**Proof.**
For any point $g \in G$ there exists a field extension $k \subset K$ and a $K$-valued point $g' \in G(K)$ mapping to $g$. If we think of $g'$ as a $K$-rational point of the group scheme $G_ K$, then we see that $\mathcal{O}_{G, g} \to \mathcal{O}_{G_ K, g'}$ is a faithfully flat local ring map (as $G_ K \to G$ is flat, and a local flat ring map is faithfully flat, see Algebra, Lemma 10.39.17). The result for $\mathcal{O}_{G_ K, g'}$ implies the result for $\mathcal{O}_{G, g}$, see Algebra, Lemma 10.30.5. Hence in order to prove (1) it suffices to prove it for $k$-rational points $g$ of $G$. In this case translation by $g$ defines an automorphism $G \to G$ which maps $e$ to $g$. Hence $\mathcal{O}_{G, g} \cong \mathcal{O}_{G, e}$. In this way we see that (2) implies (1), since irreducible components passing through $e$ correspond one to one with minimal prime ideals of $\mathcal{O}_{G, e}$.

In order to prove (2) and (3) it suffices to prove (2) when $k$ is algebraically closed. In this case, let $Z_1$, $Z_2$ be two irreducible components of $G$ passing through $e$. Since $k$ is algebraically closed the closed subscheme $Z_1 \times _ k Z_2 \subset G \times _ k G$ is irreducible too, see Varieties, Lemma 33.8.4. Hence $m(Z_1 \times _ k Z_2)$ is contained in an irreducible component of $G$. On the other hand it contains $Z_1$ and $Z_2$ since $m|_{e \times G} = \text{id}_ G$ and $m|_{G \times e} = \text{id}_ G$. We conclude $Z_1 = Z_2$ as desired.
$\square$

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