Munkres's book, $ {\pi}_1 (X, x_0) $ is abelian if and only if In the Munkres' book of topology have this question:
$ {\pi}_1 (X, x_0) $ is abelian if and only if for every pair $\alpha $ and $\beta $ of paths from $x_0$ to $x_1$, we have $ \widehat{\alpha} = \widehat{\beta} $. where $X$ is a path-connected space.
to prove in the direction $(\Rightarrow)$ I think of using, for example
$ \widehat{\alpha}([f]) = \widehat{\beta}([f]) $.
$\Leftrightarrow$
$[\bar{\alpha}]*[f]*[\alpha] = [\bar{\beta}]*[f]*[\beta] $
$\Leftrightarrow$
$([\bar{\alpha}]*[\beta])*([\bar{\beta}]*[f]*[\alpha]) = ([\bar{\beta}]*[f]*[\bar{\alpha}])*([\alpha]*[\beta]) $
but $[\bar{\alpha}]*[\beta]\notin {\pi}_1 (X, x_0) $ and $[\bar{\beta}]*[f]*[\alpha] \notin {\pi}_1 (X, x_0) $
and thus I can not use commutativity to complete the proff. I tried other ways to arrange the terms but I always fall into the same problem. 
still, I can not carry out the operation $[\alpha]*[\beta]$ because $\alpha(1)
\neq \beta(0)$.
 A: Suppose $\pi_1(X,x_0)$ is abelian (and thus $\pi_1(X,x_1)$ as well). 
Then \begin{eqnarray}
[\overline{\alpha}]*[f]*[\alpha] &=& [\overline{\alpha}]*[f]*([\beta]*[\overline{\beta}])*[\alpha]\\
&=& ([\overline{\alpha}]*[f]*[\beta])*([\overline{\beta}]*[\alpha])\\
&=& ([\overline{\beta}]*[\alpha])*([\overline{\alpha}]*[f]*[\beta])\\
&=& ([\overline{\beta}]*[f]*[\beta]).
\end{eqnarray}
Here I used that $([\overline{\alpha}]*[f]*[\beta])$ is a loop at $x_1$ and $([\overline{\beta}]*[\alpha])$ is also a loop at $x_1$.
EDIT: as requested in the comments, we will prove the reverse implication as well.
Let $f$ and $g$ be two loops with base point $x_0$ and let $\alpha$ be a path from $x_0$ to $x_1$. Then $\beta=f*\alpha$ is a path from $x_0$ to $x_1$ as well. By assumption, $\widehat{\alpha}(g)=\widehat{\beta}(g)$. Hence 
\begin{eqnarray}
 [\overline{\alpha}]*[g]*[\alpha] &=& [\overline{\beta}]*[g]*[\beta]\\
 [\beta]*[\overline{\alpha}]*[g]*[\alpha] &=& [g]*[\beta]\\
 [f]*[g]*[\alpha] &=& [g]*[\beta]\\
 [f]*[g] &=& [g]*[\beta]*[\overline{\alpha}]\\
 [f]*[g] &=& [g]*[f]\\
\end{eqnarray}
This completes the proof.
A: Apart from answer given by Mathematician 42, if you want to work within $\pi_1(X, x_0)$, then assuming $\pi_1(X, x_0)$ is abelian, for $[f]\in\pi_1(X, x_0)$, 
$$\begin{eqnarray}
\widehat{\alpha}([f])&=& [\overline{\alpha}]*[f]*[\alpha]\\
&=& ([\overline{\beta}]*[\beta])*([\overline{\alpha}]*[f]*[\alpha])*([\overline{\beta}]*[\beta])\\
&=& [\overline{\beta}]*([\beta]*[\overline{\alpha}])*[f]*([\alpha]*[\overline{\beta}]*[\beta])\\
&=& [\overline{\beta}]*[f]*([\beta]*[\overline{\alpha}])*([\alpha]*[\overline{\beta}]*[\beta])  \;\;\;\;\{\textrm{because } [f], [\beta]*[\overline{\alpha}]\in\pi_1(X, x_0)\}\\
&=& [\overline{\beta}]*[f]*[\beta]\\
&=& \widehat{\beta}([f]).
\end{eqnarray}$$
