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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.

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A.Magnus
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    possible duplicate of The fundamental group of a topological group is abelian – fixedp Mar 05 '15 at 15:10
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    @fixedp : I just put in Post Script, see above. Thanks. – A.Magnus Mar 05 '15 at 15:40
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    There is no page on any stack exchange site where this question is correctly treated. People only seems to rephrase $\pi^1$'s commutativity without proving anything. So this question does not diserve to be closed. – Olórin Mar 05 '15 at 16:15
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    @RobertGreen The old question can still be answered... I don't understand why some people are so intent on splitting the answers to the same question in different places. – Najib Idrissi Mar 05 '15 at 16:22
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    @A.Magnus Path-connected is irrelevant here. The fundamental group only "sees" the path component of the base point. And that part is important: the fundamental group is defined for pointed spaces. – Najib Idrissi Mar 05 '15 at 16:24
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    @NajibIdrissi I don't think that the rule is do not ask something that as already been asked without a proper answer, and your strategy to tackle such an old question is really not optimal, as it is really not easy to revive an older question. – Olórin Mar 05 '15 at 16:25
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    @RobertGreen What do you mean "it is not easy"? Fill in the answer box at the bottom, click "Post Your Answer", and you've just revived the question. And that question had already been asked not once, not twice, not three times, but four times before! Possibly even more! http://math.stackexchange.com/q/686496 http://math.stackexchange.com/q/727999 http://math.stackexchange.com/q/276396 http://math.stackexchange.com/q/741391 – Najib Idrissi Mar 05 '15 at 16:27
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    @NajibIdrissi My dear, that's not the role of the OP. He's asking a question that has not received, wherever it has been already asked, a proper answer. This is perfectly kosher. So that the question does not diserve to be closed. That's the only thing I am saying. Now, answering to it in another place, pointing to it to the OP, and saying that the question has already been asked somewhere and asking to close OP's question would be particularly low-lying. – Olórin Mar 05 '15 at 16:32
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    @RobertGreen "My dear"? If I count correctly, that question had already received five answers (plus one that's become obsolete due to link rot) beforehand. Are you seriously claiming none of them were adequate? (In particular this one should be okay with your "explicitness" criterion). And it is absolutely the role of the OP to search if their question has already been answered. PS: this isn't the place to discuss this. I've started a discussion here: http://chat.stackexchange.com/rooms/2165/reopen-undelete-close-delete – Najib Idrissi Mar 05 '15 at 16:35

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

Anubhav Mukherjee
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    Thanks! Couple of questions: (a) What do you mean by $\approx$ symbol? (b) I did not see how to jump from $(e_{x_0} * f)\cdot (g * e_{x_0})$ to $g * f$, I mean, suddenly reversing the order from $f, g$ to $g, f$, could you please? Thanks again. – A.Magnus Mar 05 '15 at 16:06
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    $\approx$ means homotopically equivalent relative to the base point – Anubhav Mukherjee Mar 05 '15 at 16:35
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    for your 2nd doubt...just point wise multiplication will give you the equality – Anubhav Mukherjee Mar 05 '15 at 16:36
  • Sorry to chase after you again: (a) The * is an operator for homotopy class, whereas $\cdot$ is operator for group, am I correct? (b) When you wrote $fg =(fe_{x_0}) \cdot (e_{x_0}*g)$, is that an established theorem, or one that need to be proven first in an assignment? Thanks again. – A.Magnus Mar 05 '15 at 17:50
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    I dont know wheather it is a theorem or not...but you can easily prove r.h.s =l.h.s by the definitions...I mean calculate point wise...and you can get the answer...and a) is correct :) – Anubhav Mukherjee Mar 05 '15 at 17:54
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    Yes, the book says that $[f][f'] = [f f']$, so obviously $[f][e_{x_0}] = [f e_{x_0}]$, I guess. But what I don't understand is the sudden appearance of $\cdot$ in the RHS. Is that what you refer as pointwise product? Can you elaborate a bit about it? Or any link perhaps? Remember you are talking to a slowpoke here, so patience is in order here :-) Thanks again. – A.Magnus Mar 05 '15 at 18:03
  • yes sir you can ask as many doubts as you want... can you just tell me which step you are talking about? – Anubhav Mukherjee Mar 05 '15 at 18:25
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    Thanks. I am looking at your $fg =(fe_{x_0})\cdot (e_{x_0} g)$, how do you get that? I already got from you that is homotopy operator and $\cdot$ is group operator. If you need more space, since we are advised to go to chat, you might want to add your answer on your answer page instead - just a thought to make it more convenience to you. Thanks again. – A.Magnus Mar 05 '15 at 18:28
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    $fg(t) = f(2t)$ when $0\leq t \leq 1/2 $and$ g(2t-1)$ $when 1/2\leq t\leq 1$ $(fe_{x_0}).(e_{x_0}g) (t) = fe_{x_0}(t).e_{x_0}g(t)$=$fg(t)$ .. since $x_0$ is the identity element and $e_{x_0}(t) =x_0$ – Anubhav Mukherjee Mar 05 '15 at 18:33
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    I see, so you were referring to product of $f$ and $g$. Can you explain how do you get $(fe_{x_0}).(e_{x_0}g) (t) = fe_{x_0}(t).e_{x_0}g(t)$? According to my slow mind, your $(fe_{x_0})\cdot (e_{x_0}g) (t)$ should equals to $[(fe_{x_0})\cdot e_{x_0}(t)] [(f*e_{x_0}) \cdot g(t)$], similar to operation of "distribution"? Thanks again. – A.Magnus Mar 05 '15 at 19:54
  • consider $f_1 = fe_{x_0}$ and $f_2=e_{x_0} g$ now compute $f_1.f_2$ explicitely... i mean point wise – Anubhav Mukherjee Mar 05 '15 at 20:05
  • Would the homotopy $H(t,s) = f((1-t)s)⋅g((1-t)s)⋅(f∗g)(ts)$ also work directly? – Elie Bergman Jul 08 '17 at 10:57
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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.

Olórin
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  • Proof of (*) : a morphism of reduced homogeneous spaces over a smooth group scheme is faithfully flat, by generic flatness and a translation argument. As a result, it’s a quotient map. Since the fibers are finite, one can check that the map is closed by a similar translation argument. Then, proper plus finite fibers implies finite by Zariski’s Main Theorem. – Olórin Mar 05 '15 at 15:38
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    Thanks, but is there anyway you can complement with a solution without the big tools? I am still an introductory slowpoke :-) Thanks again. – A.Magnus Mar 05 '15 at 15:51
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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.

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