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The Eilenberg-MacLane space $K(G,n)$ is defined for non-abelian $G$ for $n=1$ and abelian $G$ for $n>1$. I know there is a theorem that states that the higher homotopy group $\pi_n(X)$ is abelian for $n>1$. But I want to understand what goes wrong if one naively try to following the usually procedure of constructing Eilenberg-MacLane space:

  1. Take a wedge product of $N$ n-spheres, where $N$ is the number of generators in $G$.
  2. For each relation in $G$, attach a $n+1$-disk to trivialize the word corresponding to the relation.
  3. Attach higher-dimensional disks to trivialize generators in $\pi_j(K(G,n))$ for $j>n$.

At which step does it fail such that $K(G,n)$ cannot exist for nonabelian group and $n>1$?

Leo L.
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    Before constructing anything, you should define it first. How would you define nonabelian higher homotopy groups? What you wrote in your attempted construction is too sloppy. – Moishe Kohan Jul 13 '24 at 22:35
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    What do you think $\pi_n$ of a wedge of $N$ $n$-spheres is? – Lee Mosher Jul 13 '24 at 22:36
  • @LeeMosher I thought it would be the free product of $N$ copies of $\mathbb{Z}$. – Leo L. Jul 13 '24 at 22:43
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    @LeoL. Well, it's not; the free product of groups is not abelian (save for edge cases). – Ben Steffan Jul 13 '24 at 22:47
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    How about the following: You can "lift" the construction for $n = 1$ to higher $n$ by replacing the wedge of $S^1$s representing generators with a wedge of $S^n$'s and lift each relation you're imposing in the obvious way: instead of wrapping an appropriate number of times around a copy of $S^1$ you just wrap around $S^n$ instead, using of course that $\pi_n S^n \cong \pi_1 S^1$, in the appropriate order. It would be a good exercise to show that this yields the obvious thing (I'll leave it to you to figure out what that is). – Ben Steffan Jul 13 '24 at 22:54

1 Answers1

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This construction works fine, it just produces $K(G/[G, G], n)$.

For $n \ge 2$ the wedge product $\bigvee_{i=1}^N S^n$ satisfies

$$\pi_n \left( \bigvee_{i=1}^N S^n \right) \cong \bigoplus_{i=1}^N \mathbb{Z};$$

this computation can be done using, in order, Seifert-van Kampen (which tells you that this wedge product is simply connected), then the Hurewicz theorem (which reduces the calculation to calculating $H_n$), then Mayer-Vietoris. In particular $\pi_n$ is not the free group $F_N$, and of course it could not have been, because $\pi_n$ must be abelian. So your construction first forces the generators of $G$ (in some presentation of $G$ by generators and relations) to commute, then applies the relations, and this computes the abelianization $G/[G, G]$.

It sounds like you simply don't really believe that $\pi_n$ is abelian for $n \ge 2$; if so it would be worth studying the Eckmann-Hilton argument thoroughly. It is really very simple and very visual, see e.g. the ASCII art here.

Qiaochu Yuan
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