Why does Zorn's Lemma fail to produce a largest group?
$begingroup$
Zorn's Lemma states that if every chain $mathcal{C}$ in a partially ordered set $mathcal{S}$ has an upper bound in $mathcal{S}$, then there is at least one maximal element in $mathcal{S}$.
Why can't we apply Zorn's Lemma in the following case?
Let $mathcal{S}$ be the set of all groups. Define a partial order $preceq$ as follows: for $H, G in mathcal{S}$, define $H prec G$ if and only if $H$ is a subgroup of $G$. Then every chain $mathcal{C}=(G_{alpha})_{alpha in A}$ in $mathcal{S}$ has an upper bound $cup_{alpha in A} G_{alpha}$ in $mathcal{S}$. But certainly there is no maximal element in $mathcal{S}$.
Could anyone tell me what is wrong with this argument?
group-theory elementary-set-theory axiom-of-choice
edited Feb 7 at 15:53
Asaf Karagila♦
306k33437767
asked Feb 7 at 11:33
ScarletScarlet
11914
$endgroup$
add a comment |
$begingroup$
Zorn's Lemma states that if every chain $mathcal{C}$ in a partially ordered set $mathcal{S}$ has an upper bound in $mathcal{S}$, then there is at least one maximal element in $mathcal{S}$.
Why can't we apply Zorn's Lemma in the following case?
Let $mathcal{S}$ be the set of all groups. Define a partial order $preceq$ as follows: for $H, G in mathcal{S}$, define $H prec G$ if and only if $H$ is a subgroup of $G$. Then every chain $mathcal{C}=(G_{alpha})_{alpha in A}$ in $mathcal{S}$ has an upper bound $cup_{alpha in A} G_{alpha}$ in $mathcal{S}$. But certainly there is no maximal element in $mathcal{S}$.
Could anyone tell me what is wrong with this argument?
group-theory elementary-set-theory axiom-of-choice
edited Feb 7 at 15:53
Asaf Karagila♦
306k33437767
asked Feb 7 at 11:33
ScarletScarlet
11914
$endgroup$
1
$begingroup$
You can define partial orders on certain subclasses of groups. A fruitful one is the "largeness" ordering for finitely generated groups: $Gsucceq H$ ("$G$ is larger than $H$") if there exists a finite-index subgroup $Kleq_fG$ such that $Ktwoheadrightarrow H$. Exercise: Prove that free group $F_2$ is maximal here. Then a group $G$ is "large" or "large in the sense of Pride" (after Steve Pride, who first studied this ordering) if it is maximal in this ordering. For fruitfulness, here is a picture of Ian Agol announcing his proof of the virtual Haken conjecture.
$endgroup$
– user1729
Feb 7 at 15:30
13
$begingroup$
Note that groups are really superfluous here: the same idea suggests that Zorn's lemma should yield a maximal set. This helps make it clearer that the class/set issue is absolutely fundamental.
$endgroup$
– Noah Schweber
Feb 7 at 15:34
$begingroup$
You might be able to apply the Hausdorff maximal principle instead, if you can generalize it to classes.
$endgroup$
– PyRulez
Feb 7 at 18:03
8
$begingroup$
I lost you at "set of all groups".
$endgroup$
– WillO
Feb 8 at 0:31
add a comment |
$begingroup$
Zorn's Lemma states that if every chain $mathcal{C}$ in a partially ordered set $mathcal{S}$ has an upper bound in $mathcal{S}$, then there is at least one maximal element in $mathcal{S}$.
Why can't we apply Zorn's Lemma in the following case?
Let $mathcal{S}$ be the set of all groups. Define a partial order $preceq$ as follows: for $H, G in mathcal{S}$, define $H prec G$ if and only if $H$ is a subgroup of $G$. Then every chain $mathcal{C}=(G_{alpha})_{alpha in A}$ in $mathcal{S}$ has an upper bound $cup_{alpha in A} G_{alpha}$ in $mathcal{S}$. But certainly there is no maximal element in $mathcal{S}$.
Could anyone tell me what is wrong with this argument?
group-theory elementary-set-theory axiom-of-choice
edited Feb 7 at 15:53
Asaf Karagila♦
306k33437767
asked Feb 7 at 11:33
ScarletScarlet
11914
$endgroup$
Zorn's Lemma states that if every chain $mathcal{C}$ in a partially ordered set $mathcal{S}$ has an upper bound in $mathcal{S}$, then there is at least one maximal element in $mathcal{S}$.
Why can't we apply Zorn's Lemma in the following case?
Let $mathcal{S}$ be the set of all groups. Define a partial order $preceq$ as follows: for $H, G in mathcal{S}$, define $H prec G$ if and only if $H$ is a subgroup of $G$. Then every chain $mathcal{C}=(G_{alpha})_{alpha in A}$ in $mathcal{S}$ has an upper bound $cup_{alpha in A} G_{alpha}$ in $mathcal{S}$. But certainly there is no maximal element in $mathcal{S}$.
Could anyone tell me what is wrong with this argument?
group-theory elementary-set-theory axiom-of-choice
group-theory elementary-set-theory axiom-of-choice
edited Feb 7 at 15:53
Asaf Karagila♦
306k33437767
asked Feb 7 at 11:33
ScarletScarlet
11914
edited Feb 7 at 15:53
Asaf Karagila♦
306k33437767
asked Feb 7 at 11:33
ScarletScarlet
11914
edited Feb 7 at 15:53
Asaf Karagila♦
306k33437767
edited Feb 7 at 15:53
Asaf Karagila♦
306k33437767
edited Feb 7 at 15:53
Asaf Karagila♦
306k33437767
306k33437767
asked Feb 7 at 11:33
ScarletScarlet
11914
asked Feb 7 at 11:33
ScarletScarlet
11914
asked Feb 7 at 11:33
ScarletScarlet
11914
11914
1
$begingroup$
You can define partial orders on certain subclasses of groups. A fruitful one is the "largeness" ordering for finitely generated groups: $Gsucceq H$ ("$G$ is larger than $H$") if there exists a finite-index subgroup $Kleq_fG$ such that $Ktwoheadrightarrow H$. Exercise: Prove that free group $F_2$ is maximal here. Then a group $G$ is "large" or "large in the sense of Pride" (after Steve Pride, who first studied this ordering) if it is maximal in this ordering. For fruitfulness, here is a picture of Ian Agol announcing his proof of the virtual Haken conjecture.
$endgroup$
– user1729
Feb 7 at 15:30
13
$begingroup$
Note that groups are really superfluous here: the same idea suggests that Zorn's lemma should yield a maximal set. This helps make it clearer that the class/set issue is absolutely fundamental.
$endgroup$
– Noah Schweber
Feb 7 at 15:34
$begingroup$
You might be able to apply the Hausdorff maximal principle instead, if you can generalize it to classes.
$endgroup$
– PyRulez
Feb 7 at 18:03
8
$begingroup$
I lost you at "set of all groups".
$endgroup$
– WillO
Feb 8 at 0:31
add a comment |
1
$begingroup$
You can define partial orders on certain subclasses of groups. A fruitful one is the "largeness" ordering for finitely generated groups: $Gsucceq H$ ("$G$ is larger than $H$") if there exists a finite-index subgroup $Kleq_fG$ such that $Ktwoheadrightarrow H$. Exercise: Prove that free group $F_2$ is maximal here. Then a group $G$ is "large" or "large in the sense of Pride" (after Steve Pride, who first studied this ordering) if it is maximal in this ordering. For fruitfulness, here is a picture of Ian Agol announcing his proof of the virtual Haken conjecture.
$endgroup$
– user1729
Feb 7 at 15:30
13
$begingroup$
Note that groups are really superfluous here: the same idea suggests that Zorn's lemma should yield a maximal set. This helps make it clearer that the class/set issue is absolutely fundamental.
$endgroup$
– Noah Schweber
Feb 7 at 15:34
$begingroup$
You might be able to apply the Hausdorff maximal principle instead, if you can generalize it to classes.
$endgroup$
– PyRulez
Feb 7 at 18:03
8
$begingroup$
I lost you at "set of all groups".
$endgroup$
– WillO
Feb 8 at 0:31
1
1
$begingroup$
You can define partial orders on certain subclasses of groups. A fruitful one is the "largeness" ordering for finitely generated groups: $Gsucceq H$ ("$G$ is larger than $H$") if there exists a finite-index subgroup $Kleq_fG$ such that $Ktwoheadrightarrow H$. Exercise: Prove that free group $F_2$ is maximal here. Then a group $G$ is "large" or "large in the sense of Pride" (after Steve Pride, who first studied this ordering) if it is maximal in this ordering. For fruitfulness, here is a picture of Ian Agol announcing his proof of the virtual Haken conjecture.
$endgroup$
– user1729
Feb 7 at 15:30
$begingroup$
You can define partial orders on certain subclasses of groups. A fruitful one is the "largeness" ordering for finitely generated groups: $Gsucceq H$ ("$G$ is larger than $H$") if there exists a finite-index subgroup $Kleq_fG$ such that $Ktwoheadrightarrow H$. Exercise: Prove that free group $F_2$ is maximal here. Then a group $G$ is "large" or "large in the sense of Pride" (after Steve Pride, who first studied this ordering) if it is maximal in this ordering. For fruitfulness, here is a picture of Ian Agol announcing his proof of the virtual Haken conjecture.
$endgroup$
– user1729
Feb 7 at 15:30
13
13
$begingroup$
Note that groups are really superfluous here: the same idea suggests that Zorn's lemma should yield a maximal set. This helps make it clearer that the class/set issue is absolutely fundamental.
$endgroup$
– Noah Schweber
Feb 7 at 15:34
$begingroup$
Note that groups are really superfluous here: the same idea suggests that Zorn's lemma should yield a maximal set. This helps make it clearer that the class/set issue is absolutely fundamental.
$endgroup$
– Noah Schweber
Feb 7 at 15:34
$begingroup$
You might be able to apply the Hausdorff maximal principle instead, if you can generalize it to classes.
$endgroup$
– PyRulez
Feb 7 at 18:03
$begingroup$
You might be able to apply the Hausdorff maximal principle instead, if you can generalize it to classes.
$endgroup$
– PyRulez
Feb 7 at 18:03
8
8
$begingroup$
I lost you at "set of all groups".
$endgroup$
– WillO
Feb 8 at 0:31
$begingroup$
I lost you at "set of all groups".
$endgroup$
– WillO
Feb 8 at 0:31
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
There is no set of all groups, but Zorn's Lemma can be phrased for partially ordered classes as well, but we need to require that any chain has an upper bound, also proper class chains. However in that case it is easy to define a proper class chain which has no upper bound.
Simply take for each ordinal $alpha$ the free group generated by $alpha$. There are natural injections given by the natural injections between two ordinals. However this chain does not have an upper bound, since an upper bound would be a group, which is a set, that embeds all ordinals. This is a contradiction to Hartogs theorem, of course.
(You can think about this in the following analog: an infinite chain of finite sets, or finite groups, will not have an upper bound which is also finite.)
edited Feb 7 at 11:59
lisyarus
10.7k21433
answered Feb 7 at 11:49
Asaf Karagila♦Asaf Karagila
306k33437767
$endgroup$
$begingroup$
I guess you could also take for each $alpha$ the set of all finite subsets of $alpha$ with the symmetric difference operation, and then your natural injections are just inclusions.
$endgroup$
– bof
Feb 7 at 12:34
$begingroup$
True. That's also an option. The natural injections here are also just inclusions, though.
$endgroup$
– Asaf Karagila♦
Feb 7 at 13:46
$begingroup$
Doesn't that depend on what you mean by "the" free group generated by $alpha$? I don't know any group theory, but I vaguely recall seeing free groups constructed as subdirect products. I guess you could also construct them as sets of words. Is that how you get the inclusions? Still seems more complicated than just using Boolean groups, if all you want is a proper class chain of groups.
$endgroup$
– bof
Feb 7 at 14:31
$begingroup$
Yes, I meant as a group of words. But I agree, your solution is simpler.
$endgroup$
– Asaf Karagila♦
Feb 7 at 14:46
1
$begingroup$
And not only is there no set of all groups, there's no set of groups which contains at least one group in each isomorphism class.
$endgroup$
– Daniel Schepler
Feb 7 at 22:36
|
show 6 more comments
$begingroup$
Because there is no such thing as “the set of all groups”.
answered Feb 7 at 11:34
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
1
$begingroup$
Some information : math.stackexchange.com/questions/226413/…
$endgroup$
– Thomas Lesgourgues
Feb 7 at 11:48
$begingroup$
+1 jose carlos sir
$endgroup$
– jasmine
Feb 14 at 12:00
add a comment |
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2 Answers
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2 Answers
2
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votes
$begingroup$
There is no set of all groups, but Zorn's Lemma can be phrased for partially ordered classes as well, but we need to require that any chain has an upper bound, also proper class chains. However in that case it is easy to define a proper class chain which has no upper bound.
Simply take for each ordinal $alpha$ the free group generated by $alpha$. There are natural injections given by the natural injections between two ordinals. However this chain does not have an upper bound, since an upper bound would be a group, which is a set, that embeds all ordinals. This is a contradiction to Hartogs theorem, of course.
(You can think about this in the following analog: an infinite chain of finite sets, or finite groups, will not have an upper bound which is also finite.)
edited Feb 7 at 11:59
lisyarus
10.7k21433
answered Feb 7 at 11:49
Asaf Karagila♦Asaf Karagila
306k33437767
$endgroup$
$begingroup$
I guess you could also take for each $alpha$ the set of all finite subsets of $alpha$ with the symmetric difference operation, and then your natural injections are just inclusions.
$endgroup$
– bof
Feb 7 at 12:34
$begingroup$
True. That's also an option. The natural injections here are also just inclusions, though.
$endgroup$
– Asaf Karagila♦
Feb 7 at 13:46
$begingroup$
Doesn't that depend on what you mean by "the" free group generated by $alpha$? I don't know any group theory, but I vaguely recall seeing free groups constructed as subdirect products. I guess you could also construct them as sets of words. Is that how you get the inclusions? Still seems more complicated than just using Boolean groups, if all you want is a proper class chain of groups.
$endgroup$
– bof
Feb 7 at 14:31
$begingroup$
Yes, I meant as a group of words. But I agree, your solution is simpler.
$endgroup$
– Asaf Karagila♦
Feb 7 at 14:46
1
$begingroup$
And not only is there no set of all groups, there's no set of groups which contains at least one group in each isomorphism class.
$endgroup$
– Daniel Schepler
Feb 7 at 22:36
|
show 6 more comments
$begingroup$
There is no set of all groups, but Zorn's Lemma can be phrased for partially ordered classes as well, but we need to require that any chain has an upper bound, also proper class chains. However in that case it is easy to define a proper class chain which has no upper bound.
Simply take for each ordinal $alpha$ the free group generated by $alpha$. There are natural injections given by the natural injections between two ordinals. However this chain does not have an upper bound, since an upper bound would be a group, which is a set, that embeds all ordinals. This is a contradiction to Hartogs theorem, of course.
(You can think about this in the following analog: an infinite chain of finite sets, or finite groups, will not have an upper bound which is also finite.)
edited Feb 7 at 11:59
lisyarus
10.7k21433
answered Feb 7 at 11:49
Asaf Karagila♦Asaf Karagila
306k33437767
$endgroup$
$begingroup$
I guess you could also take for each $alpha$ the set of all finite subsets of $alpha$ with the symmetric difference operation, and then your natural injections are just inclusions.
$endgroup$
– bof
Feb 7 at 12:34
$begingroup$
True. That's also an option. The natural injections here are also just inclusions, though.
$endgroup$
– Asaf Karagila♦
Feb 7 at 13:46
$begingroup$
Doesn't that depend on what you mean by "the" free group generated by $alpha$? I don't know any group theory, but I vaguely recall seeing free groups constructed as subdirect products. I guess you could also construct them as sets of words. Is that how you get the inclusions? Still seems more complicated than just using Boolean groups, if all you want is a proper class chain of groups.
$endgroup$
– bof
Feb 7 at 14:31
$begingroup$
Yes, I meant as a group of words. But I agree, your solution is simpler.
$endgroup$
– Asaf Karagila♦
Feb 7 at 14:46
1
$begingroup$
And not only is there no set of all groups, there's no set of groups which contains at least one group in each isomorphism class.
$endgroup$
– Daniel Schepler
Feb 7 at 22:36
|
show 6 more comments
$begingroup$
There is no set of all groups, but Zorn's Lemma can be phrased for partially ordered classes as well, but we need to require that any chain has an upper bound, also proper class chains. However in that case it is easy to define a proper class chain which has no upper bound.
Simply take for each ordinal $alpha$ the free group generated by $alpha$. There are natural injections given by the natural injections between two ordinals. However this chain does not have an upper bound, since an upper bound would be a group, which is a set, that embeds all ordinals. This is a contradiction to Hartogs theorem, of course.
(You can think about this in the following analog: an infinite chain of finite sets, or finite groups, will not have an upper bound which is also finite.)
edited Feb 7 at 11:59
lisyarus
10.7k21433
answered Feb 7 at 11:49
Asaf Karagila♦Asaf Karagila
306k33437767
$endgroup$
There is no set of all groups, but Zorn's Lemma can be phrased for partially ordered classes as well, but we need to require that any chain has an upper bound, also proper class chains. However in that case it is easy to define a proper class chain which has no upper bound.
Simply take for each ordinal $alpha$ the free group generated by $alpha$. There are natural injections given by the natural injections between two ordinals. However this chain does not have an upper bound, since an upper bound would be a group, which is a set, that embeds all ordinals. This is a contradiction to Hartogs theorem, of course.
(You can think about this in the following analog: an infinite chain of finite sets, or finite groups, will not have an upper bound which is also finite.)
edited Feb 7 at 11:59
lisyarus
10.7k21433
answered Feb 7 at 11:49
Asaf Karagila♦Asaf Karagila
306k33437767
edited Feb 7 at 11:59
lisyarus
10.7k21433
edited Feb 7 at 11:59
lisyarus
10.7k21433
edited Feb 7 at 11:59
lisyarus
10.7k21433
10.7k21433
answered Feb 7 at 11:49
Asaf Karagila♦Asaf Karagila
306k33437767
answered Feb 7 at 11:49
Asaf Karagila♦Asaf Karagila
306k33437767
answered Feb 7 at 11:49
Asaf Karagila♦Asaf Karagila
306k33437767
306k33437767
$begingroup$
I guess you could also take for each $alpha$ the set of all finite subsets of $alpha$ with the symmetric difference operation, and then your natural injections are just inclusions.
$endgroup$
– bof
Feb 7 at 12:34
$begingroup$
True. That's also an option. The natural injections here are also just inclusions, though.
$endgroup$
– Asaf Karagila♦
Feb 7 at 13:46
$begingroup$
Doesn't that depend on what you mean by "the" free group generated by $alpha$? I don't know any group theory, but I vaguely recall seeing free groups constructed as subdirect products. I guess you could also construct them as sets of words. Is that how you get the inclusions? Still seems more complicated than just using Boolean groups, if all you want is a proper class chain of groups.
$endgroup$
– bof
Feb 7 at 14:31
$begingroup$
Yes, I meant as a group of words. But I agree, your solution is simpler.
$endgroup$
– Asaf Karagila♦
Feb 7 at 14:46
1
$begingroup$
And not only is there no set of all groups, there's no set of groups which contains at least one group in each isomorphism class.
$endgroup$
– Daniel Schepler
Feb 7 at 22:36
|
show 6 more comments
$begingroup$
I guess you could also take for each $alpha$ the set of all finite subsets of $alpha$ with the symmetric difference operation, and then your natural injections are just inclusions.
$endgroup$
– bof
Feb 7 at 12:34
$begingroup$
True. That's also an option. The natural injections here are also just inclusions, though.
$endgroup$
– Asaf Karagila♦
Feb 7 at 13:46
$begingroup$
Doesn't that depend on what you mean by "the" free group generated by $alpha$? I don't know any group theory, but I vaguely recall seeing free groups constructed as subdirect products. I guess you could also construct them as sets of words. Is that how you get the inclusions? Still seems more complicated than just using Boolean groups, if all you want is a proper class chain of groups.
$endgroup$
– bof
Feb 7 at 14:31
$begingroup$
Yes, I meant as a group of words. But I agree, your solution is simpler.
$endgroup$
– Asaf Karagila♦
Feb 7 at 14:46
1
$begingroup$
And not only is there no set of all groups, there's no set of groups which contains at least one group in each isomorphism class.
$endgroup$
– Daniel Schepler
Feb 7 at 22:36
$begingroup$
I guess you could also take for each $alpha$ the set of all finite subsets of $alpha$ with the symmetric difference operation, and then your natural injections are just inclusions.
$endgroup$
– bof
Feb 7 at 12:34
$begingroup$
I guess you could also take for each $alpha$ the set of all finite subsets of $alpha$ with the symmetric difference operation, and then your natural injections are just inclusions.
$endgroup$
– bof
Feb 7 at 12:34
$begingroup$
True. That's also an option. The natural injections here are also just inclusions, though.
$endgroup$
– Asaf Karagila♦
Feb 7 at 13:46
$begingroup$
True. That's also an option. The natural injections here are also just inclusions, though.
$endgroup$
– Asaf Karagila♦
Feb 7 at 13:46
$begingroup$
Doesn't that depend on what you mean by "the" free group generated by $alpha$? I don't know any group theory, but I vaguely recall seeing free groups constructed as subdirect products. I guess you could also construct them as sets of words. Is that how you get the inclusions? Still seems more complicated than just using Boolean groups, if all you want is a proper class chain of groups.
$endgroup$
– bof
Feb 7 at 14:31
$begingroup$
Doesn't that depend on what you mean by "the" free group generated by $alpha$? I don't know any group theory, but I vaguely recall seeing free groups constructed as subdirect products. I guess you could also construct them as sets of words. Is that how you get the inclusions? Still seems more complicated than just using Boolean groups, if all you want is a proper class chain of groups.
$endgroup$
– bof
Feb 7 at 14:31
$begingroup$
Yes, I meant as a group of words. But I agree, your solution is simpler.
$endgroup$
– Asaf Karagila♦
Feb 7 at 14:46
$begingroup$
Yes, I meant as a group of words. But I agree, your solution is simpler.
$endgroup$
– Asaf Karagila♦
Feb 7 at 14:46
1
1
$begingroup$
And not only is there no set of all groups, there's no set of groups which contains at least one group in each isomorphism class.
$endgroup$
– Daniel Schepler
Feb 7 at 22:36
$begingroup$
And not only is there no set of all groups, there's no set of groups which contains at least one group in each isomorphism class.
$endgroup$
– Daniel Schepler
Feb 7 at 22:36
|
show 6 more comments
$begingroup$
Because there is no such thing as “the set of all groups”.
answered Feb 7 at 11:34
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
1
$begingroup$
Some information : math.stackexchange.com/questions/226413/…
$endgroup$
– Thomas Lesgourgues
Feb 7 at 11:48
$begingroup$
+1 jose carlos sir
$endgroup$
– jasmine
Feb 14 at 12:00
add a comment |
$begingroup$
Because there is no such thing as “the set of all groups”.
answered Feb 7 at 11:34
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
1
$begingroup$
Some information : math.stackexchange.com/questions/226413/…
$endgroup$
– Thomas Lesgourgues
Feb 7 at 11:48
$begingroup$
+1 jose carlos sir
$endgroup$
– jasmine
Feb 14 at 12:00
add a comment |
$begingroup$
Because there is no such thing as “the set of all groups”.
answered Feb 7 at 11:34
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
Because there is no such thing as “the set of all groups”.
answered Feb 7 at 11:34
José Carlos SantosJosé Carlos Santos
165k22132235
answered Feb 7 at 11:34
José Carlos SantosJosé Carlos Santos
165k22132235
answered Feb 7 at 11:34
José Carlos SantosJosé Carlos Santos
165k22132235
answered Feb 7 at 11:34
José Carlos SantosJosé Carlos Santos
165k22132235
165k22132235
1
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Some information : math.stackexchange.com/questions/226413/…
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– Thomas Lesgourgues
Feb 7 at 11:48
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+1 jose carlos sir
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– jasmine
Feb 14 at 12:00
add a comment |
1
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Some information : math.stackexchange.com/questions/226413/…
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– Thomas Lesgourgues
Feb 7 at 11:48
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+1 jose carlos sir
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– jasmine
Feb 14 at 12:00
1
1
$begingroup$
Some information : math.stackexchange.com/questions/226413/…
$endgroup$
– Thomas Lesgourgues
Feb 7 at 11:48
$begingroup$
Some information : math.stackexchange.com/questions/226413/…
$endgroup$
– Thomas Lesgourgues
Feb 7 at 11:48
$begingroup$
+1 jose carlos sir
$endgroup$
– jasmine
Feb 14 at 12:00
$begingroup$
+1 jose carlos sir
$endgroup$
– jasmine
Feb 14 at 12:00
add a comment |
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1
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You can define partial orders on certain subclasses of groups. A fruitful one is the "largeness" ordering for finitely generated groups: $Gsucceq H$ ("$G$ is larger than $H$") if there exists a finite-index subgroup $Kleq_fG$ such that $Ktwoheadrightarrow H$. Exercise: Prove that free group $F_2$ is maximal here. Then a group $G$ is "large" or "large in the sense of Pride" (after Steve Pride, who first studied this ordering) if it is maximal in this ordering. For fruitfulness, here is a picture of Ian Agol announcing his proof of the virtual Haken conjecture.
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– user1729
Feb 7 at 15:30
13
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Note that groups are really superfluous here: the same idea suggests that Zorn's lemma should yield a maximal set. This helps make it clearer that the class/set issue is absolutely fundamental.
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– Noah Schweber
Feb 7 at 15:34
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You might be able to apply the Hausdorff maximal principle instead, if you can generalize it to classes.
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– PyRulez
Feb 7 at 18:03
8
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I lost you at "set of all groups".
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– WillO
Feb 8 at 0:31