On proving the problem that $\pi_1(X,x_0)$ is Abelian iff for every pair $\alpha$ and $\beta$ if paths form $x_0$to$x_1$, we have $\hat \alpha=\hat\beta$ where $X$ is path-connected space ,
I'm curious of meaning for what fundamental group is Abelian. If possible, could you explain the case of product of given two distinct loop would not be commutable on the $R^2$?.
On
Consider the wedge sum of two circles (that is, two circles that are connected by one point, let's call it the base point). By Van Kampen's theorem, you can see that the fundamental group of this space is the free product $\mathbb{Z} * \mathbb{Z}$.
Obviously this is not abelian. In fact, if you consider the loop $\alpha$ which does one turn of the first circle, from the base point, and the loop $\beta$ which does one turn on the second cicrle from the base point, the images of these two loops in the fundamental group are not commuting.
A non-Abelian fundamental group $\pi_1(X, x_0)$ means that there are two loops $\alpha, \beta: [0,1] \to X$, so continuous with $\alpha(0) = \alpha(1) =\beta(0) = \beta(1) = x_0$, such that there is no ($x_0$-fixing) homotopy between $\alpha \cdot \beta$ and $\beta \cdot \alpha$, where $\alpha \cdot \beta$ is (the homotopy class of) the map that sends $t$ to $\alpha(2t)$ for $0 \le t \le \frac12$ and to $\beta(2t-1)$ for $\frac12 \le t \le 1$, as usual, and $\beta \cdot \alpha$ is the similarly glued map in the other order. This just follows from the definitions.