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Let $X$ be a topological group. Let $\tau_1$ and $\tau_2$ representing elements of $\pi_n(X)$. Is it true that

$$ [\tau_1] [\tau_2] = [\tau_1 \tau_2] $$ in $\pi_n(X)$?,

where of course "$[\tau_1] [\tau_2]$" refers to the $\pi_n(X)$ group operation, and by $\tau_1 \tau_2$ I mean the map $S^n \ni x \mapsto \tau_1(x) \tau_2(x)$.

I know from greenberg harper algebraic topology a first course, page 30, that this is true for n=1.

It seems to me that the answer is a simple "yes", but I wonder why I can't find it on the internet.

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does… provide an answer to the question? – fritz Jan 18 '13 at 17:14
I don't see how to use E-H here since you don't have two operations on one set but here each operation is on a different set. Anyway, interesting question and I'm curious about the answer in either direction. Can we say anything concrete e.g. about $SU(2)$? – Marek Jan 18 '13 at 21:52
I don't understand your objection: we do have two group operations on the same set $\pi_n(X)$: the first one is the standard one, the second one is the pointwise product. – fritz Jan 19 '13 at 0:44
Ah, I see, I misunderstood your construction. Sorry. – Marek Jan 19 '13 at 8:04

Yes, Eckmann-Hilton applies. Interestingly, this implies that all of the homotopy groups of a topological group are abelian. For higher homotopy groups, this is not so interesting, because they are always abelian, regardless of the space. For $n=1$, however, it is not content-less: it tells us that the fundamental group of a topological group is abelian. Fundamental groups are not always abelian!

Thus, for instance, it is impossible to put a topological group structure on the twice-punctured plane.

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