weak uniquely completable sets for finite groups
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Proof. We first prove that S is UC. The (0, n − 2) cell must contain either
n − 2 or n − 1. If we adjoin the triple (0, n − 2, n − 1) to S, then (1, n − 2, 0) is
forced, but no element may now be placed in the (2, n − 2) cell. Therefore we
must adjoin the triple (0, n − 2, n − 2), and then (n − 2, 0, n − 2) is forced. The set
S ∪ {(0, n − 2, n − 2), (n − 2, 0, n − 2)} is contained in Cn and contains the UC set of
Lemma 1; it is therefore UC to Cn.
We now prove that S is weak, by showing that there are no forced triples.
Consider the array of alternatives AS .
First, we show that no element occurs exactly once in any row of AS . This follows
since in the ith row of S, at least two of the cells (i, n − 3), (i, n − 2), (i, n − 1) are
empty, and the corresponding cells of AS each contain every one of the elements
0, 2, . . . , n − 4, n − 1 that does not appear in row i of S. This leaves the elements 1
and n − 2 to consider. (Note that S contains all n occurrences of n − 3.) If row i of
S does not contain a 1, then 1 appears in cells (i, n − 4) and (i, n − 3) of AS . The
element n−2 occurs only in cells (0, n−2), (0, n−1), (n−2, 0), (n−2, n−2), (n−1, 0),
(n − 1, n − 1) of AS , and so does not appear exactly once in any row of AS .
Secondly, we show that no element occurs exactly once in any column of AS .
In the jth column of S, at least two of the cells (n − 3, j), (n − 2, j), (n − 1, j) are
empty, and the corresponding cells of AS each contain every one of the elements
0, . . . , n − 4, n − 1 that does not appear in column j of S. This leaves the element
n − 2 but, as noted above, this occurs twice in columns 0, n − 2 and n − 1 of AS , and
nowhere else.
Finally, we need to show that each non-empty cell of AS contains at least two
elements. This follows since each non-empty cell of AS in row 0 or column 0 contains
both n − 2 and n − 1, and all the other non-empty cells of AS contain both 0 and
n − 1.
3. Weak UC sets for non-cyclic groups
In this section we prove that all finite groups of order greater than 5 possess
weak UC sets. We begin with an easy lemma.
Lemma 2. Let H be a subgroup of G. If H has a weak UC set, then so does G.
Proof. The Cayley table of G contains the Cayley table of H as a subsquare.
On replacing this subsquare by the partial latin square corresponding to a weak UC
set for H, we obtain a weak UC set for G.
By Cauchy’s theorem, a group whose order is divisible by a prime p greater than
5 has a cyclic subgroup of order p and so, by Lemma 2 and the Theorem, weak UC
sets exist for all such groups. This leaves only groups of orders n = 2a3b5c, where
n > 5, to consider. By the Sylow theorems, such a group has subgroups of orders 2a,
3b and 5c. We show below that the groups of orders 23, 32 and 52 all have weak UC
sets, so this leaves only the groups of orders 2a3b5c with a ∈ {0, 1, 2}, b, c ∈ {0, 1}, to
consider. Of these groups, only D3, D5, the dihedral groups of orders 6 and 10, A4,
the alternating group of order 12, and the group Hol(C5) do not have subgroups of
the kinds already considered or which are cyclic of composite order greater than 5.
(For the groups up to order 30, see [5].) For a group of order 60, we may argue as
follows. If G is a group of order 60, then let P be a Sylow 5-subgroup of G. If P is