TITLE: Is this a different proof of the fundamental group being abelian?
QUESTION [13 upvotes]: I have proved the fundamental group of a topological group is abelian. But I've found nowhere the similar proof as mine. Everywhere I looked up, it was done either exploiting categorical properties or something like taking product of two paths. 
My proof goes as follows:
Let $a$ and $b$ be two loops in a topological group $(G,\bullet )$ starting at the identity element $e$. We need to show $ a\ast b \simeq b\ast a$, where "$\ast$" is the fundamental group operation.
Now for each $t,s\in [0,1]$, define
$F_t(s)=a(st)\ast(a(t)\bullet b(s))\ast \bar a(st)$
Now {$F_t$} gives the homotopy between $b$ and $a\ast b \ast \bar a$.

The main idea is at each time $t$, we first go to $a(t)$ along $a$ and then traverse the translated path $a(t)\bullet b$ and then return back along the inverse path of the first one. Continuity of $F$ follows from pasting lemma.

This proof seems correct but why do other proofs avoid this straightforward argument?

REPLY [4 votes]: Your idea is very nice, but the definition
$$F_t(s)=a(st)\ast(a(t)\bullet b(s))\ast \bar a(st)$$
is inadequate because $a(st)$, $a(t)\bullet b(s))$, $\bar a(st)$ are single points of $G$ which cannot be composed by $*$ which is the succession of paths. What you mean is
$$F_t = a_t * (a(t) \bullet b) * \overline{a_t}$$
where $a_t(s) = a(st), \overline{a_t}(s) = a_t(1-s) = a((1-s)t)$. Note that $\overline{a_t}$ is the inverse of $a_t$, not part of the inverse $\bar a$ of $a$. In fact, $a((1-s)t) \ne  a(1-st) = \overline{a}(st)$. Explicitly you can also write
$$F_t(s) =  \begin{cases} a(3st) & 0 \le s \le 1/3 \\ a(t) \bullet b(3s - 1) & 1/3 \le s \le 2/3 \\  a((3(1-s)t)  & 2/3 \le s \le 1 \end{cases}$$