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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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