show that $H$ is open. Pick $t \in T$ and set $W = f^{-1}(\{ t\} )$. Observe that $W \subset G$ is open and closed, in particular quasi-compact. For each $w \in W$ there exist open neighbourhoods $e \in U_ w \subset G$ and $w \in U'_ w \subset W$ such that $U_ wU'_ w \subset W$. By quasi-compactness we can find $w_1, \ldots , w_ n$ such that $W = \bigcup U'_{w_ i}$. Then $U_ t = U_{w_1} \cap \ldots \cap U_{w_ n}$ is an open neighbourhood of $e$ such that $f(gw) = t$ for all $w \in W$. Since $T$ is finite we see that $\bigcap _{t \in T} U_ t \subset H$ is an open neighbourhood of $e$. Since $H \subset G$ is a subgroup it follows that $H$ is open.

Suppose that $H \subset G$ is an open subgroup. Since $G$ is quasi-compact we see that the index of $H$ in $G$ is finite. Say $G = Hg_1 \cup \ldots \cup Hg_ n$. Then $N = \bigcap _{i = 1, \ldots , n} g_ iHg_ i^{-1}$ is an open normal subgroup contained in $H$. Since $N$ also has finite index we see that $G \to G/N$ is a surjection to a finite discrete topological group.

Consider the map

\[ G \longrightarrow \mathop{\mathrm{lim}}\nolimits _{N \subset G\text{ open and normal}} G/N \]

We claim that this map is an isomorphism of topological groups. This finishes the proof of the lemma as the limit on the right is cofiltered (the intersection of two open normal subgroups is open and normal). The map is continuous as each $G \to G/N$ is continuous. The map is injective as $G$ is Hausdorff and every neighbourhood of $e$ contains an $N$ by the arguments above. The map is surjective by Lemma 5.12.6. By Lemma 5.17.8 the map is a homeomorphism.
$\square$

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