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Let $T$ be a torus $V/\Gamma$, $\gamma$ a loop on $T$ based at the origin. Then it is easy to see that $$2 \gamma = \gamma \ast \gamma \in \pi_1(T).$$

Here $2 \gamma$ is obtained by rescaling $\gamma$ using the group law, while $\ast$ denotes the operation in the fundamental group. The way I can check this is rather direct: one lifts the loop (up to based homotopy) to a segment in $V$ and uses the identification of $\pi_1(T)$ with the lattice $\Gamma$.

Is there a more conceptual way to prove this identity that will extend to more general (real or complex) Lie groups, or maybe to linear algebraic groups? Or is this fact false in more generality?

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Yay! It's the Eckmann-Hilton argument!

There are two group structures on $\pi_1(G)$ and they commute with each other. It turns out that that is sufficient to show that they are the same structure and that that structure is commutative.

For a proof of this, using interpretative dance, take a look at the movie in this seminar that I gave last semester. There's also something on YouTube by The Catsters (see the nLab page linked above).

(Forgot to actually answer your question!) This only depends on the fact that $\pi_1$ is a representable group functor and that $G$ is a group object in $hTop$. So it will extend to other group objects in $hTop$, such as those that you mention. This also explains why $\pi_k$ is abelian for $k \ge 2$ since $\pi_2(X) = \pi_1(\Omega X)$ and $\Omega X$ is a group object in $hTop$.

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  • $\begingroup$ (NB: the fact that $\pi_1$ is representable isn't the main point, it's that $\pi_1$ preserves products so takes group objects to group objects. Representable implies this.) $\endgroup$
    – Andrew Stacey
    Commented Aug 17, 2010 at 13:33
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    $\begingroup$ The link is broken. : ( $\endgroup$
    – Matthew Drury
    Commented Mar 9, 2018 at 23:04
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The Eckmann-Hilton argument is the correct answer, but it might be amusing to note that there is a very explicit homotopy as well. Suppose $\alpha_1$, $\alpha_2 \in \pi_1 G$, and define

$\alpha : I^2 \rightarrow G$

by $\alpha(t_1,t_2) = \alpha_1(t_1) \cdot \alpha_2(t_2)$, where $\cdot$ is the product in $G$. Then along the diagonal, we have $\alpha_1 \cdot \alpha_2$, the product using the group operation, while along the bottom edge followed by the right edge we have the composition $\alpha_1 * \alpha_2$, the product of loops in the fundamental group. Deforming the path shows they're homotopic. Similarly, along the left edge, followed by the top edge we get $\alpha_2 * \alpha_1$, so this product is commutative.

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  • $\begingroup$ Very nice argument! $\endgroup$
    – Andreas Thom
    Commented Aug 19, 2010 at 12:16
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There is an elementary, formal argument given in Spanier (Theorem 1.6.8, p. 43, 1966 edition):

If two composition laws have a common two-sided identity element and are mutually distributive, then they are equal, commutative and associative.

See also the following corollaries, esp. Corollary 9 on homotopy classes of maps from an H-cogroup to an H-space.

I always found the simplicity of this riveting.

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    $\begingroup$ i believe this is exactly what Andrew mentioned above... minus that reference $\endgroup$
    – Sean Tilson
    Commented Aug 19, 2010 at 22:42
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    $\begingroup$ The link in my answer is to the nLab page where an even weaker hypothesis is assumed: "If a set is equipped with two binary operations with identity elements, as long as they commute with each other in the sense that one is (with respect to the other) a homomorphism of sets with binary operations, then everything else follows.". The argument is, as you say, quite formal and works in extremely general cases. $\endgroup$
    – Andrew Stacey
    Commented Aug 20, 2010 at 7:11
  • $\begingroup$ I've always found it entertaining that one proves commutativity before associativity in this argument. $\endgroup$
    – Robert Bruner
    Commented Aug 20, 2010 at 15:38
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    $\begingroup$ You don't need to assume a common identity element; it suffices to assume that each operation has its own 2-sided identity. (Of course, it ultimately follows that the two identity elements are equal.) Also, the hypothesis in the E-H argument, that each of the operations is a homomorphism with respect to the other, is not what is usually called distributivity. (For example, the meet and join operations in a distributive lattice are each distributive over the other, but this mutual distributivity is not what is wanted for the E-H argument.) $\endgroup$
    – Andreas Blass
    Commented Aug 20, 2010 at 19:15
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@Sean: yes, but my point was that this is true in general i.e. for any two composition laws in hom(X,Y), X,Y objects in a category, not just the homotopy category of pointed topological spaces and H-cogroups to H-spaces. Another reference is Peter Hilton's Homotopy Theory and Duality (1965), p. 5 (for the pointed homotopy category).

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