Proving fundamental group is commutative if its space is a group

Question

I am working on this proof problem on fundamental group:

A fundamental group $\pi_1 (X, x_0)$ is commutative if its space $X$ is group.

Here are what I know of:

(1) The proof should begin, I think, with the properties of $X$ as group, which are closure, association, existence of inverse and neutral element.
(2) The proof must end with $[f] * [g] = [g] * [f]$, where $f, g \in \pi_1 (X, x_0)$, thus proving that the fundamental group is abelian.

But unfortunately I do not know how to connect the (1) and (2) above, therefore any help would be very much appreciated. Thank you for your time and help.

---

Oops!
Sorry for not posting this caveat: I did my due diligence by checking prior postings before I posted this, and I did come across this 2014 similar posting here. On closer examination, however, you will see that:

(1) Of the two answers, the first one either uses a very advanced theory which I am clueless, or it is a "cute abstract nonsense proof," as pointed out by two members. Because of that, I have doubt it is useful.
(2) The second answer leads to nothing, because the link is broken. Check this one out.
(3) Finally, the OP seems like asking the question in light of path-connectedness, which I think is different from mine, since I am asking for complete proof.

Because of the above reasons, I decided to post my question. Sorry again for forgetting to post this caveat in advance. :-) Thank you very much.

Answer 1

assume $x_0$ is the identity element...let $f$ and $g$ be two loops at $x_0$ then basically we can write $f*g =(f*e_{x_0}).(e_{x_0} *g) \approx (e_{x_0}*f).(g*e_{x_0}) = g*f$

Answer 2

Here's Grothendieck's non elementary answer (as an elementary answer has already been given) to your question (this doesn't happen everyday ;-)) : the functor $$\pi^1 :(\textrm{Top}') \to (\textrm{Gr})$$ from the category $(\textrm{Top}')$ of path-connected topological spaces to the category $(\textrm{Gr})$ of groups commutes with products and maps therefore group objects to group objects. Now, group objects of $(\textrm{Top}')$, path-connected topological groups, are send to group objects of $(\textrm{Gr})$, and the latter are abelian groups.

For interested people, note that this is false in the étale case, namely, the étale fundamental group of groupe scheme may not be abelian. Let $G_0$ be a smooth group scheme defined over a finite field $k$ of cardinal $q$ and let $G$ be the base-change to an algebraic closure of $k$. Then there is the Lang map $L: G \to G$ defined by $x \mapsto (Fx) x^{-1}$ where $F$ is the Frobenius. By a theorem of Lang $L$ is a surjective map. It is also finite since the fiber over the identity is finite (the $k$-rational points) and because a morphism of homogeneous spaces for an algebraic group with finite fibers is finite. () (See the proof of () in comment to this answer.) But the Frobenius induces the zero map on the tangent spaces, so that the Lang morphism $x \mapsto (Fx) x^{-1}$ is smooth as the morphism on tangent spaces is a bijection. So, as the fibers are finite, $L$ is an étale cover. Now $L$ is in fact a Galois cover whose automorphism group is $G_0(k)$ : the right translations by the $k$-rational points are automorphisms of the cover. Since this is the right number of translations for the degree of the cover, we have indeed a Galois cover with the appropriate Galois group. But, in general, this Galois group won’t be abelian. So the étale fundamental group, which surjects onto every Galois group, can’t be abelian either.

Answer 3

This question has been answered at $G$ is Topological $\implies$ $\pi_1(G,e)$ is Abelian . My answer to that puts the question in a wider context by considering the fundamental groupoid of a topological group, and the link to this paper on covering groups of non-connected topological groups.