### William A. McWorter Jr.

 Let H be the subgrid (and subquasigroup) of T induced by the grid triangle with vertices (0, 0), (a, b), and (-b, a + b), where a and b are whole numbers, not simultaneously zero. The * operation may be defined on the factor group T/H by (H + p) * (H + q) = H + p * q. which converts T/H into a quasigroup. If a - b is not divisible by 3, then the quasigroup T/H is anticommutative. First, let's settle the fact that the operation * on T/H is well-defined. Let H + p = H + p' and H + q = H + q'. Then p - p' and q - q' are in H. We show that H + p + f(q - p) = H + p' + f(q' - p').

p + f(q - p) - (p' + f(q' - p')) = p - p' + f(q - q') - f(p - p').

Since f fixes H and p - p' and q - q' are in H, the right side of the above equation is a sum of elements of H and so is in H. Hence H + p + f(q - p) = H + p' + f(q' - p').

Next, we show that every equation (H + p) * (H + x) = H + q has a unique solution H + x, for every H + p and H + q in T/H. Existence of H + x follows from the fact p * x = q has a solution r in T. For uniqueness, let H + r' be any solution of (H + p) * (H + x) = H + q. Then p + f(r - p) - q is in H and p + f(r' - p) - q is in H. So

p + f(r - p) - q - (p + f(r' - p) - q) = f(r - r') is in H.

Since f fixes H, the point r - r' is in H, and so H + r = H + r'. That (H + y) * (H + p) = H + q has a unique solution H + y in T/H follows from a similar argument. Since y * p = q has a solution s in T, (H + s) * (H + p) = H + q. If H + s' is any solution of (H + y) * (H + p) = H + q, then s + f(p - s) - q and s' + f(p - s') - q are both in H. So

s + f(p - s) - q - (s' + f(p - s') - q) = s - s' - f(s - s') = - f(f(s - s'))

is in H.

Since f fixes H, so does f -1, whence s - s' is in H, and so H + s = H + s'.

Finally, we show that T/H is anticommutative, that is, that (H + p) * (H + q)≠(H + p) * (H + q) whenever H + p≠H + q. Deny the result and assume that (H + p) * (H + q) = (H + q) * (H + p), for some cosets H + p and H + q, with H + p≠H + q. Then p - q is not in H.

(H + p) * (H + q) = (H + q) * (H + p) implies that p + f(q - p) - (q + f(p - q)) is in H.

Hence p + f(q - p) - (q + f(p - q)) = p - q - 2f(p - q) is in H.

Also, 2f(p - q - 2f(p - q)) = 2f(p - q) - 4f(f(p - q)) and

4(f(f(p - q)) - f(p - q) + p - q) = 4·0 = 0 are in H.

Thus, the sum

[2f(p - q) - 4f(f(p - q))] - [(p - q) - 2f(p - q)] + 4(f(f(p - q)) - f(p - q) + p - q) = 3(p - q)

is in H. Since p - q is not in H, this means that the factor group T/H contains the element H + p - q of order 3. By Lagrange's theorem, 3 must divide a2 + ab + b2. But a2 + ab + b2 is congruent to a2 - 2ab + b2 = (a - b)2 modulo 3, whence 3 divides a - b, against our choice of a and b. The result is proved; T/H under * is anticommutative.  