Completing partial transversals of Cayley tables of Abelian groups

In 2003 Grüttmüller proved that if n > 3 is odd, then a partial transversal of the Cayley table of Zn with length 2 is completable to a transversal. Additionally, he conjectured that a partial transversal of the Cayley table of Zn with length k is completable to a transversal if and only if n is odd and either n ∈ {k, k + 1} or n > 3k − 1. Cavenagh, Hämäläinen, and Nelson (in 2009) showed the conjecture is true when k = 3 and n is prime. In this paper, we prove Grüttmüller’s conjecture for k = 2 and k = 3 by establishing a more general result for Cayley tables of Abelian groups of odd order. Mathematics Subject Classifications: 05B15, 20K01


Introduction
It is well-known that Cayley tables of finite cyclic groups have transversals if and only if their order is odd. In fact, such Cayley tables have many transversals, most recently shown by Eberhard, Manners, and Mrazović [2], who proved that the number of transversals in a Cayley table of Z n is on the order of (e −1/2 + o(1))n! 2 /n n−1 . With this in mind, it seems plausible that for a positive integer k and a large enough n, any partial transversal of length k in a Cayley table of Z n is contained in a transversal.
For a finite group G, let C(G) denote the Cayley table of G, and let n 3 be odd. We say that a partial transversal P of C(G) is completable if there exists a transversal T ⊆ C(G) such that P ⊆ T ; we say that T is a completion of P . First observe that partial transversals of C(Z n ) with length 1, n − 1, or n, are completable. In particular, partial transversals of length 1 are completable since C(Z n ) is decomposable into transversals; note that for Latin squares in general, completing partial transversals of length 1 is not always possible [7]. Grüttmüller proved an analogous result for partial transversals of C(Z n ) with length 2, as well as other special cases.
Theorem 1 (Grüttmüller [3]). A partial transversal of C(Z n ) with length 2 is completable if and only if n is odd and n 3.
Grüttmüller also established necessary conditions for a partial transversal of C(Z n ) with length k to be completable.
Theorem 3 (Grüttmüller [4]). If k 3 and every partial transversal of C(Z n ) with length k is completable, then n is odd and either n ∈ {k, k + 1} or n 3k − 1.
Stemming from this, Grüttmüller made the following conjecture.
Conjecture 4 (Grüttmüller [4]). If k 1, then every partial transversal of C(Z n ) with length k is completable if and only if n is odd and either n 3k − 1 or n ∈ {k, k + 1}.
The only known progress on Conjecture 4 is due to Cavenagh, Hämäläinen, and Nelson.
Theorem 5 (Cavenagh et al. [1]). Let n 11 and n prime. Then any partial transversal of C(Z n ) with length 3 is completable.
In this paper we settle Grüttmüller's conjecture for the case k = 3. In fact, we prove the following stronger result. Theorem 6. For an Abelian group G of odd order n and k ∈ {2, 3}, every partial transversal of C(G) with length k is completable if and only if either n 3k − 1 or n ∈ {k, k + 1}.
Observe that for any Abelian group G of odd order n, a partial transversal of C(G) with length 1, n − 1, or n is completable for the same reasons as when G is cyclic and thus, for the same reasons, an appropriate first step in this generalization is to establish the conditions under which partial transversals of C(G) with lengths 2 and 3 are completable.
Lastly, we mention here that the work cited above was presented in terms of diagonally cyclic Latin squares. For odd n, a Latin square L ⊆ Z n × Z n × Z n is diagonally cyclic if (i + 1, j + 1, k + 1) ∈ L whenever (i, j, k) ∈ L. There is a one-to-one correspondence between diagonally cyclic Latin squares of order n and transversals in C(Z n ) as shown in the following example. Example 7. Let L be the diagonally cyclic Latin square of order 5 given in Figure 1.
Note that the first row of L determines L, and the map which associates (0, i, s i ) ∈ L with (s i − i, i, s i ) ∈ C(Z 5 ) demonstrates the correspondence between L and the highlighted transversal of C(Z 5 ).
Thus completing partial Latin squares of order n to diagonally cyclic Latin squares of order n is equivalent to completing partial transversals of C(Z n ) to transversals. With this in mind, the theorems and conjecture above can be restated using diagonally cyclic Latin squares. For more on diagonally cyclic Latin squares and related objects, see the survey by Wanless [6].

Coset Blocks and Chains
Let n be a positive odd integer and G be an Abelian group of order n, expressed additively. We treat C(G) as a Latin square, that is, as a subset of G × G × G, where (a, b, c) ∈ C(G) if and only if c = a + b in G.
Observe that when k = 2 or k = 3, Conjecture 4 is true for Abelian groups of prime order since these groups are cyclic. For the remainder of the paper, we focus on developing machinery for handling C(G) when G has composite order.
Let n be composite with prime divisor q and m be defined so that n = mq. Define so that q = 2 + 1. Let H be a subgroup of G of order m. We use 0 for the zero element of H and for each k ∈ Z and for each g ∈ G, let kg = g + g + · · · + g (k times). Note that the factor group G/H has prime order q and thus G/H is isomorphic to Z q ; let µ : Z q → G/H be such an isomorphism. Let H 0 = H and H 1 , H 2 , . . . , H q−1 denote the remaining cosets that is isomorphic (as a Latin square) to C(H). We call this Cayley table a coset block of order m in C(G) and we denote it as C i,j (H; G, µ) (or when the context makes it unambiguous, simply C i,j (H)); note that C 0,0 (H) = C(H). We denote C 0,d (H) ∪ C 1,d+1 (H) ∪ · · · ∪ C q−1,d+q−1 (H) as the dth block diagonal of C(G) with respect to (H, µ) and call the 0th block diagonal the main block diagonal.
the electronic journal of combinatorics 28(3) (2021), #P3.60 For a subset P of C(G), we let P r , P c , and P s denote the set of rows, columns, and symbols used by the triples in P . Additionally for each i ∈ Z q , let P r i = P r ∩ H i , P c i = P c ∩ H i , and P s i = P s ∩ H i . We begin by observing an invariance in C(G); its use will reduce the complexity of later arguments.
We often leverage that if φ is an automorphism of C(G) and P is a partial transversal of C(G), then P is completable if and only if φ(P ) is completable as well. In particular, we frequently assume that (0, 0, 0) belongs to P . Indeed if (a, b, a + b) ∈ P , then (0, 0, 0) ∈ δ −a,−b (P ), and thus we can work with the image of P instead. Furthermore, for an integer c relatively prime to |G|, we use that ξ c : G → G given by ξ c (g) = cg is an automorphism of G, and therefore gives rise to an automorphism of C(G).
The constructions used in this paper hinge heavily on building chains, which are now defined, with examples that follow.
Definition 9. Let G be an Abelian group of odd order mq with q prime, m 3, and an order m subgroup H. Let µ : Define σ(P ), called the swap of P , as the subset {(x i , y i+d , x i + y i+d ) : i ∈ Z q } of C(G).
Since P satisfies the conditions outlined in (1), σ(P ) is also a partial transversal of C(G) such that P r = σ(P ) r , P c = σ(P ) c , and P s = σ(P ) s . Note that P consists of q cells, each of which are representatives of distinct coset blocks on the main block diagonal of C(G) with respect to (H, µ), while those in σ(P ) represent the coset blocks on the dth block diagonal of C(G) with respect to (H, µ). Furthermore, if T is a completion of P in C(G), then (T \P ) ∪ σ(P ) is also a transversal of C(G).  Figure 2, and is illustrated in Figure 3 We now present a classification result which highlights how to construct a chain.
Lemma 11. Let G be an Abelian group of odd order mq with q prime, m 3, and an order m subgroup H. Let µ : with respect to µ is determined by an SDR of G/H, each being a symbol used by the chain, along with a row used by one cell of the chain.
These 2q−1 equations uniquely determine the partial transversal P = {(x j , y j , s 2j ) : j ∈ Z q } for which Therefore, P is a (d, H)-chain of C(G) with respect to µ.
In later arguments we need disjoint chains -chains whose rows, columns, and symbol sets are disjoint. To that end we make the following observation which allows for the modest manipulation of a chain.
Definition 12. Let G be an Abelian group of odd order mq with q prime, m 3, and an order m subgroup H. Let µ : Observe that P 2 (5) is also a (3, H)-chain of C(G) with respect to µ. See Figure 3(b).
In practice, to find a completion of a partial transversal P of C(G), we partition P into two parts P and P , then find a chain C such that C ∪ P is completable and P ⊆ σ(C). Thus if τ is a completion of C ∪ P , then (τ \C) ∪ σ(C) is a completion of P in C(G). To that end we have the following two technical lemmas which allow for the construction of chains subject to certain conditions. Lemma 14. Let G be an Abelian group of odd order mq with q prime, m 3, and an order m subgroup H. Let µ : with respect to µ which contains (x i , y i , x i +y i ), and whose swap contains (x i , y i+d , x i +y i+d ).
Proof. By Observation 8, we may assume that i = 0 and x 0 = y 0 = 0. So we must show that a (d, H)-chain of C(G) with respect to µ exists which contains (0, 0, 0) and whose swap contains (0, y d , y d ).
We begin by letting Lemma 15. Let G be an Abelian group of odd order mq with q prime, m 3, and an order m subgroup H. Let µ : Z q → G/H be an isomorphism with µ(t) = H t for t ∈ Z q . Suppose • d, i, j ∈ Z q such that d is nonzero and 2i = 2j + d, • w, z ∈ H i , w ∈ H j , and z ∈ H j+d such that w + z = w + z , and • X is a collection of triples such that w, w / ∈ X r , z, z / ∈ X c , and w + z / ∈ X s .
Then there exists a (d, H)-chain of C(G) with respect to µ which is row-, column-, and symbol-disjoint from X, contains (w, z, w + z), and whose swap contains Proof. By using an appropriate operation from Observation 8, we may assume i = 0 and w = z = 0; hence j = d and z = −w .

Completing Partial Transversals
We begin with a proof of Theorem 6 with k = 2; note that Theorem 1 follows as a special case when the group G is cyclic.
Proof of Theorem 6 with k = 2. The forward direction is straightforward, so we focus on the reverse direction. Let n be odd and n 3. Let P be a partial transversal of C(G) and, without loss of generality, we may assume P = {(0, 0, 0), (a, b, a + b)} for some nonzero a, b ∈ G with a + b = 0.
If n is prime, then we may assume G = Z n . There exists v ∈ Z n so that av = b (mod n); then {(i, vi, i + vi) : i ∈ Z n } is a completion of P . Now assume n is composite and, if G is any Abelian group of odd order n with 3 n < n, then any partial transversal of C(G ) with length 2 is completable.
Let G be an Abelian group of odd order n = mq with q prime, m 3, and an order m subgroup H. Let µ : Z q → G/H be an isomorphism with µ(i) = H i for i ∈ Z q . Let α, β ∈ Z q such that a ∈ H α and b ∈ H β , and thus (a, b, a + b) ∈ C α,β (H). We consider four cases.
The remainder of this section builds an argument for completing partial transversals of C(G) with length 3, thereby proving the rest of Theorem 6. In what follows, when a partial transversal P cannot be completed to a transversal in a Latin square L, we say P is non-completable in L, or simply non-completable.
Observation 16. Note that Theorem 6 does not guarantee that a partial transversal in C(G) with length 3 is completable if |G| = 5 or 7. In other words, there may be non-completable partial transversals of C(G) with length 3 if G is isomorphic to Z 5 or Z 7 .
In fact, there are 100 and 294 partial transversals of C(Z 5 ) and C(Z 7 ) with length 3, respectively, which are non-completable. These were found by brute-force search and, conveniently, are easily described in the following way. For p ∈ {5, 7} and |G| = p, the non-completable partial transversals of C(G) with length 3 are of the form where a, b ∈ G, and c is an integer relatively prime to p (any such integer c is sufficient, but note there are only p − 1 distinct outcomes based on different choices of c). This partial transversal can be more simply expressed as δ a,b ξ c ({(0, 0, 0), (1, 1, 2), (2, −1, 1)}).
With the above classification, we now present two completion results in particular cases.
Lemma 17. Let n ∈ {25, 35, 49}, q be the smallest prime divisor of n, n = mq, G be an Abelian group of order n, and H be the subgroup of G having order m. If γ is a partial transversal of C(H) with length 3, then γ is completable in C(G).
Proof. Suppose τ is a completion of γ in C(H). Let µ : Z q → G/H be an isomorphism with µ(i) = H i for i ∈ Z q . Then τ is a transversal of C 0,0 (H) as well. For each i ∈ {1, 2, . . . , q − 1}, let τ i be any transversal of C i,i (H). Then τ ∪ τ 1 ∪ · · · ∪ τ q−1 is a completion of γ in C(G). Now assume that γ is non-completable in C(H), and let x ∈ H be nonzero. Note that x = H, and if G is cyclic, we let x = q. Then, by Observation 16, there exist a, b ∈ H and an integer c relatively prime to m such that γ = δ a,b ξ c (P ), where P = {(0, 0, 0), (x, x, 2x), (2x, −x, x)}. Observe that δ a,b ξ c is also an invariant map of C(G), as a, b ∈ G and c is relatively prime to n, except when m = 7 and q = c = 5; in this case use c = −2. In Figure 4, we give completions of P in all five cases. Therefore, γ is completable in C(G) as well.
Lemma 18. Let n ∈ {25, 35, 49}. Let G be an Abelian group of odd order n = mq with q the smallest prime divisor of n, and H be an order m subgroup of G. Let µ : Z q → G/H be an isomorphism with µ(i) = H i for i ∈ Z q . Let T 1 , T 2 ∈ C(G) such that (0, 0, 0), T 1 , and T 2 belong to distinct block diagonals of C(G) with respect to (H, µ). If γ = {(0, 0, 0), T 1 , T 2 } is a partial transversal of C(G), then γ is completable in C(G).   . For each noncyclic group Z p × Z p , we fix y ∈ G\H; note that x, y = G and each element of G is representable as ax + by for some a, b ∈ Z p , and we use ab to represent the element ax + by in the charts above.
Computer Proof. Despite our best efforts, the only way we found to justify this is through a brute-force computer search. Let d and e be the indices of the block diagonals with respect to (H, µ) containing T 1 and T 2 , respectively. Then perform the following procedure: here. Note that this routine successfully completed each possible γ, with very infrequent restarts. This leads us to speculate there are many chains which may be used to produce a completion, while very few would cause a conflict.
We now present a lemma highlighting how choosing the subgroup H wisely can affect the location of cells in coset blocks of C(G). Afterward, we describe another brute-force method for resolving another small case, and then conclude with the remaining proof of Theorem 6.
Lemma 19. Let p be prime, G = Z p × Z p , and T ∈ C(G). Then there exists a subgroup H of G with order p such that for all µ : Z p → G/H, T belongs to the main block diagonal of C(G) with respect to (H, µ).
Proof. Let x 1 , x 2 , y 1 , y 2 ∈ Z p so that a = (x 1 , y 1 ), b = (x 2 , y 2 ), and T = (a, b, a + b). If Hence there exists a subgroup H of G for which b − a ∈ H. So a and b belong to the same coset (say H i ) of H, and hence T ∈ C i,i (H).
Observation 20. The following is a naïve brute-force algorithm to determine which partial transversals of C(G) with length 3 are completable.
This algorithm quickly becomes infeasible as |G| increases. However, using this technique we confirmed all partial transversals in C(Z 9 ) and C(Z 3 ×Z 3 ) with length 3 are completable; note the former of these two results also follows from Theorem 2. Furthermore, we used this technique to classify the non-completable partial transversals of C(Z 5 ) and C(Z 7 ) as given in Observation 16. We also provide python code for this procedure here. Now we conclude with the remaining argument for Theorem 6.
Proof of Theorem 6 with k = 3. If n = 3 the result is trivial; if n 11 is prime, then the result follows from Theorem 5; and if n = 9 the result follows from Observation 20. Now, assume n 15 is composite, and if G is any Abelian group of odd order n with 9 n < n, then any partial transversal of C(G ) with length 3 is completable in C(G ).
Let n = mq with q the smallest prime dividing n. Then m 5, and let H be an order m subgroup of G. Let µ : Z q → G/H be an isomorphism with µ(i) = H i for i ∈ Z q .
Next, suppose two cells belong to the same coset block; without loss of generality, suppose T 1 ∈ C(H) and T 2 / ∈ C(H). Since m 3, there exists a completion τ of {T 0 , T 1 } in C(H). We now consider three subcases which are similar to the proof of Theorem 6 when k = 2.
Case 1c: Suppose α 2 = 0 and β 2 is nonzero. Let T be the cell in τ whose row is a 2 . By proceeding identically to the argument used in Case 1b, we may again produce a completion of γ in C(G). An identical argument holds if α 2 = 0 and β 2 = 0 by using transposes.
For the remaining cases, we may assume no two triples in γ belong to the same coset block. We separate these instances into subcases based on how many cells of γ share a block diagonal.

Concluding Remarks
In this paper, we considered partial transversals in Cayley tables of Abelian groups of odd order, and we achieved our result through an iterative approach which may appear to only leverage that the group is solvable. However, in constructing chains, we rely on the group being Abelian. So at this time, we believe a novel method may be required to extend the result to all odd-order groups by leveraging the Feit-Thompson Theorem.
Also, while Cayley tables of cyclic groups have transversals if and only if the order is odd, the same cannot be said for arbitrary Abelian groups of finite order. Let G be a finite Abelian group. A necessary condition for C(G) to contain a transversal is that the sum of all elements of G is 0, which is true for odd order cyclic groups, whereas such a sum is |G|/2 for even-order cyclic groups. However, the necessary condition is sufficient for the existence of transversals in C(G), regardless of the parity of |G|, through an application of a result discovered by Marshall Hall Jr. [5]. Observe that if |G| is even, then the sum of all elements of G is 0 if and only if Z 2 × Z 2 is a subgroup of G. This raises the following general question.
Question 21. Let G be an Abelian group for which all elements of G sum to zero. Given a positive integer k, is there a threshold d k such that if |G| d k , then all partial transversals of length k in C(G) are completable?
Another result used in this work was the classification of non-completable partial transversals of length 3. Given they have such a nice structure, this could imply a similar structure for non-completable partial transversals in C(G) with length k as |G| approaches 3k − 1 from below. While this looks interesting, this seems to be a much harder classification problem.
Finally, we want to thank the anonymous referees of our paper who helped with clarifying exposition and pushed us to make our results stronger.