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The Bruschlinsky group of maps of a space X into S1 up to homotopy, using the multiplication on S1, is well-known to equal the first cohomology group of X (at least assuming X is a reasonably nice space).

What is known about the analogous group of homotopy classes of maps of X into S3, with the group operation defined using the Lie group multiplication of S3 ? Denote this group of homotopy classes by g(X, S3).

In particular: If X is Sn, is it possible for g(Sn, S3) to be non-abelian? More generally, are sufficient conditions on X known for g(X, S3) to be non-abelian? Is there any standard reference about such groups?

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    $\begingroup$ If X is a suspension, then g(X, S^3) is abelian by the Hilton-Eckmann argument. In particular, g(S^n, S^3) = \pi_n(S^3) is abelian for all n. $\endgroup$
    – John Rognes
    Commented Jan 19 at 23:02
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    $\begingroup$ Usually the group is denoted $\pi^3(X)$. It's abelian if $X$ is complex of dimension $\leq 5$. On the other hand, $\pi^3(S^3\times S^3)=g(S^3\times S^3,S^3)$ is nonabelian. $\endgroup$
    – Tyrone
    Commented Jan 20 at 1:37
  • $\begingroup$ @Tyrone Do theses groups have a name ? $\endgroup$
    – Olivier Bégassat
    Commented Jan 20 at 8:13
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    $\begingroup$ @JohnRognes is it obvious that the group laws coincide on $g(S^n,S^3)=\pi_n(S^3)$? The first one is induced by the law on $S^3$ while the second one relies on something (some co-law?) on $S^n$. $\endgroup$
    – YCor
    Commented Jan 20 at 9:45
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    $\begingroup$ @YCor Yes, the H-space structure on S^3 = \Omega HP^\infty and the co-H-space structure on S^n = \Sigma S^{n-1} distribute over one another, hence induce the same pairing, and this is commutative. This is the Hilton-Eckmann argument. See e.g. Thm. 1.6.8 in Spanier's Algebraic Topology. $\endgroup$
    – John Rognes
    Commented Jan 20 at 11:08

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