Let $G$ be a group and $H, H'$ be subgroups of $G$ where $H$ is normal. Under which circumstances is $H cap...
0
To be more specific, what kind of assumptions do I have to make for $H'$ to obtain this assertion? Which one are necessary and which one are sufficient?
For instance, we could say $H subset H'$, but this case seems rather boring. Do you have some better ideas? Thank you!
abstract-algebra group-theory normal-subgroups quotient-group
asked Nov 25 at 19:37
Diglett
884520
|
show 1 more comment
0
To be more specific, what kind of assumptions do I have to make for $H'$ to obtain this assertion? Which one are necessary and which one are sufficient?
For instance, we could say $H subset H'$, but this case seems rather boring. Do you have some better ideas? Thank you!
abstract-algebra group-theory normal-subgroups quotient-group
asked Nov 25 at 19:37
Diglett
884520
1
What else do you expect? In order to be normal subgroup, it would need to be subgroup first. Once it is a subgroup, it automatically is normal since it is normal in a bigger group.
– Quang Hoang
Nov 25 at 19:41
2
If $N$ is normal in $G$ and $K$ is a subgroup of $G$ then $N cap K$ is a normal subgroup of $K$. You don't need any assumptions on $K$ for that to be true.
– the_fox
Nov 25 at 19:48
I just noticed that I made a huge mistake in the headline: $H'$ does not has to be necessarily normal!
– Diglett
Nov 25 at 20:47
1
In that case, I don't see what difference $H$ being normal in $G$ makes. Anyway, you'll find that with questions like this you only get what you seek in "trivial" cases.
– the_fox
Nov 25 at 21:03
1
Please include the full text of your query in the body of your message, not just in the title. In fact, the title should be suggestive, but should not be a full two sentences!
– Arturo Magidin
Nov 25 at 22:50
|
show 1 more comment
0
0
0
To be more specific, what kind of assumptions do I have to make for $H'$ to obtain this assertion? Which one are necessary and which one are sufficient?
For instance, we could say $H subset H'$, but this case seems rather boring. Do you have some better ideas? Thank you!
abstract-algebra group-theory normal-subgroups quotient-group
asked Nov 25 at 19:37
Diglett
884520
To be more specific, what kind of assumptions do I have to make for $H'$ to obtain this assertion? Which one are necessary and which one are sufficient?
For instance, we could say $H subset H'$, but this case seems rather boring. Do you have some better ideas? Thank you!
abstract-algebra group-theory normal-subgroups quotient-group
abstract-algebra group-theory normal-subgroups quotient-group
asked Nov 25 at 19:37
Diglett
884520
asked Nov 25 at 19:37
Diglett
884520
edited Nov 25 at 20:46
asked Nov 25 at 19:37
Diglett
884520
asked Nov 25 at 19:37
Diglett
884520
asked Nov 25 at 19:37
Diglett
884520
884520
1
What else do you expect? In order to be normal subgroup, it would need to be subgroup first. Once it is a subgroup, it automatically is normal since it is normal in a bigger group.
– Quang Hoang
Nov 25 at 19:41
2
If $N$ is normal in $G$ and $K$ is a subgroup of $G$ then $N cap K$ is a normal subgroup of $K$. You don't need any assumptions on $K$ for that to be true.
– the_fox
Nov 25 at 19:48
I just noticed that I made a huge mistake in the headline: $H'$ does not has to be necessarily normal!
– Diglett
Nov 25 at 20:47
1
In that case, I don't see what difference $H$ being normal in $G$ makes. Anyway, you'll find that with questions like this you only get what you seek in "trivial" cases.
– the_fox
Nov 25 at 21:03
1
Please include the full text of your query in the body of your message, not just in the title. In fact, the title should be suggestive, but should not be a full two sentences!
– Arturo Magidin
Nov 25 at 22:50
|
show 1 more comment
1
What else do you expect? In order to be normal subgroup, it would need to be subgroup first. Once it is a subgroup, it automatically is normal since it is normal in a bigger group.
– Quang Hoang
Nov 25 at 19:41
2
If $N$ is normal in $G$ and $K$ is a subgroup of $G$ then $N cap K$ is a normal subgroup of $K$. You don't need any assumptions on $K$ for that to be true.
– the_fox
Nov 25 at 19:48
I just noticed that I made a huge mistake in the headline: $H'$ does not has to be necessarily normal!
– Diglett
Nov 25 at 20:47
1
In that case, I don't see what difference $H$ being normal in $G$ makes. Anyway, you'll find that with questions like this you only get what you seek in "trivial" cases.
– the_fox
Nov 25 at 21:03
1
Please include the full text of your query in the body of your message, not just in the title. In fact, the title should be suggestive, but should not be a full two sentences!
– Arturo Magidin
Nov 25 at 22:50
1
1
What else do you expect? In order to be normal subgroup, it would need to be subgroup first. Once it is a subgroup, it automatically is normal since it is normal in a bigger group.
– Quang Hoang
Nov 25 at 19:41
What else do you expect? In order to be normal subgroup, it would need to be subgroup first. Once it is a subgroup, it automatically is normal since it is normal in a bigger group.
– Quang Hoang
Nov 25 at 19:41
2
2
If $N$ is normal in $G$ and $K$ is a subgroup of $G$ then $N cap K$ is a normal subgroup of $K$. You don't need any assumptions on $K$ for that to be true.
– the_fox
Nov 25 at 19:48
If $N$ is normal in $G$ and $K$ is a subgroup of $G$ then $N cap K$ is a normal subgroup of $K$. You don't need any assumptions on $K$ for that to be true.
– the_fox
Nov 25 at 19:48
I just noticed that I made a huge mistake in the headline: $H'$ does not has to be necessarily normal!
– Diglett
Nov 25 at 20:47
I just noticed that I made a huge mistake in the headline: $H'$ does not has to be necessarily normal!
– Diglett
Nov 25 at 20:47
1
1
In that case, I don't see what difference $H$ being normal in $G$ makes. Anyway, you'll find that with questions like this you only get what you seek in "trivial" cases.
– the_fox
Nov 25 at 21:03
In that case, I don't see what difference $H$ being normal in $G$ makes. Anyway, you'll find that with questions like this you only get what you seek in "trivial" cases.
– the_fox
Nov 25 at 21:03
1
1
Please include the full text of your query in the body of your message, not just in the title. In fact, the title should be suggestive, but should not be a full two sentences!
– Arturo Magidin
Nov 25 at 22:50
Please include the full text of your query in the body of your message, not just in the title. In fact, the title should be suggestive, but should not be a full two sentences!
– Arturo Magidin
Nov 25 at 22:50
|
show 1 more comment
active
oldest
votes
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
});
}
});
draft saved
draft discarded
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%2f3013287%2flet-g-be-a-group-and-h-h-be-subgroups-of-g-where-h-is-normal-under-wh%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
draft saved
draft discarded
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
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%2f3013287%2flet-g-be-a-group-and-h-h-be-subgroups-of-g-where-h-is-normal-under-wh%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
1
What else do you expect? In order to be normal subgroup, it would need to be subgroup first. Once it is a subgroup, it automatically is normal since it is normal in a bigger group.
– Quang Hoang
Nov 25 at 19:41
2
If $N$ is normal in $G$ and $K$ is a subgroup of $G$ then $N cap K$ is a normal subgroup of $K$. You don't need any assumptions on $K$ for that to be true.
– the_fox
Nov 25 at 19:48
I just noticed that I made a huge mistake in the headline: $H'$ does not has to be necessarily normal!
– Diglett
Nov 25 at 20:47
1
In that case, I don't see what difference $H$ being normal in $G$ makes. Anyway, you'll find that with questions like this you only get what you seek in "trivial" cases.
– the_fox
Nov 25 at 21:03
1
Please include the full text of your query in the body of your message, not just in the title. In fact, the title should be suggestive, but should not be a full two sentences!
– Arturo Magidin
Nov 25 at 22:50