"skyscraper group scheme" Is there a skyscraper group scheme?
Let $S$ be a DVR. Is there a group scheme $\mathcal{G}$ over $S$ which is generically {1} trivial i.e, identity group, but at the closed point some nontrivial group $G$?
For example: Let $C$ be a curve over $S$ whose generic fibre is smooth of genus $g$ with no automorphism but the closed fibre is a semistable curve(which has nontrivial automorphisms). Now consider the group of $S$-automorphisms of this curve $C$ over $S$? My question: Does this curve C over S has any non trivial S-automorphism?
 A: Yes: the following examplpe is quite different from the example by Ariyan, although the generic and closed fibers are the same as his. 
Start with the constant group scheme $A:=(\mathbb{Z}/2\mathbb{Z})_S$, and consider the closed subscheme which is the union of the zero section and the closed point of the other section. This is immediately seen to be a subgroup scheme of $A$, with generic fiber $0$ and closed fiber $\mathbb{Z}/2\mathbb{Z}$. It is finite over $S$ (in particular separated) but of course not flat. 
You can construct infinitely many variants by replacing the closed point by any infinitesimal neighborhood of it.
If you take a stable curve $C$ over $S$, it it known that its automorphis functor $\mathscr{G}:={\underline{\mathrm{Aut}}}(C/S)$ is a finite unramified $S$-group scheme, hence it "looks like" the above example, rather than Ariyan's nonseparated one. In particular ($K$ being the fraction field), the restriction homomorphism $\mathscr{G}(S)\to \mathscr{G}(K)$ is bijective, so if $C_K$ has no $K$-automorphisms then if $C$ has no $S$-automorphisms.  
A: Yes.
Let $\mathbb{A}^{1,2}$ be the affine line with a double origin. Consider the natural map $\mathbb{A}^{1,2}\to \mathbb{A}^1$. (To define this map, let $0_1$ and $0_2$ be the origins in $\mathbb{A}^{1,2}$. The above map sends any $x\neq 0_1, 0_2$ to $x$. It sends $0_1$ and $0_2$ to the origin in $\mathbb{A}^1$.)
The above morphism realizes $\mathbb{A}^{1,2}$ as a quasi-finite flat group scheme over $\mathbb{A}^1$. (It is not  a separated group scheme over $\mathbb{A}^1$, of course.)
Note that the generic fibre of this group scheme is the trivial group. The fibre over the origin is the group $\mathbb{Z}/2\mathbb{Z}$.
To get to the situation you desire, let $\mathrm{Spec} \mathbb{C}[[t]]\to \mathbb{A}^{1}$ be a dominant map whose image contains the origin. Now base-change the group scheme $\mathbb{A}^{1,2}\to \mathbb{A}^1$ along this morphism to get a group scheme 
$$ G\to \mathrm{Spec} \mathbb{C}[[t]]$$ with trivial generic fibre and a non-trivial special fibre. (Here $G=\mathbb{A}^{1,2}\times_{\mathbb{A}^1} \mathrm{Spec} \mathbb{C}[[t]]$.)
