# Why $G\to G/H$ is faithfully flat?

Let $G$ be an affine algebraic group over algebraically closed field $k$. Questions: 1. $G$ is faithfully flat since it is defined over field? 2. Let $H$ be a closed subgroup of $G$, then (as I learnt from some paper) the map $\pi\colon G\to G/H$ is faithfully flat, why? reference? When it is locally trivial, especially when $G$ is linear algebraic group?
If $k$ is a field, then any non-zero $k$-vector space whatsoever is faithfully flat. In particular, any non-trivial $k$-algebra is faithfully flat. – Zhen Lin Jul 20 '12 at 6:52
Yes, its faithfully flat over $k$. I think this is the faithfully flat means in question 1. But it is not obvious that $k[G]$ is faithfully flat over $k[G]^H$ in question 2. Thank you for your reply! – Hoxide Jul 20 '12 at 6:55