I just learned about the orbit-stabilizer theorem. Some people (such as those in the answers to this older question about the same topic) seem to argue that Lagrange's theorem is a special case of the orbit-stabilizer theorem.
So, to my understanding, the orbit-stabilizer theorem says this: Each coset of the stabilizer $\mathbf{stab}x$ collects exactly those elements of $G$, which send $x \in X$ to some particular fixed element $x' \in \textbf{orb}x$. Thus, there exists a natural one-to-one correspondence between elements of $\textbf{orb}x$ and the cosets in $G/\textbf{stab}x$:
$$ |\textbf{orb}x| = |G:\textbf{stab}x|. $$
Now I do not see how this would imply Lagrange's theorem. If I take $H \leq G$ acting on $G$ by left multiplication, I do get $\textbf{orb}g = Hg$. As for the stabilizer, we know that $\textbf{stab}g = \{e\}$, for all $g \in G$. Substituting into the equation above, we get a rather obvious fact:
$$ |Hg| = |H:\{e\}| = |H|. $$
The orbit stabilizer theorem does not seem to tell us anything about the size of the set $X$ being acted upon. Is this correct, or am I misunderstanding something here? Are those people conflating the orbit-stabilizer theorem with just any fact having to do with orbits?
Best Answer
The action that gives you Lagrange's theorem is not the action of $H$ on $G$ but the action of $G$ on $G/H$ (by left multiplication). For this action there is one orbit of size $|G/H|$ and the stabilizer (at $[e]$) is $H$. Applying the orbit-stabilizer theorem gives
$$|G/H| = [G : H] = \frac{|G|}{|H|}$$
which is exactly Lagrange's theorem.