Mixing
$mathrm{Οut}(G) = [BG, BG]$?
$begingroup$
Let $G$ be a finite group and $BG$ its classifying space. Let $[BG, BG]$ denote the set of self-maps of $BG$ up to homotopy equivalence. Automorphisms of $G$ give such self-maps, and inner automorphisms are homotopy equivalent to the identity. I am looking for an answer, hopefully with a reference if positive, to the following question:
Is it true that $mathrm{Aut}(G)/mathrm{Inn}(G) equiv [BG, BG]$?
group-theory algebraic-topology homotopy-theory automorphism-group classifying-spaces
edited Jan 30 at 17:19
the_fox
2,90231538
asked Jan 30 at 17:10
darkodarko
883513
$endgroup$
add a comment |
$begingroup$
Let $G$ be a finite group and $BG$ its classifying space. Let $[BG, BG]$ denote the set of self-maps of $BG$ up to homotopy equivalence. Automorphisms of $G$ give such self-maps, and inner automorphisms are homotopy equivalent to the identity. I am looking for an answer, hopefully with a reference if positive, to the following question:
Is it true that $mathrm{Aut}(G)/mathrm{Inn}(G) equiv [BG, BG]$?
group-theory algebraic-topology homotopy-theory automorphism-group classifying-spaces
edited Jan 30 at 17:19
the_fox
2,90231538
asked Jan 30 at 17:10
darkodarko
883513
$endgroup$
add a comment |
$begingroup$
Let $G$ be a finite group and $BG$ its classifying space. Let $[BG, BG]$ denote the set of self-maps of $BG$ up to homotopy equivalence. Automorphisms of $G$ give such self-maps, and inner automorphisms are homotopy equivalent to the identity. I am looking for an answer, hopefully with a reference if positive, to the following question:
Is it true that $mathrm{Aut}(G)/mathrm{Inn}(G) equiv [BG, BG]$?
group-theory algebraic-topology homotopy-theory automorphism-group classifying-spaces
edited Jan 30 at 17:19
the_fox
2,90231538
asked Jan 30 at 17:10
darkodarko
883513
$endgroup$
Let $G$ be a finite group and $BG$ its classifying space. Let $[BG, BG]$ denote the set of self-maps of $BG$ up to homotopy equivalence. Automorphisms of $G$ give such self-maps, and inner automorphisms are homotopy equivalent to the identity. I am looking for an answer, hopefully with a reference if positive, to the following question:
Is it true that $mathrm{Aut}(G)/mathrm{Inn}(G) equiv [BG, BG]$?
group-theory algebraic-topology homotopy-theory automorphism-group classifying-spaces
group-theory algebraic-topology homotopy-theory automorphism-group classifying-spaces
edited Jan 30 at 17:19
the_fox
2,90231538
asked Jan 30 at 17:10
darkodarko
883513
edited Jan 30 at 17:19
the_fox
2,90231538
asked Jan 30 at 17:10
darkodarko
883513
edited Jan 30 at 17:19
the_fox
2,90231538
edited Jan 30 at 17:19
the_fox
2,90231538
edited Jan 30 at 17:19
the_fox
2,90231538
2,90231538
asked Jan 30 at 17:10
darkodarko
883513
asked Jan 30 at 17:10
darkodarko
883513
asked Jan 30 at 17:10
darkodarko
883513
883513
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Presumably you want all group endomorphisms of $G$, instead of just automorphisms. Then it is true that if $G$ and $H$ are finite groups, then $$[BG, BH] cong operatorname{Hom}(G,H)/operatorname{Inn}(H).$$
This is mentioned in Martino's Classifying spaces and their maps [MR1349123], where this result is attributed to Hurewicz.
answered Jan 30 at 17:43
JHFJHF
4,9611026
$endgroup$
add a comment |
$begingroup$
The answer's no : take any nontrivial $G$ for which $Out(G) = 1$, e.g. $G=mathbb{Z/2}$ or $mathfrak{S}_n, n neq 6$.
Then the isomorphism would imply $[BG,BG] = 1$, which would imply that the identity map is nullhomotopic, in other words $BG$ would be contractible, which is of course absurd.
answered Jan 30 at 17:36
MaxMax
15.9k11144
$endgroup$
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%2f3093809%2fmathrm%25ce%259futg-bg-bg%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Presumably you want all group endomorphisms of $G$, instead of just automorphisms. Then it is true that if $G$ and $H$ are finite groups, then $$[BG, BH] cong operatorname{Hom}(G,H)/operatorname{Inn}(H).$$
This is mentioned in Martino's Classifying spaces and their maps [MR1349123], where this result is attributed to Hurewicz.
answered Jan 30 at 17:43
JHFJHF
4,9611026
$endgroup$
add a comment |
$begingroup$
Presumably you want all group endomorphisms of $G$, instead of just automorphisms. Then it is true that if $G$ and $H$ are finite groups, then $$[BG, BH] cong operatorname{Hom}(G,H)/operatorname{Inn}(H).$$
This is mentioned in Martino's Classifying spaces and their maps [MR1349123], where this result is attributed to Hurewicz.
answered Jan 30 at 17:43
JHFJHF
4,9611026
$endgroup$
add a comment |
$begingroup$
Presumably you want all group endomorphisms of $G$, instead of just automorphisms. Then it is true that if $G$ and $H$ are finite groups, then $$[BG, BH] cong operatorname{Hom}(G,H)/operatorname{Inn}(H).$$
This is mentioned in Martino's Classifying spaces and their maps [MR1349123], where this result is attributed to Hurewicz.
answered Jan 30 at 17:43
JHFJHF
4,9611026
$endgroup$
Presumably you want all group endomorphisms of $G$, instead of just automorphisms. Then it is true that if $G$ and $H$ are finite groups, then $$[BG, BH] cong operatorname{Hom}(G,H)/operatorname{Inn}(H).$$
This is mentioned in Martino's Classifying spaces and their maps [MR1349123], where this result is attributed to Hurewicz.
answered Jan 30 at 17:43
JHFJHF
4,9611026
answered Jan 30 at 17:43
JHFJHF
4,9611026
answered Jan 30 at 17:43
JHFJHF
4,9611026
answered Jan 30 at 17:43
JHFJHF
4,9611026
4,9611026
add a comment |
add a comment |
$begingroup$
The answer's no : take any nontrivial $G$ for which $Out(G) = 1$, e.g. $G=mathbb{Z/2}$ or $mathfrak{S}_n, n neq 6$.
Then the isomorphism would imply $[BG,BG] = 1$, which would imply that the identity map is nullhomotopic, in other words $BG$ would be contractible, which is of course absurd.
answered Jan 30 at 17:36
MaxMax
15.9k11144
$endgroup$
add a comment |
$begingroup$
The answer's no : take any nontrivial $G$ for which $Out(G) = 1$, e.g. $G=mathbb{Z/2}$ or $mathfrak{S}_n, n neq 6$.
Then the isomorphism would imply $[BG,BG] = 1$, which would imply that the identity map is nullhomotopic, in other words $BG$ would be contractible, which is of course absurd.
answered Jan 30 at 17:36
MaxMax
15.9k11144
$endgroup$
add a comment |
$begingroup$
The answer's no : take any nontrivial $G$ for which $Out(G) = 1$, e.g. $G=mathbb{Z/2}$ or $mathfrak{S}_n, n neq 6$.
Then the isomorphism would imply $[BG,BG] = 1$, which would imply that the identity map is nullhomotopic, in other words $BG$ would be contractible, which is of course absurd.
answered Jan 30 at 17:36
MaxMax
15.9k11144
$endgroup$
The answer's no : take any nontrivial $G$ for which $Out(G) = 1$, e.g. $G=mathbb{Z/2}$ or $mathfrak{S}_n, n neq 6$.
Then the isomorphism would imply $[BG,BG] = 1$, which would imply that the identity map is nullhomotopic, in other words $BG$ would be contractible, which is of course absurd.
answered Jan 30 at 17:36
MaxMax
15.9k11144
answered Jan 30 at 17:36
MaxMax
15.9k11144
answered Jan 30 at 17:36
MaxMax
15.9k11144
answered Jan 30 at 17:36
MaxMax
15.9k11144
15.9k11144
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%2f3093809%2fmathrm%25ce%259futg-bg-bg%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
