Orbits of algebraic groups (dimension of connected components) Let $X$ be an algebraic variety with algebraic group $G$ acting on it. Let $x\in X$. I am trying to prove that all connected components of the orbit $Gx$ are of dimension $\dim G - \dim G_x$, where $G_x = \{g\in G\,\,|\,\,gx=x\}$.
Thank you in advance!

Lemma I believe I need to use:
Let $\pi:X\rightarrow Y$ be a dominant morphism, ie $\overline{\pi(X)} = Y$. Then any irreducible component of a fibre $\pi^{-1}(y)$, where $y\in Y$, has dimension at least $\dim X - \dim Y$. Moreover, there exists some nonempty open subset $O\subseteq Y$ such that $\dim\pi^{-1}(O) = \dim X -\dim Y$.

Some progress:
I have shown that $Gx$ is a locally closed subvariety. It is the image of the morphism $\phi_x:G\rightarrow X$ where $g\mapsto gx$, thus a union of locally closed sets. Therefore it contains a subset $U$ which is dense and open in $\overline{Gx}$. Since the set GU = $\bigcap_{g\in G} gU$ is contained in $Gx$, but invariant under $G$ we have $Gx = GU$, which is open in $\overline{Gx}$. 
The problem is that I am not really sure what the connected components are, so I don't know what to apply the lemma to.
 A: If $G$ is an algebraic group, then it has a connected component $G^0$ containing the identity element $e\in G$. Any connected component $G'$ of $G$ is then some translate of $G^0$ because if $g\in G'$, then $g^{-1} G' = G^0$, so $G'=gG^0$. In particular, all connected components of $G$ have the same dimension. Let $g_0:=e$ and $G=g_0G^0 \cup \cdots \cup g_r G^0$ be the decomposition of $G$ into irreducible components. 
Then, $G.x = g_0G^0.x \cup \cdots \cup g_r G^0.x$. Since $g_i G^0.x$ is isomorphic to $g_j G^0.x$ for all $i$ and $j$, all these sets are isomorphic to $G^0.x=g_0G^0.x$, which is an irreducible variety. Consequently, each of the $g_iG^0.x$ is an irreducible component of $G.x$. 
Hence, you only have to show the statement for connected algebraic groups. In other words, we assume $G^0=G$ and we show that $\dim(G.x)=\dim(G)-\dim(G_x)$. If you know that $G.x\cong G/G_x$, then you should be able to finish up now because $\phi_x^{-1}(g.x)=gG_xg^{-1}\cong G_x$, so the fibers all have the same dimension.
