TITLE: open subgroup scheme closed
QUESTION [6 upvotes]: 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?

REPLY [3 votes]: For $k=\mathbb{C}$ (or any field of characteristic $\ne2$) let $G$ be the subgroup scheme of $SL(2)\times\mathbb{A}^1$ consisting of all matrices of the form $\begin{pmatrix}x&sy\\y&x\end{pmatrix}$. The defining equation is $x^2-sy^2=1$ which shows that $G$ is an irreducible variety. For $s\ne0$ the fiber is a torus, hence connected. The fiber over $s=0$ is $\mathbf{G}_a\times\{\pm1\}$, hence disconnected. Thus $G^0$ is $G$ minus $\{s=0,x=-1\}$, hence open but not closed.