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
5,63611849
$endgroup$
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
5,63611849
$endgroup$
$begingroup$
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
5,63611849
$endgroup$
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
5,63611849
asked Jun 2 '13 at 3:37
Camilo Arosemena-SerratoCamilo Arosemena-Serrato
5,63611849
edited Jul 9 '13 at 15:42
Camilo Arosemena-Serrato
asked Jun 2 '13 at 3:37
Camilo Arosemena-SerratoCamilo Arosemena-Serrato
5,63611849
asked Jun 2 '13 at 3:37
Camilo Arosemena-SerratoCamilo Arosemena-Serrato
5,63611849
asked Jun 2 '13 at 3:37
Camilo Arosemena-SerratoCamilo Arosemena-Serrato
5,63611849
5,63611849
$begingroup$
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
1
active
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
14.7k2456
$endgroup$
$begingroup$
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 |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f409000%2fis-there-a-simple-group-of-any-infinite-size%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
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
14.7k2456
$endgroup$
$begingroup$
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
14.7k2456
$endgroup$
$begingroup$
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
14.7k2456
$endgroup$
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
14.7k2456
answered Jun 2 '13 at 5:25
Trevor WilsonTrevor Wilson
14.7k2456
answered Jun 2 '13 at 5:25
Trevor WilsonTrevor Wilson
14.7k2456
answered Jun 2 '13 at 5:25
Trevor WilsonTrevor Wilson
14.7k2456
14.7k2456
$begingroup$
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f409000%2fis-there-a-simple-group-of-any-infinite-size%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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