The Eckmann-Hilton argument is the correct answer, but it might be amusing to note that there is a very explicit homotopy as well. Suppose alpha1, alpha2inpi1G, and define
alpha:I2rightarrowG
by alpha(t1,t2)=alpha1(t1)cdotalpha2(t2), where cdot is the product in G. Then along the diagonal, we have alpha1cdotalpha2, the product using the group operation, while along the bottom edge followed by the right edge we have the composition alpha1∗alpha2, the product of loops in the fundamental group. Deforming the path shows they're homotopic. Similarly, along the left edge, followed by the top edge we get alpha2∗alpha1, so this product is commutative.
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