If $H$ is permutable and $A$ is a subgroup of $H$, is $A$ then permutable?
Let $G$ be a finite group, $Ale H$ where $H$ is a proper subgroup of $ G$.
How to show that if $H$ is permutable in $G$ (i.e. $HB = BH$ for all $Ble G$) then $A$ is permutable in $G$?
Well, I started by assuming the opposite (if $A$ is not permutable then $H$ is not permutable) and got two cases, when $B$ is $H$ and not, but I couldn't continue. Can you give me a hint, please?
group-theory finite-groups
|
show 6 more comments
Let $G$ be a finite group, $Ale H$ where $H$ is a proper subgroup of $ G$.
How to show that if $H$ is permutable in $G$ (i.e. $HB = BH$ for all $Ble G$) then $A$ is permutable in $G$?
Well, I started by assuming the opposite (if $A$ is not permutable then $H$ is not permutable) and got two cases, when $B$ is $H$ and not, but I couldn't continue. Can you give me a hint, please?
group-theory finite-groups
Well, $AH=H=HA$.
– Berci
Dec 9 '18 at 22:20
1
Anyway, it's not true in this form: consider $H=G$..
– Berci
Dec 9 '18 at 22:22
I've fixed it! Thank you!
– H.koby
Dec 9 '18 at 22:32
But I still don't get it... How I can use that when B is not H ?
– H.koby
Dec 9 '18 at 22:34
1
Here's how you can thank me properly: answer your own question by showing that $D_8$ is the smallest counterexample to your assertion. As a bonus, implement Derek's suggestion.
– the_fox
Dec 9 '18 at 23:16
|
show 6 more comments
Let $G$ be a finite group, $Ale H$ where $H$ is a proper subgroup of $ G$.
How to show that if $H$ is permutable in $G$ (i.e. $HB = BH$ for all $Ble G$) then $A$ is permutable in $G$?
Well, I started by assuming the opposite (if $A$ is not permutable then $H$ is not permutable) and got two cases, when $B$ is $H$ and not, but I couldn't continue. Can you give me a hint, please?
group-theory finite-groups
Let $G$ be a finite group, $Ale H$ where $H$ is a proper subgroup of $ G$.
How to show that if $H$ is permutable in $G$ (i.e. $HB = BH$ for all $Ble G$) then $A$ is permutable in $G$?
Well, I started by assuming the opposite (if $A$ is not permutable then $H$ is not permutable) and got two cases, when $B$ is $H$ and not, but I couldn't continue. Can you give me a hint, please?
group-theory finite-groups
group-theory finite-groups
edited Dec 10 '18 at 17:37
the_fox
2,43411431
2,43411431
asked Dec 9 '18 at 22:12
H.koby
427
427
Well, $AH=H=HA$.
– Berci
Dec 9 '18 at 22:20
1
Anyway, it's not true in this form: consider $H=G$..
– Berci
Dec 9 '18 at 22:22
I've fixed it! Thank you!
– H.koby
Dec 9 '18 at 22:32
But I still don't get it... How I can use that when B is not H ?
– H.koby
Dec 9 '18 at 22:34
1
Here's how you can thank me properly: answer your own question by showing that $D_8$ is the smallest counterexample to your assertion. As a bonus, implement Derek's suggestion.
– the_fox
Dec 9 '18 at 23:16
|
show 6 more comments
Well, $AH=H=HA$.
– Berci
Dec 9 '18 at 22:20
1
Anyway, it's not true in this form: consider $H=G$..
– Berci
Dec 9 '18 at 22:22
I've fixed it! Thank you!
– H.koby
Dec 9 '18 at 22:32
But I still don't get it... How I can use that when B is not H ?
– H.koby
Dec 9 '18 at 22:34
1
Here's how you can thank me properly: answer your own question by showing that $D_8$ is the smallest counterexample to your assertion. As a bonus, implement Derek's suggestion.
– the_fox
Dec 9 '18 at 23:16
Well, $AH=H=HA$.
– Berci
Dec 9 '18 at 22:20
Well, $AH=H=HA$.
– Berci
Dec 9 '18 at 22:20
1
1
Anyway, it's not true in this form: consider $H=G$..
– Berci
Dec 9 '18 at 22:22
Anyway, it's not true in this form: consider $H=G$..
– Berci
Dec 9 '18 at 22:22
I've fixed it! Thank you!
– H.koby
Dec 9 '18 at 22:32
I've fixed it! Thank you!
– H.koby
Dec 9 '18 at 22:32
But I still don't get it... How I can use that when B is not H ?
– H.koby
Dec 9 '18 at 22:34
But I still don't get it... How I can use that when B is not H ?
– H.koby
Dec 9 '18 at 22:34
1
1
Here's how you can thank me properly: answer your own question by showing that $D_8$ is the smallest counterexample to your assertion. As a bonus, implement Derek's suggestion.
– the_fox
Dec 9 '18 at 23:16
Here's how you can thank me properly: answer your own question by showing that $D_8$ is the smallest counterexample to your assertion. As a bonus, implement Derek's suggestion.
– the_fox
Dec 9 '18 at 23:16
|
show 6 more comments
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
});
}
});
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%2f3033098%2fif-h-is-permutable-and-a-is-a-subgroup-of-h-is-a-then-permutable%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
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%2f3033098%2fif-h-is-permutable-and-a-is-a-subgroup-of-h-is-a-then-permutable%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
Well, $AH=H=HA$.
– Berci
Dec 9 '18 at 22:20
1
Anyway, it's not true in this form: consider $H=G$..
– Berci
Dec 9 '18 at 22:22
I've fixed it! Thank you!
– H.koby
Dec 9 '18 at 22:32
But I still don't get it... How I can use that when B is not H ?
– H.koby
Dec 9 '18 at 22:34
1
Here's how you can thank me properly: answer your own question by showing that $D_8$ is the smallest counterexample to your assertion. As a bonus, implement Derek's suggestion.
– the_fox
Dec 9 '18 at 23:16