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"The fundamental group of a topological group is abelian". does this problem admit a proof by nuke. This is inspired by a a question in mathoverflow. The usual proof is by a Eckmann-Hilton argument.But can some kind of sledgehammer be used to give a direct proof.

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The proof by Eckmann-Hilton argument, i.e. using the interchange law, seems to me pretty direct. One is relating the group structure of the topological group with the composition of homotopy classes of loops. This also gives the more general fact that the fundamental groupoid of a topological group is a group object in the category of groupoids, sometimes called a $2$-group. –  Ronnie Brown Feb 2 '13 at 10:17

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up vote 14 down vote accepted

Consider the classifying space $BG$ of $G$. One has the fibration $G \hookrightarrow EG \to BG,$ and passing to the associated long exact sequence of homotopy groups, and using the contractibility of $EG$, we find that $\pi_1(G) \cong \pi_2(BG).$ Since $\pi_2$ is abelian, we see that $\pi_1(G)$ is abelian.

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nice answer.i love this –  Koushik Feb 2 '13 at 11:22

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