Does $H\leq G$ imply $[G_x: H_x]\leq [G: H]$? Let a group $G$ act on a set $S$. Let $H$ be a subgroup of $G$ and $x\in S$. I want to show that $[G_x: H_x]\leq [G: H]$, where $G_x$ is the isotropy group of $x$ in $G$. (If necessary, we assume that these indices are finite).
If $G$ is a finite group, since $|G|=|G_x||Gx|$, we have
$$[G_x: H_x]=\frac{|G|}{|H|}\frac{|Hx|}{|Gx|}\leq\frac{|G|}{|H|}=[G:H].$$
And this is how I get the desired inequality.
As for the case where $G$ is infinite, such as $SL_2(\mathbb Z)$, I wrote $G=\bigcup_{i=1}^nHg_i$ where $n=[G:H]$ and wanted to show that $G_x=\bigcup_{i=1}^mH_xg_i'$ for some $g_i'$s. But I cannot move on since I cannot connect these $g_i'$s with the isotropy subgroup $G_x$.
This problem was raised by me from my studying in Fred Diamond & Jerry Shurman: A First Course in Modular Forms (Exercise 2.4.4(a)). I am not confident about my conjecture. Any help will be appreciated.
 A: If all indices in sight are finite : 
$[G:H_x] = [G:H][H:H_x] = [G:G_x][G_x:H_x]$ so the claim is equivalent to showing that $[H:H_x]\leq [G:G_x]$. 
But for that one there's an easy proof : $H/H_x\to G/G_x, kH_x\mapsto kG_x$ is (well-defined and ) injective. 
Indeed assume $kG_x = gG_x$ with $k,g\in H$. Then $kg^{-1}\in G_x$ and $kg^{-1}\in H $ so $kg^{-1}\in H_x$ so $kH_x=gH_x$, whence injectivity. 
So we at most need $[G:H_x]$ to be finite (the finitude of the rest follows, or you can do it the other way around)
Here's a proof of the general case with no finiteness assumption, using Zorn's lemma (we can formulate it also using transfinite sequences but most people are more comfortable with Zorn than with transfinite induction - if you don't like the axiom of choice, this construction requires that $G_x$ is well-ordered, not more) :
Let $R\subset G_x$ be a maximal subset of "$H$-independent" elements, that is a subset such that for $h\neq k \in R, h^{-1}k\notin H$ and that is maximal with respect to this property (note : I did write $H$ and not $H_x$ !).
It exists because this condition is stable under chain unions, and $\{e\}$ satisfies it, so Zorn's lemma applies. Then $|R| \leq [G:H]$ for obvious reasons . I now claim that $|R| = [G_x:H_x]$. Indeed, I claim that $G_x \subset \displaystyle\bigcup_{g\in R}gH_x$. The proof is simple : let $k\in G_x$. Then either $k\in R$ in which case $k\in kH_x$ and we're done, or $k\notin R$, in which case by maximality, $R\cup\{k\}$ is not $H$-independent, so that there is $g\in R$ with $g^{-1}k\in H$. 
But $g^{-1}k\in G_x$ as well so $g^{-1}k\in G_x\cap H = H_x$ and therefore $k\in gH_x$.
Moreover, $gH_x \subset G_x$ for $g\in R$ and if $gH_x = kH_x, g,k\in R$ then $g=k$ so that $G_x =\displaystyle\coprod_{g\in R}gH_x$ and $[G_x:H_x] = |R|$, so we are done. 
The idea here was that you can't start with any decomposition of $G_x$ into $H_x$-cosets, but if you start with $H$-cosets and then move down to $H_x$-cosets, everything works. If every index is finite, the above construction boils down to : take a set of representatives of $G$ mod $H$ with the maximum amount of elements of $G_x$, say $k_1,...,k_r, g_1,...,g_l$. Then since we took the maximum amount of elements of $G_x$, anyone in $G_x$ must be in one of $k_1H,...,k_rH$, otherwise I could have added it to the list. But then the $k_i$ are also $H_x$ representatives. 
