Let $X$ be a group scheme over $S$, $g\in X(S)$. Then we can define the translation by point $g$. $l_g:X\to X$, $l_g=m\circ (id,g\circ *)$, where $*:X\to S $. And the morphism acts on $X(T)$ as translation. The following lemma is from stackproject https://stacks.math.columbia.edu/tag/047J
Lemma 39.7.7. Let $k$ be a field. Let $ψ:G′→G$ be a morphism of group schemes over $k$. If $ψ(G′)$ is open in $G$, then $ψ(G′)$ is closed in $G$.

Proof. Let $U=ψ(G′)⊂G$. Let $Z=G \backslash ψ(G′)=G\backslash U$ with the reduced induced closed subscheme structure. By Lemma 39.7.2 the image of

$Z×_kG′⟶Z×_kU⟶G$ is open (the first arrow is surjective). On the other hand, since $ψ$ is a homomorphism of group schemes, the image of $Z×_kG′→G$ is contained in $Z$ (because translation by $ψ(g′)$ preserves $U$ for all points $g′$ of $G′$; small detail omitted). Hence $Z⊂G $is an open subset (although not necessarily an open subscheme). Thus $U=ψ(G′)$ is closed.

Here it seems that it has defined tranlation by any point in $U$, but I have no idea how to do that.

Let $X$ be a group scheme over $S$, $g\in X(S)$. Then we can define the translation by point $g$. $l_g:X\to X$, $l_g=m\circ (id,g\circ *)$, where $*:X\to S $. And the morphism acts on $X(T)$ as translation. The following lemma is from stackproject https://stacks.math.columbia.edu/tag/047J
Lemma 39.7.7. Let $k$ be a field. Let $ψ:G′→G$ be a morphism of group schemes over $k$. If $ψ(G′)$ is open in $G$, then $ψ(G′)$ is closed in $G$.

Proof. Let $U=ψ(G′)⊂G$. Let $Z=G \backslash ψ(G′)=G\backslash U$ with the reduced induced closed subscheme structure. By Lemma 39.7.2 the image of

$Z×_kG′⟶Z×_kU⟶G$ is open (the first arrow is surjective). On the other hand, since $ψ$ is a homomorphism of group schemes, the image of $Z×_kG′→G$ is contained in $Z$ (because translation by $ψ(g′)$ preserves $U$ for all points $g′$ of $G′$; small detail omitted). Hence $Z⊂G $is an open subset (although not necessarily an open subscheme). Thus $U=ψ(G′)$ is closed.

Here it seems that it has defined tranlation by any point in $U$, but I have no idea how to do that. Basically I want to know why the image of $(G\backslash U)\times_k U\to G$ is contained in $G\backslash U$, where $U$ is open subgroup of $G$.