Proving the number of commuting pairs of elements in $G$ equals number of conjugacy classes in $G$ times...
$begingroup$
The first part of this problem asks to describe $operatorname{Hom}(mathbb{Z}^2,G)$ as a subset of $G times G$ which turned out that $operatorname{Hom}(mathbb{Z}^2,G)$ is the set of pairs of elements which commute in G.
Now, I am trying to understand, and later show the fact that
$#operatorname{Hom}(mathbb{Z}^2,G) = #G cdot N$
where $N$ is the number of orbits of the group action of conjugation by $G$ on itself.
My first guess was to try constructing a bijection between the two sides of the equality but I am having a hard time understanding $#G cdot N$ as a set. Is this a correct approach?
abstract-algebra group-theory group-actions
edited Dec 3 '18 at 1:44
the_fox
2,58711533
asked Dec 3 '18 at 0:58
Richard VillalobosRichard Villalobos
1567
$endgroup$
add a comment |
$begingroup$
The first part of this problem asks to describe $operatorname{Hom}(mathbb{Z}^2,G)$ as a subset of $G times G$ which turned out that $operatorname{Hom}(mathbb{Z}^2,G)$ is the set of pairs of elements which commute in G.
Now, I am trying to understand, and later show the fact that
$#operatorname{Hom}(mathbb{Z}^2,G) = #G cdot N$
where $N$ is the number of orbits of the group action of conjugation by $G$ on itself.
My first guess was to try constructing a bijection between the two sides of the equality but I am having a hard time understanding $#G cdot N$ as a set. Is this a correct approach?
abstract-algebra group-theory group-actions
edited Dec 3 '18 at 1:44
the_fox
2,58711533
asked Dec 3 '18 at 0:58
Richard VillalobosRichard Villalobos
1567
$endgroup$
add a comment |
$begingroup$
The first part of this problem asks to describe $operatorname{Hom}(mathbb{Z}^2,G)$ as a subset of $G times G$ which turned out that $operatorname{Hom}(mathbb{Z}^2,G)$ is the set of pairs of elements which commute in G.
Now, I am trying to understand, and later show the fact that
$#operatorname{Hom}(mathbb{Z}^2,G) = #G cdot N$
where $N$ is the number of orbits of the group action of conjugation by $G$ on itself.
My first guess was to try constructing a bijection between the two sides of the equality but I am having a hard time understanding $#G cdot N$ as a set. Is this a correct approach?
abstract-algebra group-theory group-actions
edited Dec 3 '18 at 1:44
the_fox
2,58711533
asked Dec 3 '18 at 0:58
Richard VillalobosRichard Villalobos
1567
$endgroup$
The first part of this problem asks to describe $operatorname{Hom}(mathbb{Z}^2,G)$ as a subset of $G times G$ which turned out that $operatorname{Hom}(mathbb{Z}^2,G)$ is the set of pairs of elements which commute in G.
Now, I am trying to understand, and later show the fact that
$#operatorname{Hom}(mathbb{Z}^2,G) = #G cdot N$
where $N$ is the number of orbits of the group action of conjugation by $G$ on itself.
My first guess was to try constructing a bijection between the two sides of the equality but I am having a hard time understanding $#G cdot N$ as a set. Is this a correct approach?
abstract-algebra group-theory group-actions
abstract-algebra group-theory group-actions
edited Dec 3 '18 at 1:44
the_fox
2,58711533
asked Dec 3 '18 at 0:58
Richard VillalobosRichard Villalobos
1567
edited Dec 3 '18 at 1:44
the_fox
2,58711533
asked Dec 3 '18 at 0:58
Richard VillalobosRichard Villalobos
1567
edited Dec 3 '18 at 1:44
the_fox
2,58711533
edited Dec 3 '18 at 1:44
the_fox
2,58711533
edited Dec 3 '18 at 1:44
the_fox
2,58711533
2,58711533
asked Dec 3 '18 at 0:58
Richard VillalobosRichard Villalobos
1567
asked Dec 3 '18 at 0:58
Richard VillalobosRichard Villalobos
1567
asked Dec 3 '18 at 0:58
Richard VillalobosRichard Villalobos
1567
1567
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Write $text{Cl}(a)$ for the conjugacy class of $ain G$, and $C(a)$ for the centralizer of $a$ in $G$.
Note that $(a,b)$ is a commuting pair if and only if $bin C(a)$. Thus the number of commuting pairs is (counting by the first element in the pair)
$$text{# pairs} = sum_{ain G} |C(a)|.$$
Now, choose representatives $a_1,dotsc, a_N$ of the $N$ conjugacy classes. If $g,hin G$ are conjugate, then $|C(g)| = |C(h)|$, so we may rewrite the equation above as
$$text{# pairs} = sum_{i=1}^N |text{Cl}(a_i)|cdot |C(a_i)|.$$
By the orbit-stabilizer theorem, $|text{Cl}(a_i)| = |G : C(a_i)|$, so the foregoing simplifies to
$$text{# pairs} = sum_{i=1}^n |G:C(a_i)|cdot |C(a_i)| = sum_{i=1}^N |G| = text{#}Gcdot N.$$
answered Dec 3 '18 at 1:58
rogerlrogerl
17.4k22746
$endgroup$
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%2f3023455%2fproving-the-number-of-commuting-pairs-of-elements-in-g-equals-number-of-conjug%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
$begingroup$
Write $text{Cl}(a)$ for the conjugacy class of $ain G$, and $C(a)$ for the centralizer of $a$ in $G$.
Note that $(a,b)$ is a commuting pair if and only if $bin C(a)$. Thus the number of commuting pairs is (counting by the first element in the pair)
$$text{# pairs} = sum_{ain G} |C(a)|.$$
Now, choose representatives $a_1,dotsc, a_N$ of the $N$ conjugacy classes. If $g,hin G$ are conjugate, then $|C(g)| = |C(h)|$, so we may rewrite the equation above as
$$text{# pairs} = sum_{i=1}^N |text{Cl}(a_i)|cdot |C(a_i)|.$$
By the orbit-stabilizer theorem, $|text{Cl}(a_i)| = |G : C(a_i)|$, so the foregoing simplifies to
$$text{# pairs} = sum_{i=1}^n |G:C(a_i)|cdot |C(a_i)| = sum_{i=1}^N |G| = text{#}Gcdot N.$$
answered Dec 3 '18 at 1:58
rogerlrogerl
17.4k22746
$endgroup$
add a comment |
$begingroup$
Write $text{Cl}(a)$ for the conjugacy class of $ain G$, and $C(a)$ for the centralizer of $a$ in $G$.
Note that $(a,b)$ is a commuting pair if and only if $bin C(a)$. Thus the number of commuting pairs is (counting by the first element in the pair)
$$text{# pairs} = sum_{ain G} |C(a)|.$$
Now, choose representatives $a_1,dotsc, a_N$ of the $N$ conjugacy classes. If $g,hin G$ are conjugate, then $|C(g)| = |C(h)|$, so we may rewrite the equation above as
$$text{# pairs} = sum_{i=1}^N |text{Cl}(a_i)|cdot |C(a_i)|.$$
By the orbit-stabilizer theorem, $|text{Cl}(a_i)| = |G : C(a_i)|$, so the foregoing simplifies to
$$text{# pairs} = sum_{i=1}^n |G:C(a_i)|cdot |C(a_i)| = sum_{i=1}^N |G| = text{#}Gcdot N.$$
answered Dec 3 '18 at 1:58
rogerlrogerl
17.4k22746
$endgroup$
add a comment |
$begingroup$
Write $text{Cl}(a)$ for the conjugacy class of $ain G$, and $C(a)$ for the centralizer of $a$ in $G$.
Note that $(a,b)$ is a commuting pair if and only if $bin C(a)$. Thus the number of commuting pairs is (counting by the first element in the pair)
$$text{# pairs} = sum_{ain G} |C(a)|.$$
Now, choose representatives $a_1,dotsc, a_N$ of the $N$ conjugacy classes. If $g,hin G$ are conjugate, then $|C(g)| = |C(h)|$, so we may rewrite the equation above as
$$text{# pairs} = sum_{i=1}^N |text{Cl}(a_i)|cdot |C(a_i)|.$$
By the orbit-stabilizer theorem, $|text{Cl}(a_i)| = |G : C(a_i)|$, so the foregoing simplifies to
$$text{# pairs} = sum_{i=1}^n |G:C(a_i)|cdot |C(a_i)| = sum_{i=1}^N |G| = text{#}Gcdot N.$$
answered Dec 3 '18 at 1:58
rogerlrogerl
17.4k22746
$endgroup$
Write $text{Cl}(a)$ for the conjugacy class of $ain G$, and $C(a)$ for the centralizer of $a$ in $G$.
Note that $(a,b)$ is a commuting pair if and only if $bin C(a)$. Thus the number of commuting pairs is (counting by the first element in the pair)
$$text{# pairs} = sum_{ain G} |C(a)|.$$
Now, choose representatives $a_1,dotsc, a_N$ of the $N$ conjugacy classes. If $g,hin G$ are conjugate, then $|C(g)| = |C(h)|$, so we may rewrite the equation above as
$$text{# pairs} = sum_{i=1}^N |text{Cl}(a_i)|cdot |C(a_i)|.$$
By the orbit-stabilizer theorem, $|text{Cl}(a_i)| = |G : C(a_i)|$, so the foregoing simplifies to
$$text{# pairs} = sum_{i=1}^n |G:C(a_i)|cdot |C(a_i)| = sum_{i=1}^N |G| = text{#}Gcdot N.$$
answered Dec 3 '18 at 1:58
rogerlrogerl
17.4k22746
edited Dec 3 '18 at 2:04
answered Dec 3 '18 at 1:58
rogerlrogerl
17.4k22746
answered Dec 3 '18 at 1:58
rogerlrogerl
17.4k22746
answered Dec 3 '18 at 1:58
rogerlrogerl
17.4k22746
17.4k22746
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.
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%2f3023455%2fproving-the-number-of-commuting-pairs-of-elements-in-g-equals-number-of-conjug%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
