Let $G/S$ be a group scheme and $H \leq G$ an open subgroup scheme. Is $H \subseteq G$ closed? I want to apply this to $G^0 \leq G$ (see SGA 3, VI_B, Théorème 3.10) for $G$ commutative.

(*) If $S = \mathrm{Spec}(K)$, this is proven in http://jmilne.org/math/CourseNotes/iAG200.pdf Proposition 1.27 by the usual argument: the complement is the disjoint open union of the cosets of $H$.

If $G/H$ is representable by a group scheme and everything is separated, apply Exercise 1(ii) of http://math.stanford.edu/~conrad/papers/gpschemehw1.pdf to $\pi: G \to G/H$. But I want to prove this *without* having to assume $G/H$ being representable. It seems that if $H \subseteq G$ is closed, $G/H$ is representable (under certain conditions) ...

(This is a cross-post from https://math.stackexchange.com/questions/1670288/open-subgroup-scheme-closed)

Edit: Assume $S$ the spectrum of a DVR. Then, by (*), $G^0_\eta \subseteq G_\eta$ and $G^0_s \subseteq G_s$ are closed. Now, if everything is separated and flat, does it follow that $G^0 \subseteq G$ is closed?