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












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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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    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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      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?






      group-theory finite-groups group-actions







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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?






      group-theory finite-groups group-actions




      group-theory finite-groups group-actions






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      edited 33 mins ago









      Shaun

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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.






          share|cite|improve this answer































            0














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



            enter image description here



            However, if someone could figure out how to prove 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.






            share|cite|improve this answer

















            • 1




              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.
              – C. Falcon
              40 mins ago












            • @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?
              – Noble Mushtak
              27 mins ago










            • In all generality, $G/operatorname{Stab}(x)$ is not a group, since $operatorname{Stab}(x)$ needs not to be normal in $G$, the map constructed in the linked proof is only a bijection, not a group morphism. Recall that $G/operatorname{Stab}(x)$ is just a set of equivalence classes and this construction can be done indepently of all results.
              – C. Falcon
              24 mins ago








            • 1




              @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.
              – Noble Mushtak
              19 mins ago













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            2 Answers
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            2 Answers
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            4














            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.






            share|cite|improve this answer




























              4














              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.






              share|cite|improve this answer


























                4












                4








                4






                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.






                share|cite|improve this answer














                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.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 31 mins ago

























                answered 43 mins ago









                Tsemo Aristide

                55.6k11444




                55.6k11444























                    0














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



                    enter image description here



                    However, if someone could figure out how to prove 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.






                    share|cite|improve this answer

















                    • 1




                      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.
                      – C. Falcon
                      40 mins ago












                    • @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?
                      – Noble Mushtak
                      27 mins ago










                    • In all generality, $G/operatorname{Stab}(x)$ is not a group, since $operatorname{Stab}(x)$ needs not to be normal in $G$, the map constructed in the linked proof is only a bijection, not a group morphism. Recall that $G/operatorname{Stab}(x)$ is just a set of equivalence classes and this construction can be done indepently of all results.
                      – C. Falcon
                      24 mins ago








                    • 1




                      @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.
                      – Noble Mushtak
                      19 mins ago


















                    0














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



                    enter image description here



                    However, if someone could figure out how to prove 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.






                    share|cite|improve this answer

















                    • 1




                      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.
                      – C. Falcon
                      40 mins ago












                    • @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?
                      – Noble Mushtak
                      27 mins ago










                    • In all generality, $G/operatorname{Stab}(x)$ is not a group, since $operatorname{Stab}(x)$ needs not to be normal in $G$, the map constructed in the linked proof is only a bijection, not a group morphism. Recall that $G/operatorname{Stab}(x)$ is just a set of equivalence classes and this construction can be done indepently of all results.
                      – C. Falcon
                      24 mins ago








                    • 1




                      @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.
                      – Noble Mushtak
                      19 mins ago
















                    0












                    0








                    0






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



                    enter image description here



                    However, if someone could figure out how to prove 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.






                    share|cite|improve this answer












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



                    enter image description here



                    However, if someone could figure out how to prove 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.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 43 mins ago









                    Noble Mushtak

                    14.3k1734




                    14.3k1734








                    • 1




                      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.
                      – C. Falcon
                      40 mins ago












                    • @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?
                      – Noble Mushtak
                      27 mins ago










                    • In all generality, $G/operatorname{Stab}(x)$ is not a group, since $operatorname{Stab}(x)$ needs not to be normal in $G$, the map constructed in the linked proof is only a bijection, not a group morphism. Recall that $G/operatorname{Stab}(x)$ is just a set of equivalence classes and this construction can be done indepently of all results.
                      – C. Falcon
                      24 mins ago








                    • 1




                      @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.
                      – Noble Mushtak
                      19 mins ago
















                    • 1




                      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.
                      – C. Falcon
                      40 mins ago












                    • @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?
                      – Noble Mushtak
                      27 mins ago










                    • In all generality, $G/operatorname{Stab}(x)$ is not a group, since $operatorname{Stab}(x)$ needs not to be normal in $G$, the map constructed in the linked proof is only a bijection, not a group morphism. Recall that $G/operatorname{Stab}(x)$ is just a set of equivalence classes and this construction can be done indepently of all results.
                      – C. Falcon
                      24 mins ago








                    • 1




                      @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.
                      – Noble Mushtak
                      19 mins ago










                    1




                    1




                    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.
                    – C. Falcon
                    40 mins ago






                    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.
                    – C. Falcon
                    40 mins ago














                    @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?
                    – Noble Mushtak
                    27 mins ago




                    @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?
                    – Noble Mushtak
                    27 mins ago












                    In all generality, $G/operatorname{Stab}(x)$ is not a group, since $operatorname{Stab}(x)$ needs not to be normal in $G$, the map constructed in the linked proof is only a bijection, not a group morphism. Recall that $G/operatorname{Stab}(x)$ is just a set of equivalence classes and this construction can be done indepently of all results.
                    – C. Falcon
                    24 mins ago






                    In all generality, $G/operatorname{Stab}(x)$ is not a group, since $operatorname{Stab}(x)$ needs not to be normal in $G$, the map constructed in the linked proof is only a bijection, not a group morphism. Recall that $G/operatorname{Stab}(x)$ is just a set of equivalence classes and this construction can be done indepently of all results.
                    – C. Falcon
                    24 mins ago






                    1




                    1




                    @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.
                    – Noble Mushtak
                    19 mins ago






                    @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.
                    – Noble Mushtak
                    19 mins ago












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