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Is Lagrange's theorem used to prove that the length of the orbit times the order of the stabilizer is the order of the group, or is Lagrange's theorem a corollary of the orbit-stabilizer theorem?

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2 Answers 2

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Usually, Lagrange's theorem is used to prove the orbit-stabilizer theorem, not the other way around. See the following proof from "Abstract Algebra: Theory and Applications":

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However, if someone could figure out how to prove the orbit-stabilizer theorem without using Lagrange's Theorem, then you could prove Lagrange's Theorem as a corollary of the orbit-stabilizer theorem, as Tsemo Aristide showed.

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    $\begingroup$ See there, Lagrange's theorem is not needed to prove the orbit-stabilizer lemma, which is a really straightforward result: the bijection is explicit and canonical. $\endgroup$
    – C. Falcon
    Dec 30, 2018 at 15:21
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    $\begingroup$ @C.Falcon But doesn't that proof rely on the existence of $G / \text{Stab}(x)$ as a quotient group, which relies on Lagrange's Theorem? $\endgroup$
    – Noble Mushtak
    Dec 30, 2018 at 15:34
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    $\begingroup$ @C.Falcon I see. However, even in that case, you need to know $G/\text{Stab}(x)$ is a partition of $G$ in order to make the equivalence relation, and proving that all of the cosets of a subgroup partition the original group is basically Lagrange's Theorem. There's no way to go from the equation they prove, which is $\text{Orb}(x)=[G : \text{Stab}(x)]$, to the actual theorem, $|\text{Orb}(x)|\cdot |\text{Stab}(x)|=|G|$, without using Lagrange's Theorem somehow. $\endgroup$
    – Noble Mushtak
    Dec 30, 2018 at 15:42
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    $\begingroup$ The relationship is similar to that of Rolle's Theorem and the Mean Value Theorem. Rolle's Theorem is a special case which is used to prove the more general result. $\endgroup$
    – Will R
    Dec 30, 2018 at 19:21
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    $\begingroup$ @WillR: The two answers given already seem to contradict each other, and your comment provides resolution; would you like to upgrade it to an answer? $\endgroup$
    – J. W. Tanner
    Jan 28, 2019 at 0:43
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Lagrange is a corollary, if $H$ is a subgroup of $G$, $H$ acts on $G$ by left multiplication, the orbit of $1$ is $H$ so $|H|Or(G/H)=|G|$ where $Or(G/H)$ is the cardinal of the orbit space.

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  • $\begingroup$ Applying the orbit-stabilizer theorem to the action of $H$ on $G$ would only tell you that $|H|=\left|H/St(1)\right|$. It wouldn't give you any information on $G$ since the action of $H$ on itself does not depend on any embedding of $H$ into any group. $\endgroup$
    – Arnaud D.
    Dec 30, 2018 at 17:18
  • $\begingroup$ so lagrange is a corollary because Or(G/H) = [G:H] ? if so then how do you know Or(G/H) = [G:H]? if not then what do you mean? $\endgroup$
    – BCLC
    Aug 14, 2021 at 3:45

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