Non empty set and group
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Let $E$ be a non empty set. How to prove that there exists $star:Etimes Erightarrow E$ for which $(E,star)$ is a group?
group-theory
asked Sep 8 at 9:44
Eric
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up vote
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Let $E$ be a non empty set. How to prove that there exists $star:Etimes Erightarrow E$ for which $(E,star)$ is a group?
group-theory
asked Sep 8 at 9:44
Eric
21
What are your thoughts about that?
– Taroccoesbrocco
Sep 8 at 9:48
This is equivalent to the axiom of choice. See mathoverflow.net/questions/12973/…
– user21793
Sep 8 at 10:24
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Let $E$ be a non empty set. How to prove that there exists $star:Etimes Erightarrow E$ for which $(E,star)$ is a group?
group-theory
asked Sep 8 at 9:44
Eric
21
Let $E$ be a non empty set. How to prove that there exists $star:Etimes Erightarrow E$ for which $(E,star)$ is a group?
group-theory
group-theory
asked Sep 8 at 9:44
Eric
21
asked Sep 8 at 9:44
Eric
21
asked Sep 8 at 9:44
Eric
21
asked Sep 8 at 9:44
Eric
21
asked Sep 8 at 9:44
Eric
21
21
What are your thoughts about that?
– Taroccoesbrocco
Sep 8 at 9:48
This is equivalent to the axiom of choice. See mathoverflow.net/questions/12973/…
– user21793
Sep 8 at 10:24
add a comment |Â
What are your thoughts about that?
– Taroccoesbrocco
Sep 8 at 9:48
This is equivalent to the axiom of choice. See mathoverflow.net/questions/12973/…
– user21793
Sep 8 at 10:24
What are your thoughts about that?
– Taroccoesbrocco
Sep 8 at 9:48
What are your thoughts about that?
– Taroccoesbrocco
Sep 8 at 9:48
This is equivalent to the axiom of choice. See mathoverflow.net/questions/12973/…
– user21793
Sep 8 at 10:24
This is equivalent to the axiom of choice. See mathoverflow.net/questions/12973/…
– user21793
Sep 8 at 10:24
add a comment |Â
1 Answer
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If $E$ is finite, you can just identify $E$ with $Bbb Z/nBbb Z$, where $n=|E|$.
If $E$ is countably infinite, you can similarly identify it with $Bbb Z$ (or $Bbb Q$ for that matter).
More generally, for any infinite cardinality of $E$, you can consider the set of groups with carrier sets $subseteq E$, ordered by inclusion. Using Zorn's lemma, there is a maximal such group $M$. Conclude that $|M|>|Esetminus M|$ (and hence $|M|=|E|$, as desired) because otherwise you could build $Mhookrightarrow MoplusBbb Z/2Bbb Zsubseteq E$, contradicting maximality.
answered Sep 8 at 9:50
Hagen von Eitzen
267k21260483
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
If $E$ is finite, you can just identify $E$ with $Bbb Z/nBbb Z$, where $n=|E|$.
If $E$ is countably infinite, you can similarly identify it with $Bbb Z$ (or $Bbb Q$ for that matter).
More generally, for any infinite cardinality of $E$, you can consider the set of groups with carrier sets $subseteq E$, ordered by inclusion. Using Zorn's lemma, there is a maximal such group $M$. Conclude that $|M|>|Esetminus M|$ (and hence $|M|=|E|$, as desired) because otherwise you could build $Mhookrightarrow MoplusBbb Z/2Bbb Zsubseteq E$, contradicting maximality.
answered Sep 8 at 9:50
Hagen von Eitzen
267k21260483
add a comment |Â
up vote
5
down vote
If $E$ is finite, you can just identify $E$ with $Bbb Z/nBbb Z$, where $n=|E|$.
If $E$ is countably infinite, you can similarly identify it with $Bbb Z$ (or $Bbb Q$ for that matter).
More generally, for any infinite cardinality of $E$, you can consider the set of groups with carrier sets $subseteq E$, ordered by inclusion. Using Zorn's lemma, there is a maximal such group $M$. Conclude that $|M|>|Esetminus M|$ (and hence $|M|=|E|$, as desired) because otherwise you could build $Mhookrightarrow MoplusBbb Z/2Bbb Zsubseteq E$, contradicting maximality.
answered Sep 8 at 9:50
Hagen von Eitzen
267k21260483
add a comment |Â
up vote
5
down vote
up vote
5
down vote
If $E$ is finite, you can just identify $E$ with $Bbb Z/nBbb Z$, where $n=|E|$.
If $E$ is countably infinite, you can similarly identify it with $Bbb Z$ (or $Bbb Q$ for that matter).
More generally, for any infinite cardinality of $E$, you can consider the set of groups with carrier sets $subseteq E$, ordered by inclusion. Using Zorn's lemma, there is a maximal such group $M$. Conclude that $|M|>|Esetminus M|$ (and hence $|M|=|E|$, as desired) because otherwise you could build $Mhookrightarrow MoplusBbb Z/2Bbb Zsubseteq E$, contradicting maximality.
answered Sep 8 at 9:50
Hagen von Eitzen
267k21260483
If $E$ is finite, you can just identify $E$ with $Bbb Z/nBbb Z$, where $n=|E|$.
If $E$ is countably infinite, you can similarly identify it with $Bbb Z$ (or $Bbb Q$ for that matter).
More generally, for any infinite cardinality of $E$, you can consider the set of groups with carrier sets $subseteq E$, ordered by inclusion. Using Zorn's lemma, there is a maximal such group $M$. Conclude that $|M|>|Esetminus M|$ (and hence $|M|=|E|$, as desired) because otherwise you could build $Mhookrightarrow MoplusBbb Z/2Bbb Zsubseteq E$, contradicting maximality.
answered Sep 8 at 9:50
Hagen von Eitzen
267k21260483
answered Sep 8 at 9:50
Hagen von Eitzen
267k21260483
answered Sep 8 at 9:50
Hagen von Eitzen
267k21260483
answered Sep 8 at 9:50
Hagen von Eitzen
267k21260483
267k21260483
add a comment |Â
add a comment |Â
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What are your thoughts about that?
– Taroccoesbrocco
Sep 8 at 9:48
This is equivalent to the axiom of choice. See mathoverflow.net/questions/12973/…
– user21793
Sep 8 at 10:24