Question about group operation in Fundamental group

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I want to show that if the fundamental group $\pi_1(X,x_0)$ is abelian where $X$ is path-connected, then for any two paths $h_1,h_2: I\to X$ from $x_0$ to $x_1$, $\beta_{h_1}\equiv \beta_{h_2}$ where $\beta_h([f])=[h^{-1}*f*h]$. It seems that most of the solutions looks like the following: $\beta_{h_1}([f])=[h_1^{-1} *f*h_1]=[h_1^{-1}*f*h_2*h_2^{-1}*h_1]$ and as $X$ is path-connected, $\textit{we can say that $\pi_1(X,x_1)$ is also abelian}$. Hence, $[h_2^{-1}*h_1*h_1^{-1}*f*h_2]=[h_2^{-1}*f*h_2]=\beta_{h_2}([f])$. But in some solution, one proved this statement as the following: $[f]*[h_1*h_2^{-1}]=[h_1*h_2^{-1}]*[f]\iff [h_1*h_2^{-1}]^{-1}*[f]*[h_1*h_2^{-1}]=[f]\iff [h_2*h_1^{-1}]*[f]*[h_1*h_2^{-1}]=[f]\iff [h_1^{-1}]*[f]*[h_1]=[h_2^{-1}]*[f]*[h_2]$. Something like this form. I mean they do the concatenation operation even if $[h_1]$ is not an element of whatever fundamental group. So my question is, $\textit{is this operation possible?}$. In other words, even if we argue in fundamental group, is conatenation operation valid when the operation is well-defined not only as group operation but also as operation between paths.

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enter image description hereenter image description hereenter image description here The concatenation operation is, in general, is defined for two paths such that the initial point of one path coincides with the terminal point the other.
As you can see from the picture the paths has right identity and left identity as well as right and left inverses. Hence even though your [$h_1$] doesn't belong to the fundamental group it can be thought of belonging to the groupoid and hence those operations are valid.