TITLE: Fundamental group is abelian iff the fundamental group isomorphisms (a-hat) coincide
QUESTION [19 upvotes]: I want to show that if $X$ is a non-empty path connected space, then the fundamental group is abelian if and only if given any points $y, z\in X$ and paths $\alpha, \beta$ from $y$ to $z$, $\hat{\alpha} = \hat{\beta}$, where $(\hat{\alpha}([f]) = [\bar{\alpha}]*[f]*[\alpha]$ and similarly for $\beta$. I have one part of the iff:
For some $x\in X$, suppose $\pi_1(X, x)$ is abelian. Take $y, z\in X$ and paths $\alpha, \beta$ from $y$ to $z$. Then by isomorphism between the fundamental group at different base points of the same space, $\pi_1(X, z)$ is abelian. Let $f$ be a loop at $y$. Then $[\bar{\alpha}] * [f] * [\beta]$ is a loop at $z$, as is $[\bar{\beta}]*[\alpha]$, so
\begin{equation}
 \begin{split}
  \hat{\alpha}([f]) &= [\bar{\alpha}]*[f]*[\alpha]\\
    &= ([\bar{\alpha}] * [f] * [\beta])*([\bar{\beta}]*[\alpha])\\
    &= ([\bar{\beta}]*[\alpha])*([\bar{\alpha}] * [f] * [\beta])\\
    &= [\bar{\beta}]*[f]*[\beta]\\
    &= \hat{\beta}([f])
 \end{split}
\end{equation}
However, I'm having trouble with the other direction, which is to show that if $f, g$ are loops, then $[f]*[g] = [g]*[f]$. Any ideas?

REPLY [3 votes]: Here's another way, quite similar to gnometorule's answer, but a bit different.
Consider any $[f],[g] \in \pi_1(X,x)$.
Your sufficient condition in particular implies $\hat{f} = \hat{g}$, so
\begin{align}
    [f] = \hat{f}([f]) = \hat{g}([f]) = [\bar{g}] * [f] *[g]
    \ ,
\end{align}
i.e. $[f]$ and $[g]$ commute.