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 + pH + 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.
Copyright © 1996-2001 Alexander Bogomolny
|