## Transcribed Text

1. Let (1; 2; 1) be any quasigroup of order 2, (Q; 2) an idempotent commutative quasi-
group of order 3, and let S = 1; 2Q. Dene a binary operation
on S by: (a; c)
(b; d) =
(a 1 b; c 2 d): Then (S;
) is a commutative quasigroup of order 6.
(a) Construct this commutative quasigroup of order 6 with holes.
(b) Rename the symbols with 1, 2, 3, 4, 5, 6 so that the holes are f1; 2; 3; 4g and f5; 6g.
2. Construct STS(13) using Bose Construction. Use the STS(13), you constructed, to
produce a commutative quasigroup with holes of order 12.
3. Use the Quasigroup with Holes construction and the quasigroup with hole of order
6 to nd an STS(21).
Example: For each h 2 H we need to choose a quasigroup (h; ). For each h = x; y 2 H,
we have two choices:
x y
x x y
y y x
or
x y
x y x
y x y
The following are quasigroups with holes H of orders 6 and 8. The quasigroup in the cells
h h are bordered by heavy lines and the symbols they contain are in bold type. These
examples are also commutative.
1 2 3 4 5 6
1 1 2 5 6 3 4
2 2 1 6 5 4 3
3 5 6 3 4 1 2
4 6 5 4 3 2 1
5 3 4 1 2 5 6
6 4 3 2 1 6 5
1 2 3 4 5 6 7 8
1 1 2 5 6 7 8 3 4
2 2 1 8 7 3 4 6 5
3 5 8 3 4 2 7 1 6
4 6 7 4 3 8 1 5 2
5 7 3 2 8 5 6 4 1
6 8 4 7 1 6 5 2 3
7 3 6 1 5 4 2 7 8
8 4 5 6 2 1 3 8 7
We now solve the existence problem for commutative quasigroups with holes H. It turns
out that they are easy to construct if their order is 2 (mod 4).

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