# open subgroup scheme closed

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).

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 $H$.

Edit: If $G/H$ is representable by a group scheme, $H$ is normal in $G$ and everything is separated, apply Exercise 1(ii) of http://math.stanford.edu/~conrad/papers/gpschemehw1.pdf to $\pi: G \to G/H$.

Edit 2: But I want to prove this without the representability hypothesis on $G/H$.