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Aug 18, 2018 at 15:59 comment added Najib Idrissi I'd like to say something which is well-known by algebraic topologists. This commutativity actually hides some asymmetry. If you look at the usual proof that e.g. $\pi_2(X)$ is abelian, then to prove $ab=ba$ you need e.g. to "move" $b$ over $a$. But if you do it twice, moving $a$ over $b$ back, you have made a loop in the double loop space $\Omega^2 X$, i.e. you get an element of $\pi_3(X)$ that tells you that while $a$ and $b$ commute, you have two choices of making them commute ($a$ over $b$ or $b$ over $a$), and they are not identical. See: Whitehead products (and $E_n$ operads ☺).
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S Dec 14, 2013 at 0:46 history answered Tina CC BY-SA 3.0
S Dec 14, 2013 at 0:46 history made wiki Post Made Community Wiki by Tina