1
$\begingroup$

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 $\mathrm{Aut}(G)/\mathrm{Inn}(G) \equiv [BG, BG]$?

$\endgroup$
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.