Is there a simple group of any (infinite) size?












15














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







share|cite|improve this question
























  • 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.
    – Ryan Reich
    Jun 2 '13 at 5:58










  • yes, that seems what has to be done, but the problem is that Löwenheim-Skolem fails in second-order logic
    – Camilo Arosemena-Serrato
    Jun 2 '13 at 13:20






  • 1




    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.)
    – Andrés E. Caicedo
    Jun 3 '13 at 22:32


















15














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







share|cite|improve this question
























  • 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.
    – Ryan Reich
    Jun 2 '13 at 5:58










  • yes, that seems what has to be done, but the problem is that Löwenheim-Skolem fails in second-order logic
    – Camilo Arosemena-Serrato
    Jun 2 '13 at 13:20






  • 1




    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.)
    – Andrés E. Caicedo
    Jun 3 '13 at 22:32
















15












15








15


5





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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jul 9 '13 at 15:42

























asked Jun 2 '13 at 3:37









Camilo Arosemena-Serrato

5,63611848




5,63611848












  • 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.
    – Ryan Reich
    Jun 2 '13 at 5:58










  • yes, that seems what has to be done, but the problem is that Löwenheim-Skolem fails in second-order logic
    – Camilo Arosemena-Serrato
    Jun 2 '13 at 13:20






  • 1




    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.)
    – Andrés E. Caicedo
    Jun 3 '13 at 22:32




















  • 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.
    – Ryan Reich
    Jun 2 '13 at 5:58










  • yes, that seems what has to be done, but the problem is that Löwenheim-Skolem fails in second-order logic
    – Camilo Arosemena-Serrato
    Jun 2 '13 at 13:20






  • 1




    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.)
    – Andrés E. Caicedo
    Jun 3 '13 at 22:32


















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.
– Ryan Reich
Jun 2 '13 at 5:58




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.
– Ryan Reich
Jun 2 '13 at 5:58












yes, that seems what has to be done, but the problem is that Löwenheim-Skolem fails in second-order logic
– Camilo Arosemena-Serrato
Jun 2 '13 at 13:20




yes, that seems what has to be done, but the problem is that Löwenheim-Skolem fails in second-order logic
– Camilo Arosemena-Serrato
Jun 2 '13 at 13:20




1




1




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.)
– Andrés E. Caicedo
Jun 3 '13 at 22:32






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.)
– Andrés E. Caicedo
Jun 3 '13 at 22:32












1 Answer
1






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18














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 $kappa$.






share|cite|improve this answer





















  • Hi Trevor. Didn't know about Groupprops, thanks for the link!
    – Andrés E. Caicedo
    Jun 2 '13 at 6:58










  • @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.
    – Trevor Wilson
    Jun 2 '13 at 7:05













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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 $kappa$.






share|cite|improve this answer





















  • Hi Trevor. Didn't know about Groupprops, thanks for the link!
    – Andrés E. Caicedo
    Jun 2 '13 at 6:58










  • @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.
    – Trevor Wilson
    Jun 2 '13 at 7:05


















18














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 $kappa$.






share|cite|improve this answer





















  • Hi Trevor. Didn't know about Groupprops, thanks for the link!
    – Andrés E. Caicedo
    Jun 2 '13 at 6:58










  • @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.
    – Trevor Wilson
    Jun 2 '13 at 7:05
















18












18








18






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 $kappa$.






share|cite|improve this answer












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 $kappa$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jun 2 '13 at 5:25









Trevor Wilson

14.7k2456




14.7k2456












  • Hi Trevor. Didn't know about Groupprops, thanks for the link!
    – Andrés E. Caicedo
    Jun 2 '13 at 6:58










  • @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.
    – Trevor Wilson
    Jun 2 '13 at 7:05




















  • Hi Trevor. Didn't know about Groupprops, thanks for the link!
    – Andrés E. Caicedo
    Jun 2 '13 at 6:58










  • @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.
    – Trevor Wilson
    Jun 2 '13 at 7:05


















Hi Trevor. Didn't know about Groupprops, thanks for the link!
– Andrés E. Caicedo
Jun 2 '13 at 6:58




Hi Trevor. Didn't know about Groupprops, thanks for the link!
– Andrés E. Caicedo
Jun 2 '13 at 6:58












@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.
– Trevor Wilson
Jun 2 '13 at 7:05






@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.
– Trevor Wilson
Jun 2 '13 at 7:05




















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