# The fundamental group of a topological group is abelian [duplicate]

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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## marked as duplicate by Daniel Fischer♦Mar 5 at 21:00

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 '14 at 23:03
Thank you very much! –  square Feb 22 '14 at 23:11
How can one know what the fundamental group is without knowing some basic results on homotopy? –  PVAL Feb 23 '14 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 '14 at 1:53

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).