Is there a simple group of any (infinite) size?
$begingroup$
I'm trying to show that for any infinite cardinal $kappa$ there is a simple group $G$ of size $kappa$, I tried to use the compactness theorem and then ascending Löwenheim-Skolem, but this is impossible as the following argument shows:
Suppose $F$ is a set of sentences in a first-order language containing the language of groups such that for any group $G$ we have $GvDash F$ iff $G$ is a simple group, let $varphi$ be a sentence such that $GvDash varphi $ iff $G$ is abelian. Then as for each prime $p$, $mathbb Z_p$ is simple and abelian, by the compactness theorem there is an infinite abelian simple group $G$, and thus using Löwenheim-Skolem we obtain an uncountable simple abelian group $G$. Pick a non-zero element $ain G$, then $langle a rangle$ is a proper normal subgroup of $G$. Contradiction.
Note that the contradiction is not obtained by the existence of an infinite abelian simple group, something which I don't whether it's true or not, but by the assumption of the axiomatization of the simple groups, and the use of Lowënheim-Skolem theorem.
Now, is there a way to prove this assertion?
Thanks.
group-theory logic simple-groups
asked Jun 2 '13 at 3:37
Camilo Arosemena-SerratoCamilo Arosemena-Serrato
5,64611849
$endgroup$
add a comment |
$begingroup$
I'm trying to show that for any infinite cardinal $kappa$ there is a simple group $G$ of size $kappa$, I tried to use the compactness theorem and then ascending Löwenheim-Skolem, but this is impossible as the following argument shows:
Suppose $F$ is a set of sentences in a first-order language containing the language of groups such that for any group $G$ we have $GvDash F$ iff $G$ is a simple group, let $varphi$ be a sentence such that $GvDash varphi $ iff $G$ is abelian. Then as for each prime $p$, $mathbb Z_p$ is simple and abelian, by the compactness theorem there is an infinite abelian simple group $G$, and thus using Löwenheim-Skolem we obtain an uncountable simple abelian group $G$. Pick a non-zero element $ain G$, then $langle a rangle$ is a proper normal subgroup of $G$. Contradiction.
Note that the contradiction is not obtained by the existence of an infinite abelian simple group, something which I don't whether it's true or not, but by the assumption of the axiomatization of the simple groups, and the use of Lowënheim-Skolem theorem.
Now, is there a way to prove this assertion?
Thanks.
group-theory logic simple-groups
asked Jun 2 '13 at 3:37
Camilo Arosemena-SerratoCamilo Arosemena-Serrato
5,64611849
$endgroup$
$begingroup$
Pardon my ignorance of logic, but is there such a set of sentences $F$? It would appear necessary to quantify over all subgroups of $G$, which is a second-order operation.
$endgroup$
– Ryan Reich
Jun 2 '13 at 5:58
$begingroup$
yes, that seems what has to be done, but the problem is that Löwenheim-Skolem fails in second-order logic
$endgroup$
– Camilo Arosemena-Serrato
Jun 2 '13 at 13:20
1
$begingroup$
A deleted answer included the following link with nice additional information: mathoverflow.net/questions/32908/… (In particular, Simon's answer shows that simplicity is not first-order.)
$endgroup$
– Andrés E. Caicedo
Jun 3 '13 at 22:32
add a comment |
$begingroup$
I'm trying to show that for any infinite cardinal $kappa$ there is a simple group $G$ of size $kappa$, I tried to use the compactness theorem and then ascending Löwenheim-Skolem, but this is impossible as the following argument shows:
Suppose $F$ is a set of sentences in a first-order language containing the language of groups such that for any group $G$ we have $GvDash F$ iff $G$ is a simple group, let $varphi$ be a sentence such that $GvDash varphi $ iff $G$ is abelian. Then as for each prime $p$, $mathbb Z_p$ is simple and abelian, by the compactness theorem there is an infinite abelian simple group $G$, and thus using Löwenheim-Skolem we obtain an uncountable simple abelian group $G$. Pick a non-zero element $ain G$, then $langle a rangle$ is a proper normal subgroup of $G$. Contradiction.
Note that the contradiction is not obtained by the existence of an infinite abelian simple group, something which I don't whether it's true or not, but by the assumption of the axiomatization of the simple groups, and the use of Lowënheim-Skolem theorem.
Now, is there a way to prove this assertion?
Thanks.
group-theory logic simple-groups
asked Jun 2 '13 at 3:37
Camilo Arosemena-SerratoCamilo Arosemena-Serrato
5,64611849
$endgroup$
I'm trying to show that for any infinite cardinal $kappa$ there is a simple group $G$ of size $kappa$, I tried to use the compactness theorem and then ascending Löwenheim-Skolem, but this is impossible as the following argument shows:
Suppose $F$ is a set of sentences in a first-order language containing the language of groups such that for any group $G$ we have $GvDash F$ iff $G$ is a simple group, let $varphi$ be a sentence such that $GvDash varphi $ iff $G$ is abelian. Then as for each prime $p$, $mathbb Z_p$ is simple and abelian, by the compactness theorem there is an infinite abelian simple group $G$, and thus using Löwenheim-Skolem we obtain an uncountable simple abelian group $G$. Pick a non-zero element $ain G$, then $langle a rangle$ is a proper normal subgroup of $G$. Contradiction.
Note that the contradiction is not obtained by the existence of an infinite abelian simple group, something which I don't whether it's true or not, but by the assumption of the axiomatization of the simple groups, and the use of Lowënheim-Skolem theorem.
Now, is there a way to prove this assertion?
Thanks.
group-theory logic simple-groups
group-theory logic simple-groups
asked Jun 2 '13 at 3:37
Camilo Arosemena-SerratoCamilo Arosemena-Serrato
5,64611849
asked Jun 2 '13 at 3:37
Camilo Arosemena-SerratoCamilo Arosemena-Serrato
5,64611849
edited Jul 9 '13 at 15:42
Camilo Arosemena-Serrato
asked Jun 2 '13 at 3:37
Camilo Arosemena-SerratoCamilo Arosemena-Serrato
5,64611849
asked Jun 2 '13 at 3:37
Camilo Arosemena-SerratoCamilo Arosemena-Serrato
5,64611849
asked Jun 2 '13 at 3:37
Camilo Arosemena-SerratoCamilo Arosemena-Serrato
5,64611849
5,64611849
$begingroup$
Pardon my ignorance of logic, but is there such a set of sentences $F$? It would appear necessary to quantify over all subgroups of $G$, which is a second-order operation.
$endgroup$
– Ryan Reich
Jun 2 '13 at 5:58
$begingroup$
yes, that seems what has to be done, but the problem is that Löwenheim-Skolem fails in second-order logic
$endgroup$
– Camilo Arosemena-Serrato
Jun 2 '13 at 13:20
1
$begingroup$
A deleted answer included the following link with nice additional information: mathoverflow.net/questions/32908/… (In particular, Simon's answer shows that simplicity is not first-order.)
$endgroup$
– Andrés E. Caicedo
Jun 3 '13 at 22:32
add a comment |
$begingroup$
Pardon my ignorance of logic, but is there such a set of sentences $F$? It would appear necessary to quantify over all subgroups of $G$, which is a second-order operation.
$endgroup$
– Ryan Reich
Jun 2 '13 at 5:58
$begingroup$
yes, that seems what has to be done, but the problem is that Löwenheim-Skolem fails in second-order logic
$endgroup$
– Camilo Arosemena-Serrato
Jun 2 '13 at 13:20
1
$begingroup$
A deleted answer included the following link with nice additional information: mathoverflow.net/questions/32908/… (In particular, Simon's answer shows that simplicity is not first-order.)
$endgroup$
– Andrés E. Caicedo
Jun 3 '13 at 22:32
$begingroup$
Pardon my ignorance of logic, but is there such a set of sentences $F$? It would appear necessary to quantify over all subgroups of $G$, which is a second-order operation.
$endgroup$
– Ryan Reich
Jun 2 '13 at 5:58
$begingroup$
Pardon my ignorance of logic, but is there such a set of sentences $F$? It would appear necessary to quantify over all subgroups of $G$, which is a second-order operation.
$endgroup$
– Ryan Reich
Jun 2 '13 at 5:58
$begingroup$
yes, that seems what has to be done, but the problem is that Löwenheim-Skolem fails in second-order logic
$endgroup$
– Camilo Arosemena-Serrato
Jun 2 '13 at 13:20
$begingroup$
yes, that seems what has to be done, but the problem is that Löwenheim-Skolem fails in second-order logic
$endgroup$
– Camilo Arosemena-Serrato
Jun 2 '13 at 13:20
1
1
$begingroup$
A deleted answer included the following link with nice additional information: mathoverflow.net/questions/32908/… (In particular, Simon's answer shows that simplicity is not first-order.)
$endgroup$
– Andrés E. Caicedo
Jun 3 '13 at 22:32
$begingroup$
A deleted answer included the following link with nice additional information: mathoverflow.net/questions/32908/… (In particular, Simon's answer shows that simplicity is not first-order.)
$endgroup$
– Andrés E. Caicedo
Jun 3 '13 at 22:32
add a comment |
1 Answer
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oldest
votes
$begingroup$
For any (finite or infinite) cardinal $kappa$, if $kappa ge 5$ then the finitary alternating group $A(kappa)$ is simple. This is the group consisting of permutations of $kappa$ that have finite support and are even. (If $kappa$ is finite then this is just the ordinary alternating group $A_kappa$.) If $kappa$ is infinite, then $A(kappa)$ has cardinality $kappa$.
answered Jun 2 '13 at 5:25
Trevor WilsonTrevor Wilson
14.8k2456
$endgroup$
$begingroup$
Hi Trevor. Didn't know about Groupprops, thanks for the link!
$endgroup$
– Andrés E. Caicedo
Jun 2 '13 at 6:58
$begingroup$
@Andres Hi Andres, I didn't know about it either; I just heard someone mention this fact in a talk today -- then I happened to see this question and so I googled "finitary alternating group" or something like that and it turned up.
$endgroup$
– Trevor Wilson
Jun 2 '13 at 7:05
add a comment |
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1 Answer
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$begingroup$
For any (finite or infinite) cardinal $kappa$, if $kappa ge 5$ then the finitary alternating group $A(kappa)$ is simple. This is the group consisting of permutations of $kappa$ that have finite support and are even. (If $kappa$ is finite then this is just the ordinary alternating group $A_kappa$.) If $kappa$ is infinite, then $A(kappa)$ has cardinality $kappa$.
answered Jun 2 '13 at 5:25
Trevor WilsonTrevor Wilson
14.8k2456
$endgroup$
$begingroup$
Hi Trevor. Didn't know about Groupprops, thanks for the link!
$endgroup$
– Andrés E. Caicedo
Jun 2 '13 at 6:58
$begingroup$
@Andres Hi Andres, I didn't know about it either; I just heard someone mention this fact in a talk today -- then I happened to see this question and so I googled "finitary alternating group" or something like that and it turned up.
$endgroup$
– Trevor Wilson
Jun 2 '13 at 7:05
add a comment |
$begingroup$
For any (finite or infinite) cardinal $kappa$, if $kappa ge 5$ then the finitary alternating group $A(kappa)$ is simple. This is the group consisting of permutations of $kappa$ that have finite support and are even. (If $kappa$ is finite then this is just the ordinary alternating group $A_kappa$.) If $kappa$ is infinite, then $A(kappa)$ has cardinality $kappa$.
answered Jun 2 '13 at 5:25
Trevor WilsonTrevor Wilson
14.8k2456
$endgroup$
$begingroup$
Hi Trevor. Didn't know about Groupprops, thanks for the link!
$endgroup$
– Andrés E. Caicedo
Jun 2 '13 at 6:58
$begingroup$
@Andres Hi Andres, I didn't know about it either; I just heard someone mention this fact in a talk today -- then I happened to see this question and so I googled "finitary alternating group" or something like that and it turned up.
$endgroup$
– Trevor Wilson
Jun 2 '13 at 7:05
add a comment |
$begingroup$
For any (finite or infinite) cardinal $kappa$, if $kappa ge 5$ then the finitary alternating group $A(kappa)$ is simple. This is the group consisting of permutations of $kappa$ that have finite support and are even. (If $kappa$ is finite then this is just the ordinary alternating group $A_kappa$.) If $kappa$ is infinite, then $A(kappa)$ has cardinality $kappa$.
answered Jun 2 '13 at 5:25
Trevor WilsonTrevor Wilson
14.8k2456
$endgroup$
For any (finite or infinite) cardinal $kappa$, if $kappa ge 5$ then the finitary alternating group $A(kappa)$ is simple. This is the group consisting of permutations of $kappa$ that have finite support and are even. (If $kappa$ is finite then this is just the ordinary alternating group $A_kappa$.) If $kappa$ is infinite, then $A(kappa)$ has cardinality $kappa$.
answered Jun 2 '13 at 5:25
Trevor WilsonTrevor Wilson
14.8k2456
answered Jun 2 '13 at 5:25
Trevor WilsonTrevor Wilson
14.8k2456
answered Jun 2 '13 at 5:25
Trevor WilsonTrevor Wilson
14.8k2456
answered Jun 2 '13 at 5:25
Trevor WilsonTrevor Wilson
14.8k2456
14.8k2456
$begingroup$
Hi Trevor. Didn't know about Groupprops, thanks for the link!
$endgroup$
– Andrés E. Caicedo
Jun 2 '13 at 6:58
$begingroup$
@Andres Hi Andres, I didn't know about it either; I just heard someone mention this fact in a talk today -- then I happened to see this question and so I googled "finitary alternating group" or something like that and it turned up.
$endgroup$
– Trevor Wilson
Jun 2 '13 at 7:05
add a comment |
$begingroup$
Hi Trevor. Didn't know about Groupprops, thanks for the link!
$endgroup$
– Andrés E. Caicedo
Jun 2 '13 at 6:58
$begingroup$
@Andres Hi Andres, I didn't know about it either; I just heard someone mention this fact in a talk today -- then I happened to see this question and so I googled "finitary alternating group" or something like that and it turned up.
$endgroup$
– Trevor Wilson
Jun 2 '13 at 7:05
$begingroup$
Hi Trevor. Didn't know about Groupprops, thanks for the link!
$endgroup$
– Andrés E. Caicedo
Jun 2 '13 at 6:58
$begingroup$
Hi Trevor. Didn't know about Groupprops, thanks for the link!
$endgroup$
– Andrés E. Caicedo
Jun 2 '13 at 6:58
$begingroup$
@Andres Hi Andres, I didn't know about it either; I just heard someone mention this fact in a talk today -- then I happened to see this question and so I googled "finitary alternating group" or something like that and it turned up.
$endgroup$
– Trevor Wilson
Jun 2 '13 at 7:05
$begingroup$
@Andres Hi Andres, I didn't know about it either; I just heard someone mention this fact in a talk today -- then I happened to see this question and so I googled "finitary alternating group" or something like that and it turned up.
$endgroup$
– Trevor Wilson
Jun 2 '13 at 7:05
add a comment |
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$begingroup$
Pardon my ignorance of logic, but is there such a set of sentences $F$? It would appear necessary to quantify over all subgroups of $G$, which is a second-order operation.
$endgroup$
– Ryan Reich
Jun 2 '13 at 5:58
$begingroup$
yes, that seems what has to be done, but the problem is that Löwenheim-Skolem fails in second-order logic
$endgroup$
– Camilo Arosemena-Serrato
Jun 2 '13 at 13:20
1
$begingroup$
A deleted answer included the following link with nice additional information: mathoverflow.net/questions/32908/… (In particular, Simon's answer shows that simplicity is not first-order.)
$endgroup$
– Andrés E. Caicedo
Jun 3 '13 at 22:32