Counting the subgroups of order $6$ in a group of order $42$ which has a subgroup of order $6$
Let be $G$ a group of order $42$. Suppose $G$ has a subgroup of order $6$. Compute the number of conjugates of this subgroup in $G$.
This is what I thought:
Let be $H$ the subgroup of order $6$ and $K$ the only one $7$-subgroup of Sylow.
If $G$ is abelian, then there is an only one conjugate of $H$ in $G$. If $G$ is non abelian, we proceed as follow:
Acting $K$ on $H$ by conjugation,
$$|H| equiv |N_G(H) cap K| text{mod} 7$$
by the theorem $4.1$ on page $18$ of this lecture notes, therefore $|N_G(H) cap K| = 6$. On the one hand, the number of conjugates of $H$ with respect to $K$ is $frac{|G|}{|N_G(H) cap K|} = frac{42}{6} = 7$. By the other hand, $hHh^{-1} = H$ for every $h in H$, but $eHe^{-1}$ is a conjugate of $H$ with respect to $e in K$, then $hHh^{-1} = eHe^{-1}$ for every $h in H$, therefore $H$ has exactly $7$ conjugates in $G$. $square$
I don't sure about this argument, because I stated "the number of conjugates of $H$ with respect to $K$ is $frac{|G|}{|N_G(H) cap K|}$". I know that it's true that the number of conjugates of $H$ in $G$ is $frac{|G|}{|N_G(H)|}$, but I don't sure if what I stated it's true, I couldn't realize what action group I need to define to ensure that my statement it's true. I would like to know if what I do is correct and if it is how ensure that the number of conjugates of $H$ with respect to $K$ is $frac{|G|}{|N_G(H) cap K|}$. If this is not correct, I would like a hint in order to compute the number of conjugates of $H$ in $G$.
abstract-algebra group-theory proof-verification
asked Nov 23 at 10:29
Math enthusiast
1176
|
show 1 more comment
Let be $G$ a group of order $42$. Suppose $G$ has a subgroup of order $6$. Compute the number of conjugates of this subgroup in $G$.
This is what I thought:
Let be $H$ the subgroup of order $6$ and $K$ the only one $7$-subgroup of Sylow.
If $G$ is abelian, then there is an only one conjugate of $H$ in $G$. If $G$ is non abelian, we proceed as follow:
Acting $K$ on $H$ by conjugation,
$$|H| equiv |N_G(H) cap K| text{mod} 7$$
by the theorem $4.1$ on page $18$ of this lecture notes, therefore $|N_G(H) cap K| = 6$. On the one hand, the number of conjugates of $H$ with respect to $K$ is $frac{|G|}{|N_G(H) cap K|} = frac{42}{6} = 7$. By the other hand, $hHh^{-1} = H$ for every $h in H$, but $eHe^{-1}$ is a conjugate of $H$ with respect to $e in K$, then $hHh^{-1} = eHe^{-1}$ for every $h in H$, therefore $H$ has exactly $7$ conjugates in $G$. $square$
I don't sure about this argument, because I stated "the number of conjugates of $H$ with respect to $K$ is $frac{|G|}{|N_G(H) cap K|}$". I know that it's true that the number of conjugates of $H$ in $G$ is $frac{|G|}{|N_G(H)|}$, but I don't sure if what I stated it's true, I couldn't realize what action group I need to define to ensure that my statement it's true. I would like to know if what I do is correct and if it is how ensure that the number of conjugates of $H$ with respect to $K$ is $frac{|G|}{|N_G(H) cap K|}$. If this is not correct, I would like a hint in order to compute the number of conjugates of $H$ in $G$.
abstract-algebra group-theory proof-verification
asked Nov 23 at 10:29
Math enthusiast
1176
Your answer is correct: if $G$ is nonabelian, then there are exactly $7$ cyclic subgroups of order $6$, and they are all conjugate. (There are actually three isomorphism classes of groups with this property, but they all have $7$ cyclic subgroups of order $6$)
– Derek Holt
Nov 23 at 11:58
@DerekHolt, do you know what is the homomorphism which ensure my statement of that "the number of conjugates of $H$ with respect to $K$ is $frac{|G|}{|N_G(H) cap K|}$"?
– Math enthusiast
Nov 23 at 12:01
That's not correct, because $|N_K(H)|=1$. The number of conjugates of $H$ is $|G/N_G(H)|$, but $N_K(H)=1$, so $N_G(H)= H$ has order $6$.
– Derek Holt
Nov 23 at 12:11
Are you considering $N_K(H) = N_G(H) cap K$, right? Why $|N_K(H)| = 1$? I thought $N_K(H)$ would have $6$ elements and how exactly did you conclude that $N_G(H) = H$?
– Math enthusiast
Nov 23 at 12:18
Wait a moment, I think that I understood why $N_G(H) = H$. I will post my explanation soon
– Math enthusiast
Nov 23 at 12:24
|
show 1 more comment
Let be $G$ a group of order $42$. Suppose $G$ has a subgroup of order $6$. Compute the number of conjugates of this subgroup in $G$.
This is what I thought:
Let be $H$ the subgroup of order $6$ and $K$ the only one $7$-subgroup of Sylow.
If $G$ is abelian, then there is an only one conjugate of $H$ in $G$. If $G$ is non abelian, we proceed as follow:
Acting $K$ on $H$ by conjugation,
$$|H| equiv |N_G(H) cap K| text{mod} 7$$
by the theorem $4.1$ on page $18$ of this lecture notes, therefore $|N_G(H) cap K| = 6$. On the one hand, the number of conjugates of $H$ with respect to $K$ is $frac{|G|}{|N_G(H) cap K|} = frac{42}{6} = 7$. By the other hand, $hHh^{-1} = H$ for every $h in H$, but $eHe^{-1}$ is a conjugate of $H$ with respect to $e in K$, then $hHh^{-1} = eHe^{-1}$ for every $h in H$, therefore $H$ has exactly $7$ conjugates in $G$. $square$
I don't sure about this argument, because I stated "the number of conjugates of $H$ with respect to $K$ is $frac{|G|}{|N_G(H) cap K|}$". I know that it's true that the number of conjugates of $H$ in $G$ is $frac{|G|}{|N_G(H)|}$, but I don't sure if what I stated it's true, I couldn't realize what action group I need to define to ensure that my statement it's true. I would like to know if what I do is correct and if it is how ensure that the number of conjugates of $H$ with respect to $K$ is $frac{|G|}{|N_G(H) cap K|}$. If this is not correct, I would like a hint in order to compute the number of conjugates of $H$ in $G$.
abstract-algebra group-theory proof-verification
asked Nov 23 at 10:29
Math enthusiast
1176
Let be $G$ a group of order $42$. Suppose $G$ has a subgroup of order $6$. Compute the number of conjugates of this subgroup in $G$.
This is what I thought:
Let be $H$ the subgroup of order $6$ and $K$ the only one $7$-subgroup of Sylow.
If $G$ is abelian, then there is an only one conjugate of $H$ in $G$. If $G$ is non abelian, we proceed as follow:
Acting $K$ on $H$ by conjugation,
$$|H| equiv |N_G(H) cap K| text{mod} 7$$
by the theorem $4.1$ on page $18$ of this lecture notes, therefore $|N_G(H) cap K| = 6$. On the one hand, the number of conjugates of $H$ with respect to $K$ is $frac{|G|}{|N_G(H) cap K|} = frac{42}{6} = 7$. By the other hand, $hHh^{-1} = H$ for every $h in H$, but $eHe^{-1}$ is a conjugate of $H$ with respect to $e in K$, then $hHh^{-1} = eHe^{-1}$ for every $h in H$, therefore $H$ has exactly $7$ conjugates in $G$. $square$
I don't sure about this argument, because I stated "the number of conjugates of $H$ with respect to $K$ is $frac{|G|}{|N_G(H) cap K|}$". I know that it's true that the number of conjugates of $H$ in $G$ is $frac{|G|}{|N_G(H)|}$, but I don't sure if what I stated it's true, I couldn't realize what action group I need to define to ensure that my statement it's true. I would like to know if what I do is correct and if it is how ensure that the number of conjugates of $H$ with respect to $K$ is $frac{|G|}{|N_G(H) cap K|}$. If this is not correct, I would like a hint in order to compute the number of conjugates of $H$ in $G$.
abstract-algebra group-theory proof-verification
abstract-algebra group-theory proof-verification
asked Nov 23 at 10:29
Math enthusiast
1176
asked Nov 23 at 10:29
Math enthusiast
1176
edited Nov 27 at 13:48
asked Nov 23 at 10:29
Math enthusiast
1176
asked Nov 23 at 10:29
Math enthusiast
1176
asked Nov 23 at 10:29
Math enthusiast
1176
1176
Your answer is correct: if $G$ is nonabelian, then there are exactly $7$ cyclic subgroups of order $6$, and they are all conjugate. (There are actually three isomorphism classes of groups with this property, but they all have $7$ cyclic subgroups of order $6$)
– Derek Holt
Nov 23 at 11:58
@DerekHolt, do you know what is the homomorphism which ensure my statement of that "the number of conjugates of $H$ with respect to $K$ is $frac{|G|}{|N_G(H) cap K|}$"?
– Math enthusiast
Nov 23 at 12:01
That's not correct, because $|N_K(H)|=1$. The number of conjugates of $H$ is $|G/N_G(H)|$, but $N_K(H)=1$, so $N_G(H)= H$ has order $6$.
– Derek Holt
Nov 23 at 12:11
Are you considering $N_K(H) = N_G(H) cap K$, right? Why $|N_K(H)| = 1$? I thought $N_K(H)$ would have $6$ elements and how exactly did you conclude that $N_G(H) = H$?
– Math enthusiast
Nov 23 at 12:18
Wait a moment, I think that I understood why $N_G(H) = H$. I will post my explanation soon
– Math enthusiast
Nov 23 at 12:24
|
show 1 more comment
Your answer is correct: if $G$ is nonabelian, then there are exactly $7$ cyclic subgroups of order $6$, and they are all conjugate. (There are actually three isomorphism classes of groups with this property, but they all have $7$ cyclic subgroups of order $6$)
– Derek Holt
Nov 23 at 11:58
@DerekHolt, do you know what is the homomorphism which ensure my statement of that "the number of conjugates of $H$ with respect to $K$ is $frac{|G|}{|N_G(H) cap K|}$"?
– Math enthusiast
Nov 23 at 12:01
That's not correct, because $|N_K(H)|=1$. The number of conjugates of $H$ is $|G/N_G(H)|$, but $N_K(H)=1$, so $N_G(H)= H$ has order $6$.
– Derek Holt
Nov 23 at 12:11
Are you considering $N_K(H) = N_G(H) cap K$, right? Why $|N_K(H)| = 1$? I thought $N_K(H)$ would have $6$ elements and how exactly did you conclude that $N_G(H) = H$?
– Math enthusiast
Nov 23 at 12:18
Wait a moment, I think that I understood why $N_G(H) = H$. I will post my explanation soon
– Math enthusiast
Nov 23 at 12:24
Your answer is correct: if $G$ is nonabelian, then there are exactly $7$ cyclic subgroups of order $6$, and they are all conjugate. (There are actually three isomorphism classes of groups with this property, but they all have $7$ cyclic subgroups of order $6$)
– Derek Holt
Nov 23 at 11:58
Your answer is correct: if $G$ is nonabelian, then there are exactly $7$ cyclic subgroups of order $6$, and they are all conjugate. (There are actually three isomorphism classes of groups with this property, but they all have $7$ cyclic subgroups of order $6$)
– Derek Holt
Nov 23 at 11:58
@DerekHolt, do you know what is the homomorphism which ensure my statement of that "the number of conjugates of $H$ with respect to $K$ is $frac{|G|}{|N_G(H) cap K|}$"?
– Math enthusiast
Nov 23 at 12:01
@DerekHolt, do you know what is the homomorphism which ensure my statement of that "the number of conjugates of $H$ with respect to $K$ is $frac{|G|}{|N_G(H) cap K|}$"?
– Math enthusiast
Nov 23 at 12:01
That's not correct, because $|N_K(H)|=1$. The number of conjugates of $H$ is $|G/N_G(H)|$, but $N_K(H)=1$, so $N_G(H)= H$ has order $6$.
– Derek Holt
Nov 23 at 12:11
That's not correct, because $|N_K(H)|=1$. The number of conjugates of $H$ is $|G/N_G(H)|$, but $N_K(H)=1$, so $N_G(H)= H$ has order $6$.
– Derek Holt
Nov 23 at 12:11
Are you considering $N_K(H) = N_G(H) cap K$, right? Why $|N_K(H)| = 1$? I thought $N_K(H)$ would have $6$ elements and how exactly did you conclude that $N_G(H) = H$?
– Math enthusiast
Nov 23 at 12:18
Are you considering $N_K(H) = N_G(H) cap K$, right? Why $|N_K(H)| = 1$? I thought $N_K(H)$ would have $6$ elements and how exactly did you conclude that $N_G(H) = H$?
– Math enthusiast
Nov 23 at 12:18
Wait a moment, I think that I understood why $N_G(H) = H$. I will post my explanation soon
– Math enthusiast
Nov 23 at 12:24
Wait a moment, I think that I understood why $N_G(H) = H$. I will post my explanation soon
– Math enthusiast
Nov 23 at 12:24
|
show 1 more comment
1 Answer
1
active
oldest
votes
It was proved on the comments that if $G$ is non abelian and $H leq G$ such that $|H| = 6$, then $N_G(H) = H$. Thus, the number of conjugates of $H$ is $[G : N_G(H)] = frac{|G|}{|N_G(H)|} = frac{42}{6} = 7$.
answered Nov 27 at 13:43
Math enthusiast
1176
add a comment |
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%2f3010207%2fcounting-the-subgroups-of-order-6-in-a-group-of-order-42-which-has-a-subgrou%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
It was proved on the comments that if $G$ is non abelian and $H leq G$ such that $|H| = 6$, then $N_G(H) = H$. Thus, the number of conjugates of $H$ is $[G : N_G(H)] = frac{|G|}{|N_G(H)|} = frac{42}{6} = 7$.
answered Nov 27 at 13:43
Math enthusiast
1176
add a comment |
It was proved on the comments that if $G$ is non abelian and $H leq G$ such that $|H| = 6$, then $N_G(H) = H$. Thus, the number of conjugates of $H$ is $[G : N_G(H)] = frac{|G|}{|N_G(H)|} = frac{42}{6} = 7$.
answered Nov 27 at 13:43
Math enthusiast
1176
add a comment |
It was proved on the comments that if $G$ is non abelian and $H leq G$ such that $|H| = 6$, then $N_G(H) = H$. Thus, the number of conjugates of $H$ is $[G : N_G(H)] = frac{|G|}{|N_G(H)|} = frac{42}{6} = 7$.
answered Nov 27 at 13:43
Math enthusiast
1176
It was proved on the comments that if $G$ is non abelian and $H leq G$ such that $|H| = 6$, then $N_G(H) = H$. Thus, the number of conjugates of $H$ is $[G : N_G(H)] = frac{|G|}{|N_G(H)|} = frac{42}{6} = 7$.
answered Nov 27 at 13:43
Math enthusiast
1176
answered Nov 27 at 13:43
Math enthusiast
1176
answered Nov 27 at 13:43
Math enthusiast
1176
answered Nov 27 at 13:43
Math enthusiast
1176
1176
add a comment |
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.
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%2f3010207%2fcounting-the-subgroups-of-order-6-in-a-group-of-order-42-which-has-a-subgrou%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
Your answer is correct: if $G$ is nonabelian, then there are exactly $7$ cyclic subgroups of order $6$, and they are all conjugate. (There are actually three isomorphism classes of groups with this property, but they all have $7$ cyclic subgroups of order $6$)
– Derek Holt
Nov 23 at 11:58
@DerekHolt, do you know what is the homomorphism which ensure my statement of that "the number of conjugates of $H$ with respect to $K$ is $frac{|G|}{|N_G(H) cap K|}$"?
– Math enthusiast
Nov 23 at 12:01
That's not correct, because $|N_K(H)|=1$. The number of conjugates of $H$ is $|G/N_G(H)|$, but $N_K(H)=1$, so $N_G(H)= H$ has order $6$.
– Derek Holt
Nov 23 at 12:11
Are you considering $N_K(H) = N_G(H) cap K$, right? Why $|N_K(H)| = 1$? I thought $N_K(H)$ would have $6$ elements and how exactly did you conclude that $N_G(H) = H$?
– Math enthusiast
Nov 23 at 12:18
Wait a moment, I think that I understood why $N_G(H) = H$. I will post my explanation soon
– Math enthusiast
Nov 23 at 12:24