For every $\alpha,\beta$ , $\hat{\alpha}=\hat{\beta}$ $\Leftrightarrow$ for every $\alpha,\beta$ and $f$ , a loop based at $x_{0}$ , $\hat{\alpha}([f])=\hat{\beta}([f])$ $\Leftrightarrow$ $[\overline{\alpha}]\star[f]\star[\alpha]=[\overline{\beta}]\star[f]\star[\beta]$ $\Leftrightarrow$ $[f]\star[\alpha]\star[\overline{\beta}]=[\alpha]\star[\overline{\beta}]\star[f]$ $\Leftrightarrow$ $[f]\star[\alpha\star\overline{\beta}]=[\alpha\star\overline{\beta}]\star[f]$ . Note that $\alpha\star\overline{\beta}$ is an arbitrary loop based at $x_{0}$ that passes through $x_{1}$ . So, if the group of the path homotopy classes of loops based at $x_{0}$ is abelian, then the right hand side expression holds for arbitrary $\alpha$ , $\beta$ and $f$ , therefore, for every $\alpha,\beta$ , $\hat{\alpha}=\hat{\beta}$ . Vice versa, if for every $\alpha,\beta$ , $\hat{\alpha}=\hat{\beta}$ , then we have shown that the group is commutative when at least one of the terms is a path homotopy class of a loop passing through $x_{1}$ . So, take arbitrary $[f],[g]∈\pi_{1}(X,x_{0})$ and take any path $\alpha$ from $x_{0}$ to $x_{1}$ . Then, $g\star\alpha\star\overline{\alpha}$ is a loop based at $x_{0}$ passing through $x_{1}$ . Then, $[f]\star[g\star\alpha\star\overline{\alpha}]=[g\star\alpha\star\overline{\alpha}]\star[f]$ , but $[g\star\alpha\star\overline{\alpha}]=[g]$ .