Constructing a group from any set of groups [duplicate]
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This question already has an answer here:
Why is the collection of all groups a proper class rather than a set?
1 answer
How can we construct a group from any non-empty set of groups that is not in the original set (i.e. not isomorphic to any of its elements)?
group-theory
edited Dec 18 '18 at 20:58
Andrés E. Caicedo
65.5k8158249
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marked as duplicate by Matthew Towers, Arturo Magidin group-theory
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Dec 18 '18 at 22:13
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Why is the collection of all groups a proper class rather than a set?
1 answer
How can we construct a group from any non-empty set of groups that is not in the original set (i.e. not isomorphic to any of its elements)?
group-theory
edited Dec 18 '18 at 20:58
Andrés E. Caicedo
65.5k8158249
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marked as duplicate by Matthew Towers, Arturo Magidin group-theory
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Dec 18 '18 at 22:13
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
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Interestingly enough, essentially the same question was asked almost simultaneously with yours. Perhaps someone is having a final?
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– Arturo Magidin
Dec 18 '18 at 22:13
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wow... wish I waited before asking haha. finals? how i miss those days
$endgroup$
– user371732
Dec 19 '18 at 20:02
add a comment |
$begingroup$
This question already has an answer here:
Why is the collection of all groups a proper class rather than a set?
1 answer
How can we construct a group from any non-empty set of groups that is not in the original set (i.e. not isomorphic to any of its elements)?
group-theory
edited Dec 18 '18 at 20:58
Andrés E. Caicedo
65.5k8158249
$endgroup$
This question already has an answer here:
Why is the collection of all groups a proper class rather than a set?
1 answer
How can we construct a group from any non-empty set of groups that is not in the original set (i.e. not isomorphic to any of its elements)?
This question already has an answer here:
Why is the collection of all groups a proper class rather than a set?
1 answer
group-theory
group-theory
edited Dec 18 '18 at 20:58
Andrés E. Caicedo
65.5k8158249
edited Dec 18 '18 at 20:58
Andrés E. Caicedo
65.5k8158249
edited Dec 18 '18 at 20:58
Andrés E. Caicedo
65.5k8158249
edited Dec 18 '18 at 20:58
Andrés E. Caicedo
65.5k8158249
edited Dec 18 '18 at 20:58
Andrés E. Caicedo
65.5k8158249
65.5k8158249
asked Dec 18 '18 at 20:14
user371732
marked as duplicate by Matthew Towers, Arturo Magidin group-theory
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Dec 18 '18 at 22:13
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Matthew Towers, Arturo Magidin group-theory
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Dec 18 '18 at 22:13
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
Interestingly enough, essentially the same question was asked almost simultaneously with yours. Perhaps someone is having a final?
$endgroup$
– Arturo Magidin
Dec 18 '18 at 22:13
$begingroup$
wow... wish I waited before asking haha. finals? how i miss those days
$endgroup$
– user371732
Dec 19 '18 at 20:02
add a comment |
1
$begingroup$
Interestingly enough, essentially the same question was asked almost simultaneously with yours. Perhaps someone is having a final?
$endgroup$
– Arturo Magidin
Dec 18 '18 at 22:13
$begingroup$
wow... wish I waited before asking haha. finals? how i miss those days
$endgroup$
– user371732
Dec 19 '18 at 20:02
1
1
$begingroup$
Interestingly enough, essentially the same question was asked almost simultaneously with yours. Perhaps someone is having a final?
$endgroup$
– Arturo Magidin
Dec 18 '18 at 22:13
$begingroup$
Interestingly enough, essentially the same question was asked almost simultaneously with yours. Perhaps someone is having a final?
$endgroup$
– Arturo Magidin
Dec 18 '18 at 22:13
$begingroup$
wow... wish I waited before asking haha. finals? how i miss those days
$endgroup$
– user371732
Dec 19 '18 at 20:02
$begingroup$
wow... wish I waited before asking haha. finals? how i miss those days
$endgroup$
– user371732
Dec 19 '18 at 20:02
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Given a a set $X$ of groups take a group $G$ s.t. $|G|>|x|$ for all $xin X$. Then
$Gnotcong x $ for $xin X$, because there cannot be any bijection $Gto x$, and an isomorphism of groups must always be a bijection.
Only relevant to the previous versions of your question:
Your question needs clarification (without requiring any relation of the group to be constructed with the groups that are given your question is trivially answered). As it is stated now, I can say the following:
A group structure can be defined on any nonempty set. So given any set (whether the elements are groups or not), you can find a group structure on it: If it is finite, use the structure of $mathbb Z_n$, if countably infinite, the structure of $mathbb Z$ will do, and for all other cardinalities an application of the Löwenheim-Skolem theorem will give you the existence of the group structure.
(This was an answer pertaining to a part of the original question that was edited out:) Since you might have a finite set of infinite groups you cannot guarantee that the group you construct is isomorphic to one of the elements of your given set.
answered Dec 18 '18 at 20:26
Card_TrickCard_Trick
935
$endgroup$
$begingroup$
My bad ... edited!
$endgroup$
– user371732
Dec 18 '18 at 20:30
$begingroup$
well I don't want any group, I want a group guaranteed to NOT be isomorphic to any of the groups in the set. So I can't use a fixed group as an answer like Z or Zn
$endgroup$
– user371732
Dec 18 '18 at 20:52
$begingroup$
why is it that G is not isomorphic to any element in X?
$endgroup$
– user371732
Dec 18 '18 at 21:25
1
$begingroup$
so if you have a set X of cardinality c, and you construct a group of cardinality d>c, then you're implying none of the elements in X has cardinality d. how is this true?
$endgroup$
– user371732
Dec 18 '18 at 22:10
$begingroup$
My bad, this was obviously wrong. I changed it.
$endgroup$
– Card_Trick
Dec 18 '18 at 22:11
add a comment |
$begingroup$
If the given set of groups is the empty set, it is impossible to find a Group that is (isomorphic to) an element of this set.
With the problematic case of the empty et of Groups excluded, the Axiom of Choice allows us to pick an element of any (non-empty) set of groups.
answered Dec 18 '18 at 20:19
Hagen von EitzenHagen von Eitzen
1643
$endgroup$
$begingroup$
My bad ... edited!
$endgroup$
– user371732
Dec 18 '18 at 20:33
$begingroup$
A group can't be empty because it must contain an identity element.
$endgroup$
– Gödel
Dec 18 '18 at 20:53
add a comment |
$begingroup$
Let $G$ be a set of groups and suppose that it is denumerable with $mathbb{Z}in G$. So, we can suppose that $G={G_n:ninomega}$. Define $+:Gtimes Grightarrow G$ such that $+(G_n,G_m)=G_{n+m}$. It's clear that $langle G,+rangle$ is a group and $Gsimeqmathbb{Z}$.
edited Dec 18 '18 at 20:31
Card_Trick
935
answered Dec 18 '18 at 20:26
GödelGödel
1,440319
$endgroup$
$begingroup$
My bad ... edited!
$endgroup$
– user371732
Dec 18 '18 at 20:32
add a comment |
$begingroup$
Since you have a set of groups, the set of their cardinalities is bounded above. Choose a cardinal that is strictly larger than that bound. Now construct a group $G$ whose underlying set has cardinality at least as large as that chosen cardinal (for example the group of permutations of that cardinal). It follows that $G$ is not isomorphic to any of your original groups.
answered Dec 18 '18 at 20:51
Lee MosherLee Mosher
49.7k33686
$endgroup$
$begingroup$
why is the new group not isomorphic to any of the original's elements?
$endgroup$
– user371732
Dec 18 '18 at 21:27
$begingroup$
are you saying i should pick a cardinal greater than the cardinality of the set itself or greater than the max order of a group in it?
$endgroup$
– user371732
Dec 18 '18 at 21:33
$begingroup$
I rewrote that sentence to clarify.
$endgroup$
– Lee Mosher
Dec 19 '18 at 0:41
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Given a a set $X$ of groups take a group $G$ s.t. $|G|>|x|$ for all $xin X$. Then
$Gnotcong x $ for $xin X$, because there cannot be any bijection $Gto x$, and an isomorphism of groups must always be a bijection.
Only relevant to the previous versions of your question:
Your question needs clarification (without requiring any relation of the group to be constructed with the groups that are given your question is trivially answered). As it is stated now, I can say the following:
A group structure can be defined on any nonempty set. So given any set (whether the elements are groups or not), you can find a group structure on it: If it is finite, use the structure of $mathbb Z_n$, if countably infinite, the structure of $mathbb Z$ will do, and for all other cardinalities an application of the Löwenheim-Skolem theorem will give you the existence of the group structure.
(This was an answer pertaining to a part of the original question that was edited out:) Since you might have a finite set of infinite groups you cannot guarantee that the group you construct is isomorphic to one of the elements of your given set.
answered Dec 18 '18 at 20:26
Card_TrickCard_Trick
935
$endgroup$
$begingroup$
My bad ... edited!
$endgroup$
– user371732
Dec 18 '18 at 20:30
$begingroup$
well I don't want any group, I want a group guaranteed to NOT be isomorphic to any of the groups in the set. So I can't use a fixed group as an answer like Z or Zn
$endgroup$
– user371732
Dec 18 '18 at 20:52
$begingroup$
why is it that G is not isomorphic to any element in X?
$endgroup$
– user371732
Dec 18 '18 at 21:25
1
$begingroup$
so if you have a set X of cardinality c, and you construct a group of cardinality d>c, then you're implying none of the elements in X has cardinality d. how is this true?
$endgroup$
– user371732
Dec 18 '18 at 22:10
$begingroup$
My bad, this was obviously wrong. I changed it.
$endgroup$
– Card_Trick
Dec 18 '18 at 22:11
add a comment |
$begingroup$
Given a a set $X$ of groups take a group $G$ s.t. $|G|>|x|$ for all $xin X$. Then
$Gnotcong x $ for $xin X$, because there cannot be any bijection $Gto x$, and an isomorphism of groups must always be a bijection.
Only relevant to the previous versions of your question:
Your question needs clarification (without requiring any relation of the group to be constructed with the groups that are given your question is trivially answered). As it is stated now, I can say the following:
A group structure can be defined on any nonempty set. So given any set (whether the elements are groups or not), you can find a group structure on it: If it is finite, use the structure of $mathbb Z_n$, if countably infinite, the structure of $mathbb Z$ will do, and for all other cardinalities an application of the Löwenheim-Skolem theorem will give you the existence of the group structure.
(This was an answer pertaining to a part of the original question that was edited out:) Since you might have a finite set of infinite groups you cannot guarantee that the group you construct is isomorphic to one of the elements of your given set.
answered Dec 18 '18 at 20:26
Card_TrickCard_Trick
935
$endgroup$
$begingroup$
My bad ... edited!
$endgroup$
– user371732
Dec 18 '18 at 20:30
$begingroup$
well I don't want any group, I want a group guaranteed to NOT be isomorphic to any of the groups in the set. So I can't use a fixed group as an answer like Z or Zn
$endgroup$
– user371732
Dec 18 '18 at 20:52
$begingroup$
why is it that G is not isomorphic to any element in X?
$endgroup$
– user371732
Dec 18 '18 at 21:25
1
$begingroup$
so if you have a set X of cardinality c, and you construct a group of cardinality d>c, then you're implying none of the elements in X has cardinality d. how is this true?
$endgroup$
– user371732
Dec 18 '18 at 22:10
$begingroup$
My bad, this was obviously wrong. I changed it.
$endgroup$
– Card_Trick
Dec 18 '18 at 22:11
add a comment |
$begingroup$
Given a a set $X$ of groups take a group $G$ s.t. $|G|>|x|$ for all $xin X$. Then
$Gnotcong x $ for $xin X$, because there cannot be any bijection $Gto x$, and an isomorphism of groups must always be a bijection.
Only relevant to the previous versions of your question:
Your question needs clarification (without requiring any relation of the group to be constructed with the groups that are given your question is trivially answered). As it is stated now, I can say the following:
A group structure can be defined on any nonempty set. So given any set (whether the elements are groups or not), you can find a group structure on it: If it is finite, use the structure of $mathbb Z_n$, if countably infinite, the structure of $mathbb Z$ will do, and for all other cardinalities an application of the Löwenheim-Skolem theorem will give you the existence of the group structure.
(This was an answer pertaining to a part of the original question that was edited out:) Since you might have a finite set of infinite groups you cannot guarantee that the group you construct is isomorphic to one of the elements of your given set.
answered Dec 18 '18 at 20:26
Card_TrickCard_Trick
935
$endgroup$
Given a a set $X$ of groups take a group $G$ s.t. $|G|>|x|$ for all $xin X$. Then
$Gnotcong x $ for $xin X$, because there cannot be any bijection $Gto x$, and an isomorphism of groups must always be a bijection.
Only relevant to the previous versions of your question:
Your question needs clarification (without requiring any relation of the group to be constructed with the groups that are given your question is trivially answered). As it is stated now, I can say the following:
A group structure can be defined on any nonempty set. So given any set (whether the elements are groups or not), you can find a group structure on it: If it is finite, use the structure of $mathbb Z_n$, if countably infinite, the structure of $mathbb Z$ will do, and for all other cardinalities an application of the Löwenheim-Skolem theorem will give you the existence of the group structure.
(This was an answer pertaining to a part of the original question that was edited out:) Since you might have a finite set of infinite groups you cannot guarantee that the group you construct is isomorphic to one of the elements of your given set.
answered Dec 18 '18 at 20:26
Card_TrickCard_Trick
935
edited Dec 18 '18 at 22:11
answered Dec 18 '18 at 20:26
Card_TrickCard_Trick
935
answered Dec 18 '18 at 20:26
Card_TrickCard_Trick
935
answered Dec 18 '18 at 20:26
Card_TrickCard_Trick
935
935
$begingroup$
My bad ... edited!
$endgroup$
– user371732
Dec 18 '18 at 20:30
$begingroup$
well I don't want any group, I want a group guaranteed to NOT be isomorphic to any of the groups in the set. So I can't use a fixed group as an answer like Z or Zn
$endgroup$
– user371732
Dec 18 '18 at 20:52
$begingroup$
why is it that G is not isomorphic to any element in X?
$endgroup$
– user371732
Dec 18 '18 at 21:25
1
$begingroup$
so if you have a set X of cardinality c, and you construct a group of cardinality d>c, then you're implying none of the elements in X has cardinality d. how is this true?
$endgroup$
– user371732
Dec 18 '18 at 22:10
$begingroup$
My bad, this was obviously wrong. I changed it.
$endgroup$
– Card_Trick
Dec 18 '18 at 22:11
add a comment |
$begingroup$
My bad ... edited!
$endgroup$
– user371732
Dec 18 '18 at 20:30
$begingroup$
well I don't want any group, I want a group guaranteed to NOT be isomorphic to any of the groups in the set. So I can't use a fixed group as an answer like Z or Zn
$endgroup$
– user371732
Dec 18 '18 at 20:52
$begingroup$
why is it that G is not isomorphic to any element in X?
$endgroup$
– user371732
Dec 18 '18 at 21:25
1
$begingroup$
so if you have a set X of cardinality c, and you construct a group of cardinality d>c, then you're implying none of the elements in X has cardinality d. how is this true?
$endgroup$
– user371732
Dec 18 '18 at 22:10
$begingroup$
My bad, this was obviously wrong. I changed it.
$endgroup$
– Card_Trick
Dec 18 '18 at 22:11
$begingroup$
My bad ... edited!
$endgroup$
– user371732
Dec 18 '18 at 20:30
$begingroup$
My bad ... edited!
$endgroup$
– user371732
Dec 18 '18 at 20:30
$begingroup$
well I don't want any group, I want a group guaranteed to NOT be isomorphic to any of the groups in the set. So I can't use a fixed group as an answer like Z or Zn
$endgroup$
– user371732
Dec 18 '18 at 20:52
$begingroup$
well I don't want any group, I want a group guaranteed to NOT be isomorphic to any of the groups in the set. So I can't use a fixed group as an answer like Z or Zn
$endgroup$
– user371732
Dec 18 '18 at 20:52
$begingroup$
why is it that G is not isomorphic to any element in X?
$endgroup$
– user371732
Dec 18 '18 at 21:25
$begingroup$
why is it that G is not isomorphic to any element in X?
$endgroup$
– user371732
Dec 18 '18 at 21:25
1
1
$begingroup$
so if you have a set X of cardinality c, and you construct a group of cardinality d>c, then you're implying none of the elements in X has cardinality d. how is this true?
$endgroup$
– user371732
Dec 18 '18 at 22:10
$begingroup$
so if you have a set X of cardinality c, and you construct a group of cardinality d>c, then you're implying none of the elements in X has cardinality d. how is this true?
$endgroup$
– user371732
Dec 18 '18 at 22:10
$begingroup$
My bad, this was obviously wrong. I changed it.
$endgroup$
– Card_Trick
Dec 18 '18 at 22:11
$begingroup$
My bad, this was obviously wrong. I changed it.
$endgroup$
– Card_Trick
Dec 18 '18 at 22:11
add a comment |
$begingroup$
If the given set of groups is the empty set, it is impossible to find a Group that is (isomorphic to) an element of this set.
With the problematic case of the empty et of Groups excluded, the Axiom of Choice allows us to pick an element of any (non-empty) set of groups.
answered Dec 18 '18 at 20:19
Hagen von EitzenHagen von Eitzen
1643
$endgroup$
$begingroup$
My bad ... edited!
$endgroup$
– user371732
Dec 18 '18 at 20:33
$begingroup$
A group can't be empty because it must contain an identity element.
$endgroup$
– Gödel
Dec 18 '18 at 20:53
add a comment |
$begingroup$
If the given set of groups is the empty set, it is impossible to find a Group that is (isomorphic to) an element of this set.
With the problematic case of the empty et of Groups excluded, the Axiom of Choice allows us to pick an element of any (non-empty) set of groups.
answered Dec 18 '18 at 20:19
Hagen von EitzenHagen von Eitzen
1643
$endgroup$
$begingroup$
My bad ... edited!
$endgroup$
– user371732
Dec 18 '18 at 20:33
$begingroup$
A group can't be empty because it must contain an identity element.
$endgroup$
– Gödel
Dec 18 '18 at 20:53
add a comment |
$begingroup$
If the given set of groups is the empty set, it is impossible to find a Group that is (isomorphic to) an element of this set.
With the problematic case of the empty et of Groups excluded, the Axiom of Choice allows us to pick an element of any (non-empty) set of groups.
answered Dec 18 '18 at 20:19
Hagen von EitzenHagen von Eitzen
1643
$endgroup$
If the given set of groups is the empty set, it is impossible to find a Group that is (isomorphic to) an element of this set.
With the problematic case of the empty et of Groups excluded, the Axiom of Choice allows us to pick an element of any (non-empty) set of groups.
answered Dec 18 '18 at 20:19
Hagen von EitzenHagen von Eitzen
1643
answered Dec 18 '18 at 20:19
Hagen von EitzenHagen von Eitzen
1643
answered Dec 18 '18 at 20:19
Hagen von EitzenHagen von Eitzen
1643
answered Dec 18 '18 at 20:19
Hagen von EitzenHagen von Eitzen
1643
1643
$begingroup$
My bad ... edited!
$endgroup$
– user371732
Dec 18 '18 at 20:33
$begingroup$
A group can't be empty because it must contain an identity element.
$endgroup$
– Gödel
Dec 18 '18 at 20:53
add a comment |
$begingroup$
My bad ... edited!
$endgroup$
– user371732
Dec 18 '18 at 20:33
$begingroup$
A group can't be empty because it must contain an identity element.
$endgroup$
– Gödel
Dec 18 '18 at 20:53
$begingroup$
My bad ... edited!
$endgroup$
– user371732
Dec 18 '18 at 20:33
$begingroup$
My bad ... edited!
$endgroup$
– user371732
Dec 18 '18 at 20:33
$begingroup$
A group can't be empty because it must contain an identity element.
$endgroup$
– Gödel
Dec 18 '18 at 20:53
$begingroup$
A group can't be empty because it must contain an identity element.
$endgroup$
– Gödel
Dec 18 '18 at 20:53
add a comment |
$begingroup$
Let $G$ be a set of groups and suppose that it is denumerable with $mathbb{Z}in G$. So, we can suppose that $G={G_n:ninomega}$. Define $+:Gtimes Grightarrow G$ such that $+(G_n,G_m)=G_{n+m}$. It's clear that $langle G,+rangle$ is a group and $Gsimeqmathbb{Z}$.
edited Dec 18 '18 at 20:31
Card_Trick
935
answered Dec 18 '18 at 20:26
GödelGödel
1,440319
$endgroup$
$begingroup$
My bad ... edited!
$endgroup$
– user371732
Dec 18 '18 at 20:32
add a comment |
$begingroup$
Let $G$ be a set of groups and suppose that it is denumerable with $mathbb{Z}in G$. So, we can suppose that $G={G_n:ninomega}$. Define $+:Gtimes Grightarrow G$ such that $+(G_n,G_m)=G_{n+m}$. It's clear that $langle G,+rangle$ is a group and $Gsimeqmathbb{Z}$.
edited Dec 18 '18 at 20:31
Card_Trick
935
answered Dec 18 '18 at 20:26
GödelGödel
1,440319
$endgroup$
$begingroup$
My bad ... edited!
$endgroup$
– user371732
Dec 18 '18 at 20:32
add a comment |
$begingroup$
Let $G$ be a set of groups and suppose that it is denumerable with $mathbb{Z}in G$. So, we can suppose that $G={G_n:ninomega}$. Define $+:Gtimes Grightarrow G$ such that $+(G_n,G_m)=G_{n+m}$. It's clear that $langle G,+rangle$ is a group and $Gsimeqmathbb{Z}$.
edited Dec 18 '18 at 20:31
Card_Trick
935
answered Dec 18 '18 at 20:26
GödelGödel
1,440319
$endgroup$
Let $G$ be a set of groups and suppose that it is denumerable with $mathbb{Z}in G$. So, we can suppose that $G={G_n:ninomega}$. Define $+:Gtimes Grightarrow G$ such that $+(G_n,G_m)=G_{n+m}$. It's clear that $langle G,+rangle$ is a group and $Gsimeqmathbb{Z}$.
edited Dec 18 '18 at 20:31
Card_Trick
935
answered Dec 18 '18 at 20:26
GödelGödel
1,440319
edited Dec 18 '18 at 20:31
Card_Trick
935
edited Dec 18 '18 at 20:31
Card_Trick
935
edited Dec 18 '18 at 20:31
Card_Trick
935
935
answered Dec 18 '18 at 20:26
GödelGödel
1,440319
answered Dec 18 '18 at 20:26
GödelGödel
1,440319
answered Dec 18 '18 at 20:26
GödelGödel
1,440319
1,440319
$begingroup$
My bad ... edited!
$endgroup$
– user371732
Dec 18 '18 at 20:32
add a comment |
$begingroup$
My bad ... edited!
$endgroup$
– user371732
Dec 18 '18 at 20:32
$begingroup$
My bad ... edited!
$endgroup$
– user371732
Dec 18 '18 at 20:32
$begingroup$
My bad ... edited!
$endgroup$
– user371732
Dec 18 '18 at 20:32
add a comment |
$begingroup$
Since you have a set of groups, the set of their cardinalities is bounded above. Choose a cardinal that is strictly larger than that bound. Now construct a group $G$ whose underlying set has cardinality at least as large as that chosen cardinal (for example the group of permutations of that cardinal). It follows that $G$ is not isomorphic to any of your original groups.
answered Dec 18 '18 at 20:51
Lee MosherLee Mosher
49.7k33686
$endgroup$
$begingroup$
why is the new group not isomorphic to any of the original's elements?
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– user371732
Dec 18 '18 at 21:27
$begingroup$
are you saying i should pick a cardinal greater than the cardinality of the set itself or greater than the max order of a group in it?
$endgroup$
– user371732
Dec 18 '18 at 21:33
$begingroup$
I rewrote that sentence to clarify.
$endgroup$
– Lee Mosher
Dec 19 '18 at 0:41
add a comment |
$begingroup$
Since you have a set of groups, the set of their cardinalities is bounded above. Choose a cardinal that is strictly larger than that bound. Now construct a group $G$ whose underlying set has cardinality at least as large as that chosen cardinal (for example the group of permutations of that cardinal). It follows that $G$ is not isomorphic to any of your original groups.
answered Dec 18 '18 at 20:51
Lee MosherLee Mosher
49.7k33686
$endgroup$
$begingroup$
why is the new group not isomorphic to any of the original's elements?
$endgroup$
– user371732
Dec 18 '18 at 21:27
$begingroup$
are you saying i should pick a cardinal greater than the cardinality of the set itself or greater than the max order of a group in it?
$endgroup$
– user371732
Dec 18 '18 at 21:33
$begingroup$
I rewrote that sentence to clarify.
$endgroup$
– Lee Mosher
Dec 19 '18 at 0:41
add a comment |
$begingroup$
Since you have a set of groups, the set of their cardinalities is bounded above. Choose a cardinal that is strictly larger than that bound. Now construct a group $G$ whose underlying set has cardinality at least as large as that chosen cardinal (for example the group of permutations of that cardinal). It follows that $G$ is not isomorphic to any of your original groups.
answered Dec 18 '18 at 20:51
Lee MosherLee Mosher
49.7k33686
$endgroup$
Since you have a set of groups, the set of their cardinalities is bounded above. Choose a cardinal that is strictly larger than that bound. Now construct a group $G$ whose underlying set has cardinality at least as large as that chosen cardinal (for example the group of permutations of that cardinal). It follows that $G$ is not isomorphic to any of your original groups.
answered Dec 18 '18 at 20:51
Lee MosherLee Mosher
49.7k33686
edited Dec 19 '18 at 0:41
answered Dec 18 '18 at 20:51
Lee MosherLee Mosher
49.7k33686
answered Dec 18 '18 at 20:51
Lee MosherLee Mosher
49.7k33686
answered Dec 18 '18 at 20:51
Lee MosherLee Mosher
49.7k33686
49.7k33686
$begingroup$
why is the new group not isomorphic to any of the original's elements?
$endgroup$
– user371732
Dec 18 '18 at 21:27
$begingroup$
are you saying i should pick a cardinal greater than the cardinality of the set itself or greater than the max order of a group in it?
$endgroup$
– user371732
Dec 18 '18 at 21:33
$begingroup$
I rewrote that sentence to clarify.
$endgroup$
– Lee Mosher
Dec 19 '18 at 0:41
add a comment |
$begingroup$
why is the new group not isomorphic to any of the original's elements?
$endgroup$
– user371732
Dec 18 '18 at 21:27
$begingroup$
are you saying i should pick a cardinal greater than the cardinality of the set itself or greater than the max order of a group in it?
$endgroup$
– user371732
Dec 18 '18 at 21:33
$begingroup$
I rewrote that sentence to clarify.
$endgroup$
– Lee Mosher
Dec 19 '18 at 0:41
$begingroup$
why is the new group not isomorphic to any of the original's elements?
$endgroup$
– user371732
Dec 18 '18 at 21:27
$begingroup$
why is the new group not isomorphic to any of the original's elements?
$endgroup$
– user371732
Dec 18 '18 at 21:27
$begingroup$
are you saying i should pick a cardinal greater than the cardinality of the set itself or greater than the max order of a group in it?
$endgroup$
– user371732
Dec 18 '18 at 21:33
$begingroup$
are you saying i should pick a cardinal greater than the cardinality of the set itself or greater than the max order of a group in it?
$endgroup$
– user371732
Dec 18 '18 at 21:33
$begingroup$
I rewrote that sentence to clarify.
$endgroup$
– Lee Mosher
Dec 19 '18 at 0:41
$begingroup$
I rewrote that sentence to clarify.
$endgroup$
– Lee Mosher
Dec 19 '18 at 0:41
add a comment |
1
$begingroup$
Interestingly enough, essentially the same question was asked almost simultaneously with yours. Perhaps someone is having a final?
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– Arturo Magidin
Dec 18 '18 at 22:13
$begingroup$
wow... wish I waited before asking haha. finals? how i miss those days
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– user371732
Dec 19 '18 at 20:02