For any set Q of groups, why does there exist a group G that is not isomorphic to any group in Q? [duplicate]
This question already has an answer here:
Why is the collection of all groups a proper class rather than a set?
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If any set of groups is allowed, wouldn't the set of all groups disprove the statement?
group-theory elementary-set-theory
edited Dec 18 at 20:59
Andrés E. Caicedo
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asked Dec 18 at 20:38
Rahul Chowdhury
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marked as duplicate by Asaf Karagila♦ elementary-set-theory
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Dec 18 at 21:41
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 |
This question already has an answer here:
Why is the collection of all groups a proper class rather than a set?
1 answer
If any set of groups is allowed, wouldn't the set of all groups disprove the statement?
group-theory elementary-set-theory
edited Dec 18 at 20:59
Andrés E. Caicedo
64.7k8158246
asked Dec 18 at 20:38
Rahul Chowdhury
1865
marked as duplicate by Asaf Karagila♦ elementary-set-theory
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Dec 18 at 21:41
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.
5
There is no set of all groups, just as there is no set of all sets.
– Lord Shark the Unknown
Dec 18 at 20:39
add a comment |
This question already has an answer here:
Why is the collection of all groups a proper class rather than a set?
1 answer
If any set of groups is allowed, wouldn't the set of all groups disprove the statement?
group-theory elementary-set-theory
edited Dec 18 at 20:59
Andrés E. Caicedo
64.7k8158246
asked Dec 18 at 20:38
Rahul Chowdhury
1865
This question already has an answer here:
Why is the collection of all groups a proper class rather than a set?
1 answer
If any set of groups is allowed, wouldn't the set of all groups disprove the statement?
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 elementary-set-theory
group-theory elementary-set-theory
edited Dec 18 at 20:59
Andrés E. Caicedo
64.7k8158246
asked Dec 18 at 20:38
Rahul Chowdhury
1865
edited Dec 18 at 20:59
Andrés E. Caicedo
64.7k8158246
asked Dec 18 at 20:38
Rahul Chowdhury
1865
edited Dec 18 at 20:59
Andrés E. Caicedo
64.7k8158246
edited Dec 18 at 20:59
Andrés E. Caicedo
64.7k8158246
edited Dec 18 at 20:59
Andrés E. Caicedo
64.7k8158246
64.7k8158246
asked Dec 18 at 20:38
Rahul Chowdhury
1865
asked Dec 18 at 20:38
Rahul Chowdhury
1865
asked Dec 18 at 20:38
Rahul Chowdhury
1865
1865
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Dec 18 at 21:41
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Dec 18 at 21:41
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.
5
There is no set of all groups, just as there is no set of all sets.
– Lord Shark the Unknown
Dec 18 at 20:39
add a comment |
5
There is no set of all groups, just as there is no set of all sets.
– Lord Shark the Unknown
Dec 18 at 20:39
5
5
There is no set of all groups, just as there is no set of all sets.
– Lord Shark the Unknown
Dec 18 at 20:39
There is no set of all groups, just as there is no set of all sets.
– Lord Shark the Unknown
Dec 18 at 20:39
add a comment |
1 Answer
1
active
oldest
votes
Assume that $Q$ is a set of groups. Since it is a set then there exists a cardinal number $mathcal{K}$ such that $mathcal{K}>|G|$ for any $Gin Q$. The cardinal number is given by first taking $X=bigcup_{Gin Q} G$ (yes, we can take the union because $Q$ is a set) and then $mathcal{K}=|2^{X}|$ implies that $mathcal{K}>|G|$ for any $Gin Q$. (Cantor's theorem).
Now take $S(mathcal{K})$ meaning the group of all bijections $mathcal{K}tomathcal{K}$ (or more precisely bijections $Xto X$ where $|X|=mathcal{K}$, this doesn't depend on the choice of $X$). You can easily check that $|S(mathcal{K})|geqmathcal{K}$ and thus $S(mathcal{K})$ cannot be isomorphic to any $Gin Q$.
This also shows that there is no such thing as "the set of all groups" even up to isomorphism. Just like there is no such thing as "the set of all sets". The class of all groups is not a set.
answered Dec 18 at 20:43
freakish
11.2k1628
1
Why should there exist such a cardinal number $mathcal K$?
– Omnomnomnom
Dec 18 at 20:46
3
@Omnomnomnom Take $X=bigcup_{Gin Q}G$ and then put $mathcal{K}=|2^X|$.
– freakish
Dec 18 at 20:49
Ah okay, I see now that the union only needs to be a set, not another group
– Omnomnomnom
Dec 18 at 20:50
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Assume that $Q$ is a set of groups. Since it is a set then there exists a cardinal number $mathcal{K}$ such that $mathcal{K}>|G|$ for any $Gin Q$. The cardinal number is given by first taking $X=bigcup_{Gin Q} G$ (yes, we can take the union because $Q$ is a set) and then $mathcal{K}=|2^{X}|$ implies that $mathcal{K}>|G|$ for any $Gin Q$. (Cantor's theorem).
Now take $S(mathcal{K})$ meaning the group of all bijections $mathcal{K}tomathcal{K}$ (or more precisely bijections $Xto X$ where $|X|=mathcal{K}$, this doesn't depend on the choice of $X$). You can easily check that $|S(mathcal{K})|geqmathcal{K}$ and thus $S(mathcal{K})$ cannot be isomorphic to any $Gin Q$.
This also shows that there is no such thing as "the set of all groups" even up to isomorphism. Just like there is no such thing as "the set of all sets". The class of all groups is not a set.
answered Dec 18 at 20:43
freakish
11.2k1628
1
Why should there exist such a cardinal number $mathcal K$?
– Omnomnomnom
Dec 18 at 20:46
3
@Omnomnomnom Take $X=bigcup_{Gin Q}G$ and then put $mathcal{K}=|2^X|$.
– freakish
Dec 18 at 20:49
Ah okay, I see now that the union only needs to be a set, not another group
– Omnomnomnom
Dec 18 at 20:50
add a comment |
Assume that $Q$ is a set of groups. Since it is a set then there exists a cardinal number $mathcal{K}$ such that $mathcal{K}>|G|$ for any $Gin Q$. The cardinal number is given by first taking $X=bigcup_{Gin Q} G$ (yes, we can take the union because $Q$ is a set) and then $mathcal{K}=|2^{X}|$ implies that $mathcal{K}>|G|$ for any $Gin Q$. (Cantor's theorem).
Now take $S(mathcal{K})$ meaning the group of all bijections $mathcal{K}tomathcal{K}$ (or more precisely bijections $Xto X$ where $|X|=mathcal{K}$, this doesn't depend on the choice of $X$). You can easily check that $|S(mathcal{K})|geqmathcal{K}$ and thus $S(mathcal{K})$ cannot be isomorphic to any $Gin Q$.
This also shows that there is no such thing as "the set of all groups" even up to isomorphism. Just like there is no such thing as "the set of all sets". The class of all groups is not a set.
answered Dec 18 at 20:43
freakish
11.2k1628
1
Why should there exist such a cardinal number $mathcal K$?
– Omnomnomnom
Dec 18 at 20:46
3
@Omnomnomnom Take $X=bigcup_{Gin Q}G$ and then put $mathcal{K}=|2^X|$.
– freakish
Dec 18 at 20:49
Ah okay, I see now that the union only needs to be a set, not another group
– Omnomnomnom
Dec 18 at 20:50
add a comment |
Assume that $Q$ is a set of groups. Since it is a set then there exists a cardinal number $mathcal{K}$ such that $mathcal{K}>|G|$ for any $Gin Q$. The cardinal number is given by first taking $X=bigcup_{Gin Q} G$ (yes, we can take the union because $Q$ is a set) and then $mathcal{K}=|2^{X}|$ implies that $mathcal{K}>|G|$ for any $Gin Q$. (Cantor's theorem).
Now take $S(mathcal{K})$ meaning the group of all bijections $mathcal{K}tomathcal{K}$ (or more precisely bijections $Xto X$ where $|X|=mathcal{K}$, this doesn't depend on the choice of $X$). You can easily check that $|S(mathcal{K})|geqmathcal{K}$ and thus $S(mathcal{K})$ cannot be isomorphic to any $Gin Q$.
This also shows that there is no such thing as "the set of all groups" even up to isomorphism. Just like there is no such thing as "the set of all sets". The class of all groups is not a set.
answered Dec 18 at 20:43
freakish
11.2k1628
Assume that $Q$ is a set of groups. Since it is a set then there exists a cardinal number $mathcal{K}$ such that $mathcal{K}>|G|$ for any $Gin Q$. The cardinal number is given by first taking $X=bigcup_{Gin Q} G$ (yes, we can take the union because $Q$ is a set) and then $mathcal{K}=|2^{X}|$ implies that $mathcal{K}>|G|$ for any $Gin Q$. (Cantor's theorem).
Now take $S(mathcal{K})$ meaning the group of all bijections $mathcal{K}tomathcal{K}$ (or more precisely bijections $Xto X$ where $|X|=mathcal{K}$, this doesn't depend on the choice of $X$). You can easily check that $|S(mathcal{K})|geqmathcal{K}$ and thus $S(mathcal{K})$ cannot be isomorphic to any $Gin Q$.
This also shows that there is no such thing as "the set of all groups" even up to isomorphism. Just like there is no such thing as "the set of all sets". The class of all groups is not a set.
answered Dec 18 at 20:43
freakish
11.2k1628
edited Dec 19 at 10:01
answered Dec 18 at 20:43
freakish
11.2k1628
answered Dec 18 at 20:43
freakish
11.2k1628
answered Dec 18 at 20:43
freakish
11.2k1628
11.2k1628
1
Why should there exist such a cardinal number $mathcal K$?
– Omnomnomnom
Dec 18 at 20:46
3
@Omnomnomnom Take $X=bigcup_{Gin Q}G$ and then put $mathcal{K}=|2^X|$.
– freakish
Dec 18 at 20:49
Ah okay, I see now that the union only needs to be a set, not another group
– Omnomnomnom
Dec 18 at 20:50
add a comment |
1
Why should there exist such a cardinal number $mathcal K$?
– Omnomnomnom
Dec 18 at 20:46
3
@Omnomnomnom Take $X=bigcup_{Gin Q}G$ and then put $mathcal{K}=|2^X|$.
– freakish
Dec 18 at 20:49
Ah okay, I see now that the union only needs to be a set, not another group
– Omnomnomnom
Dec 18 at 20:50
1
1
Why should there exist such a cardinal number $mathcal K$?
– Omnomnomnom
Dec 18 at 20:46
Why should there exist such a cardinal number $mathcal K$?
– Omnomnomnom
Dec 18 at 20:46
3
3
@Omnomnomnom Take $X=bigcup_{Gin Q}G$ and then put $mathcal{K}=|2^X|$.
– freakish
Dec 18 at 20:49
@Omnomnomnom Take $X=bigcup_{Gin Q}G$ and then put $mathcal{K}=|2^X|$.
– freakish
Dec 18 at 20:49
Ah okay, I see now that the union only needs to be a set, not another group
– Omnomnomnom
Dec 18 at 20:50
Ah okay, I see now that the union only needs to be a set, not another group
– Omnomnomnom
Dec 18 at 20:50
add a comment |
5
There is no set of all groups, just as there is no set of all sets.
– Lord Shark the Unknown
Dec 18 at 20:39