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At first, if a group G is an infinite loop space (all are based), then \pi_0(G) must be Abelian. Therefore, if G is discret, then it must be Abelian. In fact, any Abelian group does be infinite loop space, by the EM space construction. But we have non-Abelian examples, the infinite groups U and O are infinite loop spaces, by Bott periodicity. (Does this contradict with the statement that the coefficient of a cohomology must be an Abelian group?) Are there any other examples?

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up vote 0 down vote accepted

Edit: I just reread the question, and it says "if a group is an infinite loop space..." I realise the first paragraph of my answer is incorrect. The rest still stands.


Firstly, $\pi_0(G)$ does not have to be abelian - it is $\pi_1(G,e)$ which is abelian, as the Eckmann-Hilton argument shows.

The coefficients of cohomology do not have to be abelian groups, but I guess you are referring to the idea that the ordinary cohomology of a space (singular, or sheaf, say) only makes sense in all non-negative dimensions for abelian-group coefficients. One can define $H^1(X,G)$ ($X$ a space) for a nonabelian (topological) group, but for higher degree cohomology it is not straightforward, see this MO answer.

Now notice that one can use loop spectra as coefficients for cohomology, but one gets an extraordary cohomology theory: this is the case of $U$ and $O$, which represent spectra, and given $K$-theory and $KO$-theory respectively. (see the Wikipedia page on K-theory for example)

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Let G=\Omega^k(X). \pi_0(G)=[S^0, G]=[S^0,\Omega^k(X)]=[\Sigma^k S^0, X]=\pi_k(X). Take k=2. Any thing wrong? –  Ma Ming Oct 20 '10 at 6:46
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@Maming: no, nothing wrong: \pi_0 (G) has to be an abelian group. –  Johannes Ebert Oct 20 '10 at 7:17
    
@Johannes Ebert: Wait, here are two group structues on \pi_0(G)! Is it obious that they are the same? –  Ma Ming Oct 20 '10 at 7:31
    
Maming, that's what the Eckmann-Hilton argument tells you. –  Andrew Stacey Oct 20 '10 at 8:06
    
@Andrew Stacey: thanks, I see the point. –  Ma Ming Oct 20 '10 at 8:26
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There is a recognition principle for infinite loop spaces, which involves a lot of machinery and tells you whether a given space is equivalent to an infinite loop space. More specifically, $Y$ is equivalent to an infinite loop space if and only if ($\pi_0(Y)$ is a group and) $Y$ is an $E_\infty$-space, meaning $Y$ admits a product which is associative and commutative up to some coherent sequence of "higher homotopies". A good place to start reading about this is J. F. Adams' book "Infinite Loop Spaces".

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And perhaps after that, May's The Geometry of Iterated Loop Spaces. –  Todd Trimble Oct 20 '10 at 7:38
    
This didn't solve the problem. –  Ma Ming Oct 20 '10 at 7:41
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Maming, it didn't give you a nice, concise, precise answer to your question, but it did tell you how to go about finding more information and perhaps solve the problem for yourself. So I think that this is a great MO answer. –  Andrew Stacey Oct 20 '10 at 8:07
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I agree with Andrew. Giving good references are among the best types of answers, even if they mean you have to do some work yourself. The book by Adams is in any case fantastic and is well worth reading. –  Todd Trimble Oct 20 '10 at 14:25
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