Subgroup cardinal
$begingroup$
Given the group $(G,*)$. $H$ is a subgroup of G. How can you prove that $lvert G setminus H rvert ge lvert H rvert$ ? Can Lagrange's Theorem be helpful for the proof?
group-theory
edited Dec 1 '18 at 16:18
Andreas Caranti
56.6k34395
asked Dec 1 '18 at 14:18
GaboruGaboru
877
$endgroup$
add a comment |
$begingroup$
Given the group $(G,*)$. $H$ is a subgroup of G. How can you prove that $lvert G setminus H rvert ge lvert H rvert$ ? Can Lagrange's Theorem be helpful for the proof?
group-theory
edited Dec 1 '18 at 16:18
Andreas Caranti
56.6k34395
asked Dec 1 '18 at 14:18
GaboruGaboru
877
$endgroup$
$begingroup$
First, you need to assume that $H$ is a proper subgroup or it will not be true. Then the proof of Lagrange's theorem will indeed directly show this.
$endgroup$
– Tobias Kildetoft
Dec 1 '18 at 14:21
add a comment |
$begingroup$
Given the group $(G,*)$. $H$ is a subgroup of G. How can you prove that $lvert G setminus H rvert ge lvert H rvert$ ? Can Lagrange's Theorem be helpful for the proof?
group-theory
edited Dec 1 '18 at 16:18
Andreas Caranti
56.6k34395
asked Dec 1 '18 at 14:18
GaboruGaboru
877
$endgroup$
Given the group $(G,*)$. $H$ is a subgroup of G. How can you prove that $lvert G setminus H rvert ge lvert H rvert$ ? Can Lagrange's Theorem be helpful for the proof?
group-theory
group-theory
edited Dec 1 '18 at 16:18
Andreas Caranti
56.6k34395
asked Dec 1 '18 at 14:18
GaboruGaboru
877
edited Dec 1 '18 at 16:18
Andreas Caranti
56.6k34395
asked Dec 1 '18 at 14:18
GaboruGaboru
877
edited Dec 1 '18 at 16:18
Andreas Caranti
56.6k34395
edited Dec 1 '18 at 16:18
Andreas Caranti
56.6k34395
edited Dec 1 '18 at 16:18
Andreas Caranti
56.6k34395
56.6k34395
asked Dec 1 '18 at 14:18
GaboruGaboru
877
asked Dec 1 '18 at 14:18
GaboruGaboru
877
asked Dec 1 '18 at 14:18
GaboruGaboru
877
877
$begingroup$
First, you need to assume that $H$ is a proper subgroup or it will not be true. Then the proof of Lagrange's theorem will indeed directly show this.
$endgroup$
– Tobias Kildetoft
Dec 1 '18 at 14:21
add a comment |
$begingroup$
First, you need to assume that $H$ is a proper subgroup or it will not be true. Then the proof of Lagrange's theorem will indeed directly show this.
$endgroup$
– Tobias Kildetoft
Dec 1 '18 at 14:21
$begingroup$
First, you need to assume that $H$ is a proper subgroup or it will not be true. Then the proof of Lagrange's theorem will indeed directly show this.
$endgroup$
– Tobias Kildetoft
Dec 1 '18 at 14:21
$begingroup$
First, you need to assume that $H$ is a proper subgroup or it will not be true. Then the proof of Lagrange's theorem will indeed directly show this.
$endgroup$
– Tobias Kildetoft
Dec 1 '18 at 14:21
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You have to assume it is a proper subgroup or the statement is false. Then there is some element $g in G-H$. It is not the identity because the identity is in $H$. If you form $gh$ for any $hin H$ it will not be in $H$. If it were you could form $ghh^{-1}=g$ and say it must be in $H$. The elements $gh$ for $hin H$ are all distinct, so we have an injection from $H$ into $G-H$
answered Dec 1 '18 at 14:27
Ross MillikanRoss Millikan
296k23198371
$endgroup$
$begingroup$
This only really makes sense for finite groups.
$endgroup$
– Tobias Kildetoft
Dec 1 '18 at 14:28
$begingroup$
@TobiasKildetoft: good point. I added that.
$endgroup$
– Ross Millikan
Dec 1 '18 at 14:30
$begingroup$
To make this work for infinite groups, you just need to use the formulation in terms of cosets instead and it still works.
$endgroup$
– Tobias Kildetoft
Dec 1 '18 at 14:33
1
$begingroup$
@TobiasKildetoft: Yes, I realized that and replaced the answer.
$endgroup$
– Ross Millikan
Dec 1 '18 at 14:34
$begingroup$
How can you prove that there are 2 elements in GH which are commutative? Knowing that H is a necommutative subgroup of G.
$endgroup$
– Gaboru
Dec 1 '18 at 14:38
|
show 2 more comments
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%2f3021397%2fsubgroup-cardinal%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
$begingroup$
You have to assume it is a proper subgroup or the statement is false. Then there is some element $g in G-H$. It is not the identity because the identity is in $H$. If you form $gh$ for any $hin H$ it will not be in $H$. If it were you could form $ghh^{-1}=g$ and say it must be in $H$. The elements $gh$ for $hin H$ are all distinct, so we have an injection from $H$ into $G-H$
answered Dec 1 '18 at 14:27
Ross MillikanRoss Millikan
296k23198371
$endgroup$
$begingroup$
This only really makes sense for finite groups.
$endgroup$
– Tobias Kildetoft
Dec 1 '18 at 14:28
$begingroup$
@TobiasKildetoft: good point. I added that.
$endgroup$
– Ross Millikan
Dec 1 '18 at 14:30
$begingroup$
To make this work for infinite groups, you just need to use the formulation in terms of cosets instead and it still works.
$endgroup$
– Tobias Kildetoft
Dec 1 '18 at 14:33
1
$begingroup$
@TobiasKildetoft: Yes, I realized that and replaced the answer.
$endgroup$
– Ross Millikan
Dec 1 '18 at 14:34
$begingroup$
How can you prove that there are 2 elements in GH which are commutative? Knowing that H is a necommutative subgroup of G.
$endgroup$
– Gaboru
Dec 1 '18 at 14:38
|
show 2 more comments
$begingroup$
You have to assume it is a proper subgroup or the statement is false. Then there is some element $g in G-H$. It is not the identity because the identity is in $H$. If you form $gh$ for any $hin H$ it will not be in $H$. If it were you could form $ghh^{-1}=g$ and say it must be in $H$. The elements $gh$ for $hin H$ are all distinct, so we have an injection from $H$ into $G-H$
answered Dec 1 '18 at 14:27
Ross MillikanRoss Millikan
296k23198371
$endgroup$
$begingroup$
This only really makes sense for finite groups.
$endgroup$
– Tobias Kildetoft
Dec 1 '18 at 14:28
$begingroup$
@TobiasKildetoft: good point. I added that.
$endgroup$
– Ross Millikan
Dec 1 '18 at 14:30
$begingroup$
To make this work for infinite groups, you just need to use the formulation in terms of cosets instead and it still works.
$endgroup$
– Tobias Kildetoft
Dec 1 '18 at 14:33
1
$begingroup$
@TobiasKildetoft: Yes, I realized that and replaced the answer.
$endgroup$
– Ross Millikan
Dec 1 '18 at 14:34
$begingroup$
How can you prove that there are 2 elements in GH which are commutative? Knowing that H is a necommutative subgroup of G.
$endgroup$
– Gaboru
Dec 1 '18 at 14:38
|
show 2 more comments
$begingroup$
You have to assume it is a proper subgroup or the statement is false. Then there is some element $g in G-H$. It is not the identity because the identity is in $H$. If you form $gh$ for any $hin H$ it will not be in $H$. If it were you could form $ghh^{-1}=g$ and say it must be in $H$. The elements $gh$ for $hin H$ are all distinct, so we have an injection from $H$ into $G-H$
answered Dec 1 '18 at 14:27
Ross MillikanRoss Millikan
296k23198371
$endgroup$
You have to assume it is a proper subgroup or the statement is false. Then there is some element $g in G-H$. It is not the identity because the identity is in $H$. If you form $gh$ for any $hin H$ it will not be in $H$. If it were you could form $ghh^{-1}=g$ and say it must be in $H$. The elements $gh$ for $hin H$ are all distinct, so we have an injection from $H$ into $G-H$
answered Dec 1 '18 at 14:27
Ross MillikanRoss Millikan
296k23198371
edited Dec 1 '18 at 14:34
answered Dec 1 '18 at 14:27
Ross MillikanRoss Millikan
296k23198371
answered Dec 1 '18 at 14:27
Ross MillikanRoss Millikan
296k23198371
answered Dec 1 '18 at 14:27
Ross MillikanRoss Millikan
296k23198371
296k23198371
$begingroup$
This only really makes sense for finite groups.
$endgroup$
– Tobias Kildetoft
Dec 1 '18 at 14:28
$begingroup$
@TobiasKildetoft: good point. I added that.
$endgroup$
– Ross Millikan
Dec 1 '18 at 14:30
$begingroup$
To make this work for infinite groups, you just need to use the formulation in terms of cosets instead and it still works.
$endgroup$
– Tobias Kildetoft
Dec 1 '18 at 14:33
1
$begingroup$
@TobiasKildetoft: Yes, I realized that and replaced the answer.
$endgroup$
– Ross Millikan
Dec 1 '18 at 14:34
$begingroup$
How can you prove that there are 2 elements in GH which are commutative? Knowing that H is a necommutative subgroup of G.
$endgroup$
– Gaboru
Dec 1 '18 at 14:38
|
show 2 more comments
$begingroup$
This only really makes sense for finite groups.
$endgroup$
– Tobias Kildetoft
Dec 1 '18 at 14:28
$begingroup$
@TobiasKildetoft: good point. I added that.
$endgroup$
– Ross Millikan
Dec 1 '18 at 14:30
$begingroup$
To make this work for infinite groups, you just need to use the formulation in terms of cosets instead and it still works.
$endgroup$
– Tobias Kildetoft
Dec 1 '18 at 14:33
1
$begingroup$
@TobiasKildetoft: Yes, I realized that and replaced the answer.
$endgroup$
– Ross Millikan
Dec 1 '18 at 14:34
$begingroup$
How can you prove that there are 2 elements in GH which are commutative? Knowing that H is a necommutative subgroup of G.
$endgroup$
– Gaboru
Dec 1 '18 at 14:38
$begingroup$
This only really makes sense for finite groups.
$endgroup$
– Tobias Kildetoft
Dec 1 '18 at 14:28
$begingroup$
This only really makes sense for finite groups.
$endgroup$
– Tobias Kildetoft
Dec 1 '18 at 14:28
$begingroup$
@TobiasKildetoft: good point. I added that.
$endgroup$
– Ross Millikan
Dec 1 '18 at 14:30
$begingroup$
@TobiasKildetoft: good point. I added that.
$endgroup$
– Ross Millikan
Dec 1 '18 at 14:30
$begingroup$
To make this work for infinite groups, you just need to use the formulation in terms of cosets instead and it still works.
$endgroup$
– Tobias Kildetoft
Dec 1 '18 at 14:33
$begingroup$
To make this work for infinite groups, you just need to use the formulation in terms of cosets instead and it still works.
$endgroup$
– Tobias Kildetoft
Dec 1 '18 at 14:33
1
1
$begingroup$
@TobiasKildetoft: Yes, I realized that and replaced the answer.
$endgroup$
– Ross Millikan
Dec 1 '18 at 14:34
$begingroup$
@TobiasKildetoft: Yes, I realized that and replaced the answer.
$endgroup$
– Ross Millikan
Dec 1 '18 at 14:34
$begingroup$
How can you prove that there are 2 elements in GH which are commutative? Knowing that H is a necommutative subgroup of G.
$endgroup$
– Gaboru
Dec 1 '18 at 14:38
$begingroup$
How can you prove that there are 2 elements in GH which are commutative? Knowing that H is a necommutative subgroup of G.
$endgroup$
– Gaboru
Dec 1 '18 at 14:38
|
show 2 more comments
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%2f3021397%2fsubgroup-cardinal%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
$begingroup$
First, you need to assume that $H$ is a proper subgroup or it will not be true. Then the proof of Lagrange's theorem will indeed directly show this.
$endgroup$
– Tobias Kildetoft
Dec 1 '18 at 14:21