# The fundamental group of a topological group is abelian

I want to show the fundamental group of a topological group is abelian. In fact, the question says the topological group is path connected. I do not know where I should use path-connectedness. I think, it is still true if we do not suppose path-connectedness. Right?

I do not know homotopy. I have just learned the fundamental group.

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The fundamental group depends only on the path component of the base point, so the path-connectedness just says that there are no irrelevant path components, it's nothing essential. –  Daniel Fischer Feb 22 at 23:03
Thank you very much! –  square Feb 22 at 23:11
How can one know what the fundamental group is without knowing some basic results on homotopy? –  PVAL Feb 23 at 0:10
PVAL I know some basic results on homotopy. I have studied chapter0 and beginning of chapter1 Hatcher. I mean I have not studied chapter4. –  square Mar 4 at 1:53

## 2 Answers

The usual proof is to show that for all loops $\alpha,\beta \colon [0,1] \to G$ of the topological group $G$, the concatenation $\alpha \cdot \beta$ is homotop to $t \mapsto \alpha(t)\beta(t)$ and to $t \mapsto \beta(t)\alpha(t)$. It requires to exhibit formulae.

Howerver, my favourite proof of this result is the following one, from Grothendieck (I think) : the fundamental group functor $\pi_1 \colon \mathsf{pcTop} \to \mathsf{Grp}$ from the category of path-connected topological spaces to the category of groups respects products (classical lemma), so sends group objects to group objects ; the group objects of $\mathsf{pcTop}$, which are the path-connected topological groups (by definition), are send to group objects of $\mathsf{Grp}$, which are the abelian groups (easy exercise).

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Hey, cute abstract nonsense proof! Thanks for posting it! (+1) –  Bruno Stonek Feb 23 at 9:41
Thanks a lot. @Pece –  square Mar 4 at 1:56

The following link has a really nice proof that I like:

http://www.koschei.net/blog/archives/000434.html

The great part is that they show concatenation of paths and path multiplication are homotopic without super complicated formulas. It is possible to write down formulas (I've done it) but it can take hours of trial and error to get there (in my experience). Once you have this the argument is simple.

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Brian Klatt: Thank you so much. I didn't know how to use path-connectedness assumption. –  square Mar 4 at 1:46