Mixing
A subgroup $H$ of $G$ must contain the identity from $G$
$begingroup$
I was given the following definition of a subgroup: Let $G$ be a group and let $H$ be a subset of $G$. We say that $H$ is a subgroup of $G$ if $H$ is a group under the same operation from $G$.
I was told that this easily implies that the identity from $G$ must also be in $H$, but I can't see this. I get that the subset must then be closed on the same operation. I get that the operation is associative. I get that there must be some identity, but I can't see how the identity in $H$ must be the same as the identity in $G$.
abstract-algebra group-theory definition
asked Jan 29 at 15:11
John DoeJohn Doe
25121346
$endgroup$
add a comment |
$begingroup$
I was given the following definition of a subgroup: Let $G$ be a group and let $H$ be a subset of $G$. We say that $H$ is a subgroup of $G$ if $H$ is a group under the same operation from $G$.
I was told that this easily implies that the identity from $G$ must also be in $H$, but I can't see this. I get that the subset must then be closed on the same operation. I get that the operation is associative. I get that there must be some identity, but I can't see how the identity in $H$ must be the same as the identity in $G$.
abstract-algebra group-theory definition
asked Jan 29 at 15:11
John DoeJohn Doe
25121346
$endgroup$
add a comment |
$begingroup$
I was given the following definition of a subgroup: Let $G$ be a group and let $H$ be a subset of $G$. We say that $H$ is a subgroup of $G$ if $H$ is a group under the same operation from $G$.
I was told that this easily implies that the identity from $G$ must also be in $H$, but I can't see this. I get that the subset must then be closed on the same operation. I get that the operation is associative. I get that there must be some identity, but I can't see how the identity in $H$ must be the same as the identity in $G$.
abstract-algebra group-theory definition
asked Jan 29 at 15:11
John DoeJohn Doe
25121346
$endgroup$
I was given the following definition of a subgroup: Let $G$ be a group and let $H$ be a subset of $G$. We say that $H$ is a subgroup of $G$ if $H$ is a group under the same operation from $G$.
I was told that this easily implies that the identity from $G$ must also be in $H$, but I can't see this. I get that the subset must then be closed on the same operation. I get that the operation is associative. I get that there must be some identity, but I can't see how the identity in $H$ must be the same as the identity in $G$.
abstract-algebra group-theory definition
abstract-algebra group-theory definition
asked Jan 29 at 15:11
John DoeJohn Doe
25121346
asked Jan 29 at 15:11
John DoeJohn Doe
25121346
asked Jan 29 at 15:11
John DoeJohn Doe
25121346
asked Jan 29 at 15:11
John DoeJohn Doe
25121346
asked Jan 29 at 15:11
John DoeJohn Doe
25121346
25121346
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Assume $e_H$ is an identity in $H.$ Then $e_H*e_H=e_H.$ Multiplying both sides with the inverse of $e_H$ in $G$ we get $e_H=e_G,$ where $e_G$ is the identity in $H.$
edited Jan 29 at 15:18
kccu
10.8k11229
answered Jan 29 at 15:14
Reiner MartinReiner Martin
3,509414
$endgroup$
1
$begingroup$
Nice way of seeing it.
$endgroup$
– John Doe
Jan 29 at 15:17
add a comment |
$begingroup$
Since $H$ is not empty, $hin H$; now $hh^{-1}=1_H=1_G$.
edited Jan 29 at 17:44
Shaun
9,870113684
answered Jan 29 at 15:12
Tsemo AristideTsemo Aristide
60.1k11446
$endgroup$
3
$begingroup$
Hmm, I think there is a problem in this answer. How do you know that $h^{-1}$ is in $H$?
$endgroup$
– Mark
Jan 29 at 15:18
$begingroup$
Seems like missing the point. Since $H$ is a group it has an identity - the question is why $1_h=1_G$. Not that that's hard, of course...
$endgroup$
– David C. Ullrich
Jan 29 at 15:19
2
$begingroup$
@Mark $H$ is a subgroup, so we know inverses in $H$ exist, but we don't know a priori that the inverse of $h$ in $H$ is the same as the inverse of $h$ in $G$.
$endgroup$
– kccu
Jan 29 at 15:20
1
$begingroup$
@kccu, yes, this is exactly what I wrote in the comment. $h^{-1}$ is the inverse of $h$ in $G$. It is not that trivial that it equals to the inverse in $H$. (easy to prove that, but it still has to be proved)
$endgroup$
– Mark
Jan 29 at 15:21
$begingroup$
I think that for $h$ having an inverse you should first show that $e$ is in $H$ or assume since it is a group not the other way around.
$endgroup$
– Dani
Jan 29 at 15:24
add a comment |
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%2f3092290%2fa-subgroup-h-of-g-must-contain-the-identity-from-g%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Assume $e_H$ is an identity in $H.$ Then $e_H*e_H=e_H.$ Multiplying both sides with the inverse of $e_H$ in $G$ we get $e_H=e_G,$ where $e_G$ is the identity in $H.$
edited Jan 29 at 15:18
kccu
10.8k11229
answered Jan 29 at 15:14
Reiner MartinReiner Martin
3,509414
$endgroup$
1
$begingroup$
Nice way of seeing it.
$endgroup$
– John Doe
Jan 29 at 15:17
add a comment |
$begingroup$
Assume $e_H$ is an identity in $H.$ Then $e_H*e_H=e_H.$ Multiplying both sides with the inverse of $e_H$ in $G$ we get $e_H=e_G,$ where $e_G$ is the identity in $H.$
edited Jan 29 at 15:18
kccu
10.8k11229
answered Jan 29 at 15:14
Reiner MartinReiner Martin
3,509414
$endgroup$
1
$begingroup$
Nice way of seeing it.
$endgroup$
– John Doe
Jan 29 at 15:17
add a comment |
$begingroup$
Assume $e_H$ is an identity in $H.$ Then $e_H*e_H=e_H.$ Multiplying both sides with the inverse of $e_H$ in $G$ we get $e_H=e_G,$ where $e_G$ is the identity in $H.$
edited Jan 29 at 15:18
kccu
10.8k11229
answered Jan 29 at 15:14
Reiner MartinReiner Martin
3,509414
$endgroup$
Assume $e_H$ is an identity in $H.$ Then $e_H*e_H=e_H.$ Multiplying both sides with the inverse of $e_H$ in $G$ we get $e_H=e_G,$ where $e_G$ is the identity in $H.$
edited Jan 29 at 15:18
kccu
10.8k11229
answered Jan 29 at 15:14
Reiner MartinReiner Martin
3,509414
edited Jan 29 at 15:18
kccu
10.8k11229
edited Jan 29 at 15:18
kccu
10.8k11229
edited Jan 29 at 15:18
kccu
10.8k11229
10.8k11229
answered Jan 29 at 15:14
Reiner MartinReiner Martin
3,509414
answered Jan 29 at 15:14
Reiner MartinReiner Martin
3,509414
answered Jan 29 at 15:14
Reiner MartinReiner Martin
3,509414
3,509414
1
$begingroup$
Nice way of seeing it.
$endgroup$
– John Doe
Jan 29 at 15:17
add a comment |
1
$begingroup$
Nice way of seeing it.
$endgroup$
– John Doe
Jan 29 at 15:17
1
1
$begingroup$
Nice way of seeing it.
$endgroup$
– John Doe
Jan 29 at 15:17
$begingroup$
Nice way of seeing it.
$endgroup$
– John Doe
Jan 29 at 15:17
add a comment |
$begingroup$
Since $H$ is not empty, $hin H$; now $hh^{-1}=1_H=1_G$.
edited Jan 29 at 17:44
Shaun
9,870113684
answered Jan 29 at 15:12
Tsemo AristideTsemo Aristide
60.1k11446
$endgroup$
3
$begingroup$
Hmm, I think there is a problem in this answer. How do you know that $h^{-1}$ is in $H$?
$endgroup$
– Mark
Jan 29 at 15:18
$begingroup$
Seems like missing the point. Since $H$ is a group it has an identity - the question is why $1_h=1_G$. Not that that's hard, of course...
$endgroup$
– David C. Ullrich
Jan 29 at 15:19
2
$begingroup$
@Mark $H$ is a subgroup, so we know inverses in $H$ exist, but we don't know a priori that the inverse of $h$ in $H$ is the same as the inverse of $h$ in $G$.
$endgroup$
– kccu
Jan 29 at 15:20
1
$begingroup$
@kccu, yes, this is exactly what I wrote in the comment. $h^{-1}$ is the inverse of $h$ in $G$. It is not that trivial that it equals to the inverse in $H$. (easy to prove that, but it still has to be proved)
$endgroup$
– Mark
Jan 29 at 15:21
$begingroup$
I think that for $h$ having an inverse you should first show that $e$ is in $H$ or assume since it is a group not the other way around.
$endgroup$
– Dani
Jan 29 at 15:24
add a comment |
$begingroup$
Since $H$ is not empty, $hin H$; now $hh^{-1}=1_H=1_G$.
edited Jan 29 at 17:44
Shaun
9,870113684
answered Jan 29 at 15:12
Tsemo AristideTsemo Aristide
60.1k11446
$endgroup$
3
$begingroup$
Hmm, I think there is a problem in this answer. How do you know that $h^{-1}$ is in $H$?
$endgroup$
– Mark
Jan 29 at 15:18
$begingroup$
Seems like missing the point. Since $H$ is a group it has an identity - the question is why $1_h=1_G$. Not that that's hard, of course...
$endgroup$
– David C. Ullrich
Jan 29 at 15:19
2
$begingroup$
@Mark $H$ is a subgroup, so we know inverses in $H$ exist, but we don't know a priori that the inverse of $h$ in $H$ is the same as the inverse of $h$ in $G$.
$endgroup$
– kccu
Jan 29 at 15:20
1
$begingroup$
@kccu, yes, this is exactly what I wrote in the comment. $h^{-1}$ is the inverse of $h$ in $G$. It is not that trivial that it equals to the inverse in $H$. (easy to prove that, but it still has to be proved)
$endgroup$
– Mark
Jan 29 at 15:21
$begingroup$
I think that for $h$ having an inverse you should first show that $e$ is in $H$ or assume since it is a group not the other way around.
$endgroup$
– Dani
Jan 29 at 15:24
add a comment |
$begingroup$
Since $H$ is not empty, $hin H$; now $hh^{-1}=1_H=1_G$.
edited Jan 29 at 17:44
Shaun
9,870113684
answered Jan 29 at 15:12
Tsemo AristideTsemo Aristide
60.1k11446
$endgroup$
Since $H$ is not empty, $hin H$; now $hh^{-1}=1_H=1_G$.
edited Jan 29 at 17:44
Shaun
9,870113684
answered Jan 29 at 15:12
Tsemo AristideTsemo Aristide
60.1k11446
edited Jan 29 at 17:44
Shaun
9,870113684
edited Jan 29 at 17:44
Shaun
9,870113684
edited Jan 29 at 17:44
Shaun
9,870113684
9,870113684
answered Jan 29 at 15:12
Tsemo AristideTsemo Aristide
60.1k11446
answered Jan 29 at 15:12
Tsemo AristideTsemo Aristide
60.1k11446
answered Jan 29 at 15:12
Tsemo AristideTsemo Aristide
60.1k11446
60.1k11446
3
$begingroup$
Hmm, I think there is a problem in this answer. How do you know that $h^{-1}$ is in $H$?
$endgroup$
– Mark
Jan 29 at 15:18
$begingroup$
Seems like missing the point. Since $H$ is a group it has an identity - the question is why $1_h=1_G$. Not that that's hard, of course...
$endgroup$
– David C. Ullrich
Jan 29 at 15:19
2
$begingroup$
@Mark $H$ is a subgroup, so we know inverses in $H$ exist, but we don't know a priori that the inverse of $h$ in $H$ is the same as the inverse of $h$ in $G$.
$endgroup$
– kccu
Jan 29 at 15:20
1
$begingroup$
@kccu, yes, this is exactly what I wrote in the comment. $h^{-1}$ is the inverse of $h$ in $G$. It is not that trivial that it equals to the inverse in $H$. (easy to prove that, but it still has to be proved)
$endgroup$
– Mark
Jan 29 at 15:21
$begingroup$
I think that for $h$ having an inverse you should first show that $e$ is in $H$ or assume since it is a group not the other way around.
$endgroup$
– Dani
Jan 29 at 15:24
add a comment |
3
$begingroup$
Hmm, I think there is a problem in this answer. How do you know that $h^{-1}$ is in $H$?
$endgroup$
– Mark
Jan 29 at 15:18
$begingroup$
Seems like missing the point. Since $H$ is a group it has an identity - the question is why $1_h=1_G$. Not that that's hard, of course...
$endgroup$
– David C. Ullrich
Jan 29 at 15:19
2
$begingroup$
@Mark $H$ is a subgroup, so we know inverses in $H$ exist, but we don't know a priori that the inverse of $h$ in $H$ is the same as the inverse of $h$ in $G$.
$endgroup$
– kccu
Jan 29 at 15:20
1
$begingroup$
@kccu, yes, this is exactly what I wrote in the comment. $h^{-1}$ is the inverse of $h$ in $G$. It is not that trivial that it equals to the inverse in $H$. (easy to prove that, but it still has to be proved)
$endgroup$
– Mark
Jan 29 at 15:21
$begingroup$
I think that for $h$ having an inverse you should first show that $e$ is in $H$ or assume since it is a group not the other way around.
$endgroup$
– Dani
Jan 29 at 15:24
3
3
$begingroup$
Hmm, I think there is a problem in this answer. How do you know that $h^{-1}$ is in $H$?
$endgroup$
– Mark
Jan 29 at 15:18
$begingroup$
Hmm, I think there is a problem in this answer. How do you know that $h^{-1}$ is in $H$?
$endgroup$
– Mark
Jan 29 at 15:18
$begingroup$
Seems like missing the point. Since $H$ is a group it has an identity - the question is why $1_h=1_G$. Not that that's hard, of course...
$endgroup$
– David C. Ullrich
Jan 29 at 15:19
$begingroup$
Seems like missing the point. Since $H$ is a group it has an identity - the question is why $1_h=1_G$. Not that that's hard, of course...
$endgroup$
– David C. Ullrich
Jan 29 at 15:19
2
2
$begingroup$
@Mark $H$ is a subgroup, so we know inverses in $H$ exist, but we don't know a priori that the inverse of $h$ in $H$ is the same as the inverse of $h$ in $G$.
$endgroup$
– kccu
Jan 29 at 15:20
$begingroup$
@Mark $H$ is a subgroup, so we know inverses in $H$ exist, but we don't know a priori that the inverse of $h$ in $H$ is the same as the inverse of $h$ in $G$.
$endgroup$
– kccu
Jan 29 at 15:20
1
1
$begingroup$
@kccu, yes, this is exactly what I wrote in the comment. $h^{-1}$ is the inverse of $h$ in $G$. It is not that trivial that it equals to the inverse in $H$. (easy to prove that, but it still has to be proved)
$endgroup$
– Mark
Jan 29 at 15:21
$begingroup$
@kccu, yes, this is exactly what I wrote in the comment. $h^{-1}$ is the inverse of $h$ in $G$. It is not that trivial that it equals to the inverse in $H$. (easy to prove that, but it still has to be proved)
$endgroup$
– Mark
Jan 29 at 15:21
$begingroup$
I think that for $h$ having an inverse you should first show that $e$ is in $H$ or assume since it is a group not the other way around.
$endgroup$
– Dani
Jan 29 at 15:24
$begingroup$
I think that for $h$ having an inverse you should first show that $e$ is in $H$ or assume since it is a group not the other way around.
$endgroup$
– Dani
Jan 29 at 15:24
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%2f3092290%2fa-subgroup-h-of-g-must-contain-the-identity-from-g%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
