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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?

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    $\begingroup$ $X$ (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. $\endgroup$
    – Randall
    Sep 20, 2022 at 16:35
  • $\begingroup$ 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. $\endgroup$
    – Randall
    Sep 20, 2022 at 16:42
  • $\begingroup$ @Randall Yes, I do realize. But this does not really use the orbit-stabilizer theorem does it? $\endgroup$
    – Jan Matula
    Sep 20, 2022 at 16:44
  • $\begingroup$ 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. $\endgroup$
    – Randall
    Sep 20, 2022 at 16:48
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    $\begingroup$ In that case, that is the answer to my question. Thanks a lot for clarifying this to me. $\endgroup$
    – Jan Matula
    Sep 20, 2022 at 17:32

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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.

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  • $\begingroup$ Ah makes sense, probably should have thought of that. Thanks. $\endgroup$
    – Jan Matula
    Sep 20, 2022 at 17:36
  • $\begingroup$ "the stabilizer (at $[e]$)": isn't the stabilizer at $H$? $\endgroup$
    – Kan't
    Sep 20, 2022 at 20:03
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    $\begingroup$ 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. $\endgroup$ Sep 20, 2022 at 20:04

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