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?
group-theory finite-groups group-actions
edited Dec 30 '18 at 15:28
Shaun
11.1k113688
asked Dec 30 '18 at 15:16
J. W. TannerJ. W. Tanner
5,1451520
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|
4
$begingroup$
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 Dec 30 '18 at 15:28
Shaun
11.1k113688
asked Dec 30 '18 at 15:16
J. W. TannerJ. W. Tanner
5,1451520
$endgroup$
|
4
4
4
3
$begingroup$
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 Dec 30 '18 at 15:28
Shaun
11.1k113688
asked Dec 30 '18 at 15:16
J. W. TannerJ. W. Tanner
5,1451520
$endgroup$
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 Dec 30 '18 at 15:28
Shaun
11.1k113688
asked Dec 30 '18 at 15:16
J. W. TannerJ. W. Tanner
5,1451520
edited Dec 30 '18 at 15:28
Shaun
11.1k113688
asked Dec 30 '18 at 15:16
J. W. TannerJ. W. Tanner
5,1451520
edited Dec 30 '18 at 15:28
Shaun
11.1k113688
edited Dec 30 '18 at 15:28
Shaun
11.1k113688
edited Dec 30 '18 at 15:28
Shaun
11.1k113688
11.1k113688
asked Dec 30 '18 at 15:16
J. W. TannerJ. W. Tanner
5,1451520
asked Dec 30 '18 at 15:16
J. W. TannerJ. W. Tanner
5,1451520
asked Dec 30 '18 at 15:16
J. W. TannerJ. W. Tanner
5,1451520
5,1451520
|
|
2 Answers
2
active
oldest
votes
4
$begingroup$
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.
edited Jan 28 at 0:56
J. W. Tanner
5,1451520
answered Dec 30 '18 at 15:18
Noble MushtakNoble Mushtak
15.4k1835
$endgroup$
2
$begingroup$
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 '18 at 15:21
$begingroup$
@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 '18 at 15:34
$begingroup$
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.
$endgroup$
– C. Falcon
Dec 30 '18 at 15:37
5
$begingroup$
@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 '18 at 15:42
$begingroup$
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 '18 at 19:21
|
show 3 more comments
4
$begingroup$
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 Dec 30 '18 at 15:18
Tsemo AristideTsemo Aristide
61k11446
$endgroup$
$begingroup$
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 '18 at 17:18
|
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
4
$begingroup$
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.
edited Jan 28 at 0:56
J. W. Tanner
5,1451520
answered Dec 30 '18 at 15:18
Noble MushtakNoble Mushtak
15.4k1835
$endgroup$
2
$begingroup$
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 '18 at 15:21
$begingroup$
@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 '18 at 15:34
$begingroup$
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.
$endgroup$
– C. Falcon
Dec 30 '18 at 15:37
5
$begingroup$
@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 '18 at 15:42
$begingroup$
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 '18 at 19:21
|
show 3 more comments
4
$begingroup$
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.
edited Jan 28 at 0:56
J. W. Tanner
5,1451520
answered Dec 30 '18 at 15:18
Noble MushtakNoble Mushtak
15.4k1835
$endgroup$
2
$begingroup$
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 '18 at 15:21
$begingroup$
@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 '18 at 15:34
$begingroup$
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.
$endgroup$
– C. Falcon
Dec 30 '18 at 15:37
5
$begingroup$
@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 '18 at 15:42
$begingroup$
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 '18 at 19:21
|
show 3 more comments
4
4
4
$begingroup$
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.
edited Jan 28 at 0:56
J. W. Tanner
5,1451520
answered Dec 30 '18 at 15:18
Noble MushtakNoble Mushtak
15.4k1835
$endgroup$
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.
edited Jan 28 at 0:56
J. W. Tanner
5,1451520
answered Dec 30 '18 at 15:18
Noble MushtakNoble Mushtak
15.4k1835
edited Jan 28 at 0:56
J. W. Tanner
5,1451520
edited Jan 28 at 0:56
J. W. Tanner
5,1451520
edited Jan 28 at 0:56
J. W. Tanner
5,1451520
5,1451520
answered Dec 30 '18 at 15:18
Noble MushtakNoble Mushtak
15.4k1835
answered Dec 30 '18 at 15:18
Noble MushtakNoble Mushtak
15.4k1835
answered Dec 30 '18 at 15:18
Noble MushtakNoble Mushtak
15.4k1835
15.4k1835
2
$begingroup$
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 '18 at 15:21
$begingroup$
@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 '18 at 15:34
$begingroup$
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.
$endgroup$
– C. Falcon
Dec 30 '18 at 15:37
5
$begingroup$
@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 '18 at 15:42
$begingroup$
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 '18 at 19:21
|
show 3 more comments
2
$begingroup$
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 '18 at 15:21
$begingroup$
@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 '18 at 15:34
$begingroup$
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.
$endgroup$
– C. Falcon
Dec 30 '18 at 15:37
5
$begingroup$
@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 '18 at 15:42
$begingroup$
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 '18 at 19:21
2
2
$begingroup$
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 '18 at 15:21
$begingroup$
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 '18 at 15:21
$begingroup$
@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 '18 at 15:34
$begingroup$
@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 '18 at 15:34
$begingroup$
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.
$endgroup$
– C. Falcon
Dec 30 '18 at 15:37
$begingroup$
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.
$endgroup$
– C. Falcon
Dec 30 '18 at 15:37
5
5
$begingroup$
@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 '18 at 15:42
$begingroup$
@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 '18 at 15:42
$begingroup$
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 '18 at 19:21
$begingroup$
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 '18 at 19:21
|
show 3 more comments
4
$begingroup$
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 Dec 30 '18 at 15:18
Tsemo AristideTsemo Aristide
61k11446
$endgroup$
$begingroup$
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 '18 at 17:18
|
4
$begingroup$
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 Dec 30 '18 at 15:18
Tsemo AristideTsemo Aristide
61k11446
$endgroup$
$begingroup$
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 '18 at 17:18
|
4
4
4
$begingroup$
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 Dec 30 '18 at 15:18
Tsemo AristideTsemo Aristide
61k11446
$endgroup$
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 Dec 30 '18 at 15:18
Tsemo AristideTsemo Aristide
61k11446
edited Dec 30 '18 at 15:30
answered Dec 30 '18 at 15:18
Tsemo AristideTsemo Aristide
61k11446
answered Dec 30 '18 at 15:18
Tsemo AristideTsemo Aristide
61k11446
answered Dec 30 '18 at 15:18
Tsemo AristideTsemo Aristide
61k11446
61k11446
$begingroup$
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 '18 at 17:18
|
$begingroup$
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 '18 at 17:18
$begingroup$
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 '18 at 17:18
$begingroup$
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 '18 at 17:18
|
