# What is the relationship between the orbit-stabilizer theorem and Lagrange's theorem?

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?

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":

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.

• 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. Dec 30, 2018 at 15:21
• @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? Dec 30, 2018 at 15:34
• @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. Dec 30, 2018 at 15:42
• 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. Dec 30, 2018 at 19:21
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.
• 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. Dec 30, 2018 at 17:18