I am working on solving the following question:
Let $G$ be a path-connected topological group. Let $\alpha,\beta:[0,1]\rightarrow G$ be two loops based at $e$ in $G$, and consider the map $$F:[0,1]\times[0,1]\ni(s,t)\mapsto\alpha(s)\beta(t)\in G,$$ where the multiplication of $\alpha(s)$ and $\beta(t)$ is the group multiplication. Draw a diagram to show the effect of this map on the square, and prove that the fundamental group of $G$ must be abelian (therefore a topological space with non-abelian fundamental group cannot have a compatible group structure).
It is clear to me that $\alpha(0)=\alpha(1)=\beta(0)=\beta(1)=e$, hence $F(s,0)=F(s,1)=\alpha(s)$ and $F(0,t)=F(1,t)=\beta(t)$, but I don't really understand what $\alpha(s)$ and $\beta(t)$ are (besides being group elements). It seems to me that in order to show $\langle\alpha\rangle\langle\beta\rangle=\langle\beta\rangle\langle\alpha\rangle$, I need to explicitly construct a homotopy between the loop concatenation $\alpha\beta$ and the group multiplications $\alpha(t)\beta(t)$ and $\beta(t)\alpha(t)$. However, with my lack of understanding of what exactly these are, I'm struggling to do so. I realize this question has been solved in other places on SE, but what I'm specifically asking is for help interpreting the necessary pieces.
Note that $t\mapsto F(t,0)$ is just $\alpha$ because $\beta(0)=e$. Likewise, $t\mapsto F(t,1)$ is $\alpha$ and either of $t\mapsto F(0,t)$ and $t\mapsto F(1,t)$ is $\beta$. Thus by moving along the left and then top edge of the square $[0,1]\times [0,1]$, we follows first $\alpha$, then $\beta$; whereas along the bottom and then right edge, we follow first $\beta$, then $\alpha$. These two paths within the square are clearly homotopic. Applying $F$ to such a homotopy, we obtain the desired homotopy between the two loops.