 Solution

Writing the square coordinates modulo one, the curve has the parametrizationIt is therefore the $(2,3)$ torus knot, or trefoil, on the common boundary of the two solid tori. Applying the Seifert-van Kampen theorem to the complements on the two sides gives one generator $a$ for the meridian of one solid torus and one generator $b$ for the other; the boundary slope identifies two turns of the first with three turns of the second. HenceThis group is nonabelian: imposing the additional relation $a^2=b^3=1$ gives a quotientwhich maps onto the nonabelian group $S_3$ by sending $a$ to a transposition and $b$ to a three-cycle. An abelian group cannot have a nonabelian quotient.