Quotient of algebraic group by subgroup with trivial character group Let $G$ be a linear algebraic group and $H$ be a closed subgroup. Suppose that all homomorphisms of algebraic groups $H\to\mathbb{G}_m$ are trivial. How to prove that $G/H$ is quasi-affine?

Here is a proof using the setup of Springer. 
Let $V$ be a $G$-representation and $W\subset V$ be  a line such that $H=\{g\in G\mid gW\subseteq W\}$ and $\mathfrak{h}=\{X\in\mathfrak{g}\mid XW\subseteq W\}$. Then $H$ acts (morphically) on $W$, inducing a morphism $H\to\mathbb{G}_m$, which is trivial by hypothesis. Therefore $H$ fixes every $w\in W$. In fact, as $W$ is one-dimensional, fixing one point is equivalent to fixing all points, so $H=\{g\in G\mid gw=w~\forall w\in W\}$. Fix $w\in W$. By essentially Chavalley's theorem, $Gx$ is open in its closure, hence quasi-affine. Let $o_w$ be the orbit map. Then (the first map is the obvious closed immersion)
$$
H\to G\overset{o_w}{\to}X,
$$
is constant, so has derivative zero. Thus 
$$
\mathfrak{h}\subseteq\{X\in\mathfrak{g}\mid Xw=0~\forall w\in W\}\subseteq\{X\in \mathfrak{g}\mid XW\subset W\}=\mathfrak{h},
$$
whence equality. Therefore $\mathfrak{h}=\ker(do_w)$, so
$$
0\to\mathfrak{h}\to\mathfrak{g}\to T_wX\to0
$$
is exact. (If you are in characteristic zero, this is superfluous.)
Therefore the orbit map is a bijection on both the group (obviously) and Lie algebra levels. It is then an isomorphism by theorem 5.3.2 in Springer.
