# Questioning the relationship between the orbit-stabilizer theorem and Lagrange's theorem

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?

• $X$ (in your case, $G$) is always a disjoint union of the orbits. So, this, coupled with the fact that all orbits have the same size (namely $|H|$), gives Lagrange. Sep 20, 2022 at 16:35
• Note that the disjointness of the orbits/cosets does in fact give you information about the size of the set being acted upon, especially when you know the sizes of all the orbits. Sep 20, 2022 at 16:42
• @Randall Yes, I do realize. But this does not really use the orbit-stabilizer theorem does it? Sep 20, 2022 at 16:44
• Yes, it does. It is a special case of the theorem, as you point out yourself: $|Hg|=|H:\{e\}|$. You can certainly prove it from scratch using only coset information, and that's how most people teach it. But it is a good motivator for understanding/believing in the orbit-stabilizer theorem. Sep 20, 2022 at 16:48
• In that case, that is the answer to my question. Thanks a lot for clarifying this to me. Sep 20, 2022 at 17:32

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
$$|G/H| = [G : H] = \frac{|G|}{|H|}$$
• "the stabilizer (at $[e]$)": isn't the stabilizer at $H$? Sep 20, 2022 at 20:03
• Yes, that's what I mean. By $[e]$ I mean the image of $e \in G$ under the projection $G \to G/H$. Maybe $eH$ would have been less ambiguous. Sep 20, 2022 at 20:04