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?
group-theory finite-groups group-actions
edited 59 mins ago
Shaun
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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
edited 59 mins ago
Shaun
8,691113680
asked 1 hour ago
J. W. Tanner
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New contributor
J. W. Tanner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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
edited 59 mins ago
Shaun
8,691113680
asked 1 hour ago
J. W. Tanner
112
New contributor
J. W. Tanner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
edited 59 mins ago
Shaun
8,691113680
asked 1 hour ago
J. W. Tanner
112
New contributor
J. W. Tanner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 59 mins ago
Shaun
8,691113680
asked 1 hour ago
J. W. Tanner
112
New contributor
J. W. Tanner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 59 mins ago
Shaun
8,691113680
edited 59 mins ago
Shaun
8,691113680
edited 59 mins ago
Shaun
8,691113680
8,691113680
asked 1 hour ago
J. W. Tanner
112
New contributor
J. W. Tanner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
J. W. Tanner
112
asked 1 hour ago
J. W. Tanner
112
112
New contributor
J. W. Tanner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
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Check out our Code of Conduct.
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2 Answers
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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.
answered 1 hour ago
Tsemo Aristide
55.6k11444
add a comment |
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":
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.
answered 1 hour ago
Noble Mushtak
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
1 hour 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
53 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
50 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
45 mins ago
add a comment |
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2 Answers
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2 Answers
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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.
answered 1 hour ago
Tsemo Aristide
55.6k11444
add a comment |
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.
answered 1 hour ago
Tsemo Aristide
55.6k11444
add a comment |
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.
answered 1 hour ago
Tsemo Aristide
55.6k11444
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.
answered 1 hour ago
Tsemo Aristide
55.6k11444
edited 57 mins ago
answered 1 hour ago
Tsemo Aristide
55.6k11444
answered 1 hour ago
Tsemo Aristide
55.6k11444
answered 1 hour ago
Tsemo Aristide
55.6k11444
55.6k11444
add a comment |
add a comment |
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":
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.
answered 1 hour ago
Noble Mushtak
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
1 hour 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
53 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
50 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
45 mins ago
add a comment |
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":
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.
answered 1 hour ago
Noble Mushtak
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
1 hour 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
53 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
50 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
45 mins ago
add a comment |
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":
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.
answered 1 hour ago
Noble Mushtak
14.3k1734
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":
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.
answered 1 hour ago
Noble Mushtak
14.3k1734
answered 1 hour ago
Noble Mushtak
14.3k1734
answered 1 hour ago
Noble Mushtak
14.3k1734
answered 1 hour 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
1 hour 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
53 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
50 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
45 mins ago
add a comment |
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
1 hour 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
53 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
50 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
45 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
1 hour 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
1 hour 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
53 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
53 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
50 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
50 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
45 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
45 mins ago
add a comment |
J. W. Tanner is a new contributor. Be nice, and check out our Code of Conduct.
J. W. Tanner is a new contributor. Be nice, and check out our Code of Conduct.
J. W. Tanner is a new contributor. Be nice, and check out our Code of Conduct.
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