If $gcd(|G|,lvertoperatorname{Aut}(H)rvert) =1$ then $N_G(H)=C_G(H)$
$begingroup$
Suppose that $H$ is a subgroup of $G$ such that $gcd(|G|,lvertoperatorname{Aut}(H)rvert)=1$.
Prove that $N_G(H)=C_G(H)$.
I want to solve the above problem. I think because there is a normalizer, we should think about inner homomorphism and use the condition to make trivial homomorphism.
But in this case, there is no assumption about $H$ being normal, so I can't use the inner homomorphism.
Help me solve this problem!
abstract-algebra group-theory finite-groups
edited Jan 11 at 17:40
the_fox
2,90231538
asked Nov 26 '18 at 12:44
PearlPearl
1218
$endgroup$
add a comment |
$begingroup$
Suppose that $H$ is a subgroup of $G$ such that $gcd(|G|,lvertoperatorname{Aut}(H)rvert)=1$.
Prove that $N_G(H)=C_G(H)$.
I want to solve the above problem. I think because there is a normalizer, we should think about inner homomorphism and use the condition to make trivial homomorphism.
But in this case, there is no assumption about $H$ being normal, so I can't use the inner homomorphism.
Help me solve this problem!
abstract-algebra group-theory finite-groups
edited Jan 11 at 17:40
the_fox
2,90231538
asked Nov 26 '18 at 12:44
PearlPearl
1218
$endgroup$
add a comment |
$begingroup$
Suppose that $H$ is a subgroup of $G$ such that $gcd(|G|,lvertoperatorname{Aut}(H)rvert)=1$.
Prove that $N_G(H)=C_G(H)$.
I want to solve the above problem. I think because there is a normalizer, we should think about inner homomorphism and use the condition to make trivial homomorphism.
But in this case, there is no assumption about $H$ being normal, so I can't use the inner homomorphism.
Help me solve this problem!
abstract-algebra group-theory finite-groups
edited Jan 11 at 17:40
the_fox
2,90231538
asked Nov 26 '18 at 12:44
PearlPearl
1218
$endgroup$
Suppose that $H$ is a subgroup of $G$ such that $gcd(|G|,lvertoperatorname{Aut}(H)rvert)=1$.
Prove that $N_G(H)=C_G(H)$.
I want to solve the above problem. I think because there is a normalizer, we should think about inner homomorphism and use the condition to make trivial homomorphism.
But in this case, there is no assumption about $H$ being normal, so I can't use the inner homomorphism.
Help me solve this problem!
abstract-algebra group-theory finite-groups
abstract-algebra group-theory finite-groups
edited Jan 11 at 17:40
the_fox
2,90231538
asked Nov 26 '18 at 12:44
PearlPearl
1218
edited Jan 11 at 17:40
the_fox
2,90231538
asked Nov 26 '18 at 12:44
PearlPearl
1218
edited Jan 11 at 17:40
the_fox
2,90231538
edited Jan 11 at 17:40
the_fox
2,90231538
edited Jan 11 at 17:40
the_fox
2,90231538
2,90231538
asked Nov 26 '18 at 12:44
PearlPearl
1218
asked Nov 26 '18 at 12:44
PearlPearl
1218
asked Nov 26 '18 at 12:44
PearlPearl
1218
1218
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is easy. The $N/C$ Theorem tells you that $N_G(H)/C_G(H)$ embeds isomorphically as a subgroup of $operatorname{Aut}(H)$, thus $leftlvert N_G(H)/C_G(H)rightrvert$ divides $leftlvert operatorname{Aut}(H)right rvert$. But $leftlvert N_G(H)/C_G(H)rightrvert$ also divides $|G|$ (since it divides $|N_G(H)|$ which in turn divides $|G|$), so it divides $gcd(|G|,leftlvert operatorname{Aut}(H)right rvert)=1$. Therefore $|N_G(H)| = |C_G(H)|$ and since $C_G(H) leq N_G(H)$ we get $N_G(H)=C_G(H)$.
answered Nov 26 '18 at 12:53
the_foxthe_fox
2,90231538
$endgroup$
1
$begingroup$
Wow~ thanks a lot. From you, I can learn the N/C theorem.
$endgroup$
– Pearl
Nov 26 '18 at 13:07
add a comment |
Your Answer
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%2f3014275%2fif-gcdg-lvert-operatornameauth-rvert-1-then-n-gh-c-gh%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
$begingroup$
This is easy. The $N/C$ Theorem tells you that $N_G(H)/C_G(H)$ embeds isomorphically as a subgroup of $operatorname{Aut}(H)$, thus $leftlvert N_G(H)/C_G(H)rightrvert$ divides $leftlvert operatorname{Aut}(H)right rvert$. But $leftlvert N_G(H)/C_G(H)rightrvert$ also divides $|G|$ (since it divides $|N_G(H)|$ which in turn divides $|G|$), so it divides $gcd(|G|,leftlvert operatorname{Aut}(H)right rvert)=1$. Therefore $|N_G(H)| = |C_G(H)|$ and since $C_G(H) leq N_G(H)$ we get $N_G(H)=C_G(H)$.
answered Nov 26 '18 at 12:53
the_foxthe_fox
2,90231538
$endgroup$
1
$begingroup$
Wow~ thanks a lot. From you, I can learn the N/C theorem.
$endgroup$
– Pearl
Nov 26 '18 at 13:07
add a comment |
$begingroup$
This is easy. The $N/C$ Theorem tells you that $N_G(H)/C_G(H)$ embeds isomorphically as a subgroup of $operatorname{Aut}(H)$, thus $leftlvert N_G(H)/C_G(H)rightrvert$ divides $leftlvert operatorname{Aut}(H)right rvert$. But $leftlvert N_G(H)/C_G(H)rightrvert$ also divides $|G|$ (since it divides $|N_G(H)|$ which in turn divides $|G|$), so it divides $gcd(|G|,leftlvert operatorname{Aut}(H)right rvert)=1$. Therefore $|N_G(H)| = |C_G(H)|$ and since $C_G(H) leq N_G(H)$ we get $N_G(H)=C_G(H)$.
answered Nov 26 '18 at 12:53
the_foxthe_fox
2,90231538
$endgroup$
1
$begingroup$
Wow~ thanks a lot. From you, I can learn the N/C theorem.
$endgroup$
– Pearl
Nov 26 '18 at 13:07
add a comment |
$begingroup$
This is easy. The $N/C$ Theorem tells you that $N_G(H)/C_G(H)$ embeds isomorphically as a subgroup of $operatorname{Aut}(H)$, thus $leftlvert N_G(H)/C_G(H)rightrvert$ divides $leftlvert operatorname{Aut}(H)right rvert$. But $leftlvert N_G(H)/C_G(H)rightrvert$ also divides $|G|$ (since it divides $|N_G(H)|$ which in turn divides $|G|$), so it divides $gcd(|G|,leftlvert operatorname{Aut}(H)right rvert)=1$. Therefore $|N_G(H)| = |C_G(H)|$ and since $C_G(H) leq N_G(H)$ we get $N_G(H)=C_G(H)$.
answered Nov 26 '18 at 12:53
the_foxthe_fox
2,90231538
$endgroup$
This is easy. The $N/C$ Theorem tells you that $N_G(H)/C_G(H)$ embeds isomorphically as a subgroup of $operatorname{Aut}(H)$, thus $leftlvert N_G(H)/C_G(H)rightrvert$ divides $leftlvert operatorname{Aut}(H)right rvert$. But $leftlvert N_G(H)/C_G(H)rightrvert$ also divides $|G|$ (since it divides $|N_G(H)|$ which in turn divides $|G|$), so it divides $gcd(|G|,leftlvert operatorname{Aut}(H)right rvert)=1$. Therefore $|N_G(H)| = |C_G(H)|$ and since $C_G(H) leq N_G(H)$ we get $N_G(H)=C_G(H)$.
answered Nov 26 '18 at 12:53
the_foxthe_fox
2,90231538
answered Nov 26 '18 at 12:53
the_foxthe_fox
2,90231538
answered Nov 26 '18 at 12:53
the_foxthe_fox
2,90231538
answered Nov 26 '18 at 12:53
the_foxthe_fox
2,90231538
2,90231538
1
$begingroup$
Wow~ thanks a lot. From you, I can learn the N/C theorem.
$endgroup$
– Pearl
Nov 26 '18 at 13:07
add a comment |
1
$begingroup$
Wow~ thanks a lot. From you, I can learn the N/C theorem.
$endgroup$
– Pearl
Nov 26 '18 at 13:07
1
1
$begingroup$
Wow~ thanks a lot. From you, I can learn the N/C theorem.
$endgroup$
– Pearl
Nov 26 '18 at 13:07
$begingroup$
Wow~ thanks a lot. From you, I can learn the N/C theorem.
$endgroup$
– Pearl
Nov 26 '18 at 13:07
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%2f3014275%2fif-gcdg-lvert-operatornameauth-rvert-1-then-n-gh-c-gh%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
