All group-based latin squares possess near transversals

TL;DR

This paper proves that all group-based Latin squares, specifically those equivalent to finite group Cayley tables, contain near transversals, confirming a longstanding conjecture for this class.

Contribution

It establishes that the Brualdi-Ryser-Stein conjecture holds for Latin squares derived from finite groups, a significant subclass.

Findings

All group-based Latin squares have near transversals.

The conjecture is confirmed for Latin squares equivalent to Cayley tables.

Supports the conjecture's validity in a broad class of Latin squares.

Abstract

In a latin square of order $n$, a near transversal is a collection of $n-1$ cells which intersects each row, column, and symbol class at most once. A longstanding conjecture of Brualdi, Ryser, and Stein asserts that every latin square possesses a near transversal. We show that this conjecture is true for every latin square that is main class equivalent to the Cayley table of a finite group.

Equations

$${\Lambda }_{n,m}:=M_{km}+\left(\frac{l-m}{2}\right)Y_{2k}.$$

$$X_1{X}_2:=\{x_1{x}_2:x_1\in X_1,x_2\in X_2\}.$$

$$G=K⋉H.$$

$$G=⟨b⟩⋉H.$$

$$\alpha :h\mapsto aha.$$

$$H^{\ast }:=\{h\in H:\alpha (h)=h\}$$

$$T_1$$

$$T_2$$

$$T$$

$$U_1$$

$$U_2$$

$$U$$

$$W:=T\cup U.$$

$$⟨b⟩H^{\ast }\cap ⟨b⟩F=\varnothing .$$

$$b{H}^{\ast }{b}^{-1}=H^{\ast }.$$

$$bF{b}^{-1}=F$$

$$f\mapsto f^2\text{ is a permutation of }F.$$

$$(f,f),(f,fb),(bf,fb),\left(bf,f{b}^2\right),\dots ,\left(b^{k-1}f,f{b}^{k-1}\right),\left(b^{k-1}f,f\right).$$

$$b^i{f}^2{b}^{i+ϵ}=b^j{h}^2{b}^{j+\delta }$$

$$b^i{f}^2{b}^{i+ϵ}\in b^{2i+ϵ}F\text{ and }{b}^j{h}^2{b}^{j+\delta }\in b^{2j+\delta }F.$$

$$b^i{h}^2{b}^{i+ϵ}=b^{i+k/2}{f}^2{b}^{i+ϵ+k/2}=b^i\alpha (f^2)b^{i+ϵ},$$

$$(h_0,h_0),(h_0,h_{1}b),(b{h}_1,h_{1}b),\dots ,(b^{k-1}{h}_{m-1},b^{k-1}{h}_{m-1}),(b^{k-1}{h}_{m-1},h_0)$$

$$b^i{h}_i{h}_{i+ϵ}{b}^{i+ϵ}=b^{j}a{h}_j{h}_{j+\delta }a{b}^{j+\delta }=b^j{h}_j{h}_{j+\delta }{b}^{j+\delta },$$

$$b^i{h}_i{h}_{i+ϵ}{b}^{i+ϵ}=b^j{h}_j{h}_{j+\delta }{b}^{j+\delta }.$$

$$b^i{h}_i{h}_{i+ϵ}{b}^{i+ϵ}\in b^{2i+ϵ}{H}^{\ast }\text{ and }{b}^j{h}_j{h}_{j+\delta }{b}^{j+\delta }\in b^{2j+\delta }{H}^{\ast }.$$

Full text

1 Introduction and main theorems

A latin square of order $n$ is an $n\times n$ array in which each row and column is a permutation of some set of $n$ symbols. We refer to the set of cells containing any fixed symbol as a symbol class. Let $L=[L_{i,j}]$ be a latin square of order $n$. A partial transversal of $L$ is a collection of cells which intersects each row, column, and symbol class at most once. A transversal is a partial transversal of size $n$ and a near transversal is a partial transversal of size $n-1$. Although it is straightforward to find latin squares possessing no transversals (see [12, p.​ 405]), there is no known example of a latin square which does not possess a near transversal.

Conjecture 1**.**

Every latin square possesses a near transversal.

First discussed in the literature roughly 50 years ago, Conjecture 1 has been variously attributed to Brualdi, Ryser, and Stein (see [12, Section 5]). The strongest general lower bound to date is due to Hatami and Shor [9], who showed that every latin square possesses a partial transversal of size $n-O\left(\log^{2}(n)\right)$. There have also been numerous attempts to establish this conjecture as a special case of some stronger statement, including work in terms of hypergraph matchings [1], covering radii of sets of permutations [4], and colorings of strongly regular graphs [7]. The present paper approaches Conjecture 1 from the opposite direction by proving its most widely discussed special case (see [10, p.​ 335]).

Let $G$ be a finite group. The Cayley table of $G$, denoted $L(G)$, is the latin square with rows and columns indexed by the elements of $G$ where $L(G)_{g,h}=gh$. In general, a latin square is said to be group-based if there is an interpretation of its symbols which allows it to be realized as the Cayley table of some finite group.

Theorem 2**.**

Every group-based latin square possesses a near transversal.

It is worth noting that Theorem 2 can be trivially extended to a slightly wider class of latin squares. The existence of a near transversal is not affected by relabelling the rows, columns, or symbols of $L$, nor is it affected by permuting the roles played by rows, columns, and symbols. Thus, every latin square which is main class equivalent to a group-based latin square possesses a near transversal by Theorem 2.

We prove Theorem 2 using a graph-theoretic technical lemma. Letting $\mathcal{I}$ denote the set indexing the rows and columns of the latin square $L$, the latin square graph $\Gamma(L)$ is defined on the vertex set $\{(r,c)\,:\,r,c\in\mathcal{I}\}$ with $(r,c)\sim(s,d)$ if and only if one of $r=s$, $c=d$, or $L_{r,c}=L_{s,d}$ holds. Note that there is a natural tripartition of $\Gamma(L)$’s edges into, respectively, row edges, column edges, and symbol edges. Moreover, there is a bijective correspondence between near transversals of $L$ and independent sets of size $n-1$ in $\Gamma(L)$.

Given graphs $\Gamma=(V,E)$ and $\Gamma^{\prime}=(V^{\prime},E^{\prime})$, the disjoint union of $\Gamma$ and $\Gamma^{\prime}$ is $\Gamma+\Gamma^{\prime}:=(V\sqcup V^{\prime},E\sqcup E^{\prime}).$ For a positive integer $k$, we write $k\Gamma$ for the disjoint union of $k$ copies of $\Gamma$. Given a set $W\subseteq V$, the induced subgraph of $\Gamma$ with respect to $W$ is $\Gamma[W]:=(W,\{e\in E:e\subseteq W\}).$ The Möbius ladder of order $2n$, denoted $M_{n}$, is the cubic graph formed from a cycle of length $2n$—referred to as the rim of $M_{n}$—by adding $n$ edges, one joining each pair of vertices at distance $n$ in the initial cycle. The prism graph of order $2n$, denoted $Y_{n}$, is the Cartesian product of a cycle of length $n$ and an edge; symbolically Y_{n}:=C_{n}\mathbin{\text{\scalebox{0.84}{\square}}}K_{2}.

Lemma 3**.**

Let $L$ be a group-based latin square of even order $n$, let $k$ be the greatest power of 2 dividing $n$, and let $l:=n/k$. If $L$ does not possess a transversal, then there is a positive integer $m$ dividing $l$ such that $\Gamma(L)$ has an induced subgraph isomorphic to

[TABLE]

Section 2 will be devoted to proving Lemma 3, while Section 3 mentions two possible extensions of the present work. We conclude this section by proving Theorem 2 assuming Lemma 3.

Proof of Theorem 2.

Let $L$ be a group-based latin square of order $n$. We may assume $L$ does not possess a transversal. As first shown in [3], this implies $n$ is even. We may therefore apply Lemma 3 to find an induced copy of $\Lambda_{n,m}$ in $\Gamma(L)$. Because the $\left(2k(l-m)\right)$-vertex graph $\left(\frac{l-m}{2}\right)Y_{2k}$ is bipartite, it contains an independent set of size at least $k(l-m)$. Moreover, one can find an independent set of size $km-1$ in $M_{km}$ by greedily selecting vertices in cyclic order around its rim. Thus $\Lambda_{n,m}$ contains an independent set of size $(l-m)k+km=n-1$ which corresponds to a near transversal of $L$. ∎

2 Proof of Lemma 3

Let $G$ be a group of order $n$ with identity element $e$ and let $Syl_{2}(G)$ denote the isomorphism class of $G$’s Sylow 2-subgroups. A bijection $\sigma:G\rightarrow G$ is called a complete mapping of $G$ if the set $\{L(G)_{g,\sigma(g)}\,:\,g\in G\}$ is a transversal of $L(G)$. Using the classification of finite simple groups and partial results of Hall and Paige, Bray, Evans, and Wilcox recently characterized the groups possessing complete mappings.

Theorem 4** ([5, 8, 13]).**

A finite group $G$ possesses a complete mapping if and only if $Syl_{2}(G)$ is either trivial or non-cyclic.

It is not hard to check that if $H$ is a group of odd order, then the identity map is a complete mapping. Another nicely structured complete mapping for groups of odd order was found by Beals, Gallian, Headley, and Jungreis.

Lemma 5** ([2]).**

For every group $H$ of odd order $m$, there exists an ordering $H=\{h_{0},h_{1},\ldots,h_{m-1}\}$ such that, taking indices modulo $m$, both $h_{i}\mapsto h_{i+1}$ and $h_{i}\mapsto h_{i}$ are complete mappings.

Given two subsets $X_{1},X_{2}\subseteq G$ the product set of $X_{1}$ by $X_{2}$ is

[TABLE]

We write $Xy$ for the product set $X\{y\}$. Let $K$ be a subgroup of $G$ and let $H$ be a normal subgroup of $G$. We say that $G$ is the semidirect product of $K$ and $H$, written $G=K\ltimes H$, if $K\cap H=\{e\}$, $KH=G$, and $|G|=|K||H|$. The following was noted in [8] as following from a result of Burnside.

Lemma 6** (Burnside, [8]).**

Let $G$ be a finite group and let $K$ be a Sylow 2-subgroup of $G$. If $K$ is cyclic and nontrivial, then there is a normal subgroup of odd order $H\triangleleft G$ such that

[TABLE]

To simplify notation we set $[n]:=\{0,1,\ldots,n-1\}$ for every positive integer $n$.

Proof of Lemma 3.

Let $L$ be a latin square based on a group $G$ of order $n=kl$, where $k\geq 2$ is a power of 2 and $l$ is odd. Moreover, suppose $L$ does not possess a transversal. Theorem 4 tells us that $Syl_{2}(G)=\mathbb{Z}_{k}$. It then follows from Lemma 6 that $G$ contains a normal subgroup $H$ of order $l$ and an element $b$ of order $k$ such that

[TABLE]

Let $a:=b^{k/2}$. As $H\triangleleft G$ and $a$ has order 2, $H$ has an automorphism

[TABLE]

Let

[TABLE]

and observe that $H^{\ast}$ is a subgroup of $H$. Let $m:=|H^{\ast}|$. As $m$ divides $l$ and $l$ is odd, $m$ is odd. By Lemma 5, there is an ordering $H^{\ast}=\{h_{0},h_{1},\ldots,h_{m-1}\}$ for which the map $h_{i}\mapsto h_{i+1}$ is a complete mapping. Here and throughout the rest of this proof, indices are taken modulo $m$.

Let $\Gamma:=\Gamma(L)$. Toward defining a set $W\subseteq V(\Gamma)$ which induces $\Lambda_{n,m}$, let

[TABLE]

Furthermore, let $F:=H\setminus H^{\ast}$ and let

[TABLE]

Finally, let

[TABLE]

We show $\Gamma[W]\cong\Lambda_{n,m}$ via the following series of three claims.

Claim 1**.**

$T\cap U=\emptyset$* and there is no edge between $T$ and $U$.*

As $G=\langle b\rangle\ltimes H$, every element of $G$ has a unique representation of the form $g=b^{i}h$ for $i\in[k]$ and $h\in H$. Therefore, the definition of $F$ implies

[TABLE]

But for every $(t,s)\in T$ and every $(u,v)\in U$ we have $t\in\langle b\rangle H^{\ast}$ and $u\in\langle b\rangle F$. Thus $T\cap U=\emptyset$ and there are no row edges between $T$ and $U$.

As $H\triangleleft G$ we have $bHb^{-1}=H$. Moreover, as $a=b^{k/2}$, for every $h\in H^{\ast}$ we have $\alpha(bhb^{-1})=bhb^{-1}$, so $bhb^{-1}\in H^{\ast}$. Thus

[TABLE]

It then follows from the definition of $F$ that

[TABLE]

and, as the identity map is a complete mapping of both $H$ and $H^{\ast}$,

[TABLE]

Thus for every $(u,v)\in U$, both $v$ and $uv$ are in $\langle b\rangle F$. But (2) tells us that for every $(t,s)\in T$, both $s$ and $ts$ are in $\langle b\rangle H^{\ast}$. It then follows from (1) that there are no column edges and no symbol edges between $T$ and $U$.

Claim 2**.**

$\Gamma[U]$* consists of $\frac{l-m}{2}$ disjoint copies of $Y_{2k}$.*

Observe that, when enumerating the vertices in $U$, every element of $\langle b\rangle F$ occurs exactly twice as a first coordinate and, by (3), exactly twice as a second coordinate. Thus, each vertex in $\Gamma[U]$ is incident to exactly one row edge and exactly one column edge, so that the row and column edges in $\Gamma[U]$ form a 2-factor (of $\Gamma[U]$). Specifically, they form $l-m$ disjoint $2k$-cycles $\{C_{f}\,:\,f\in F\}$, with each $C_{f}$ defined by the vertex-sequence

[TABLE]

It follows from the definitions of $H^{\ast}$ and $F$ that $\alpha|_{F}$ is a fixed-point free involution. Thus, to establish Claim 2 it suffices to show that for every $i,j\in[k]$, every $f,h\in F$, and every $\epsilon,\delta\in\{0,1\}$, the vertices $(b^{i}f,fb^{i+\epsilon})$ and $(b^{j}h,hb^{j+\delta})$ are joined by a symbol edge if and only if $j\equiv i+k/2\pmod{k}$, $h=\alpha(f)$, and $\epsilon=\delta$.

The “if” direction of this equivalence follows directly from the definition of $\alpha$. For the converse direction we assume

[TABLE]

and, as latin square graphs are loopless, $(b^{i}f,fb^{i+\epsilon})\neq(b^{j}h,hb^{j+\delta})$. It follows from (3) and (4) that

[TABLE]

Thus $\epsilon=\delta$ and $|i-j|\in\{0,k/2\}$.

Now if $i=j$, then $b^{i}f^{2}b^{i+\epsilon}=b^{i}h^{2}b^{i+\epsilon}$ and (4) implies $f=h$, contradicting the fact that $(b^{i}f,fb^{i+\epsilon})\neq(b^{j}h,hb^{j+\delta})$. It follows that $j$ is the unique element of $[k]$ satisfying $j\equiv i+k/2\pmod{k}$. Thus

[TABLE]

so $h^{2}=\alpha(f^{2})=(\alpha(f))^{2}$ and (4) implies $h=\alpha(f)$.

Claim 3**.**

$\Gamma[T]$* is isomorphic to $M_{km}$*

Observe that, when enumerating the vertices in $T$, every element of $\langle b\rangle H^{\ast}$ occurs exactly twice as a first coordinate and, by (2), exactly twice as a second coordinate. Thus, as is the case for $\Gamma[U]$, each vertex in $\Gamma[T]$ is incident to exactly one row edge and exactly one column edge. Unlike in $\Gamma[U]$, the row and column edges of $\Gamma[T]$ form a single cycle of length $2mk$. Indeed, as $m$ is odd and $|b|$ is a power of 2,

[TABLE]

is a Hamilton cycle in $\Gamma[T]$ which contains all of $\Gamma[T]$’s row and column edges.

To establish Claim 3 it suffices to show that for every $i,j\in[km]$ and every $\epsilon,\delta\in\{0,1\}$, the vertices $(b^{i}h_{i},h_{i+\epsilon}b^{i+\epsilon})$ and $(b^{j}h_{j},h_{j+\delta}b^{j+\delta})$ are joined by a symbol edge in $\Gamma[T]$ if and only if $i\equiv j+\frac{k}{2}m\pmod{km}$ and $\epsilon=\delta$.

Indeed if $i\equiv j+\frac{k}{2}m\pmod{km}$, then $i\equiv j\pmod{m}$ and, as $m$ is odd, $i\equiv j+k/2\pmod{k}$. Together with $\epsilon=\delta$, as well as Lemma 5 and the definition of $H^{\ast}$, this implies

[TABLE]

which establishes the “if” direction of the desired equivalence.

For the converse direction consider $(b^{i}h_{i},h_{i+\epsilon}b^{i+\epsilon}),(b^{j}h_{j},h_{j+\delta}b^{j+\delta})\in T$ and assume that the group elements defining this pair of distinct vertices satisfy

[TABLE]

From (2) we see that

[TABLE]

Thus $\epsilon=\delta$ and $i\equiv j\pmod{k/2}$. Now $b^{j}\in\{b^{i},b^{i+k/2}\}$ and as $H^{\ast}$ is pointwise fixed by the automorphism $\alpha:h\mapsto b^{k/2}hb^{k/2}$, both possible values of $b^{j}$ yield $h_{i}h_{i+\epsilon}=h_{j}h_{j+\epsilon}$. Lemma 5 then implies $h_{i}=h_{j}$, so $i\equiv j\pmod{m}$.

Suppose $b^{j}=b^{i}$, which is equivalent to $i\equiv j\pmod{k}$. As $\gcd(k,m)=1$ and $i,j\in[km]$, this implies $i=j$, contradicting the fact that $(b^{i}h_{i},h_{i+\epsilon}b^{i+\epsilon})$ and $(b^{j}h_{j},h_{j+\delta}b^{j+\delta})$ are distinct vertices. Therefore $j\equiv i+k/2\pmod{k}$ and, as $k/2$ and $m$ are coprime, we conclude that $i\equiv j+\frac{k}{2}m\pmod{km}$. ∎

3 Concluding remarks

The most obvious extension of the present paper is the general case of Conjecture 1. However, a proof of this conjecture would likely differ substantially from the argument presented above. Indeed, latin square graphs are in general not vertex-transitive, calling into question whether general latin square graphs can be shown to possess the sort of “nice” induced subgraphs found in Lemma 3.

There is, however, an extension of Theorem 2 to which the above techniques may be applicable. A partial transversal is non-extendable if it is not contained in any larger partial transversal. The following conjecture was noted by Evans [6, p.​ 470] as a special case of a conjecture of Keedwell concerning sequenceable groups.

Conjecture 7** (Keedwell).**

For every finite non-Abelian group $G$, the latin square $L(G)$ possesses a non-extendable near transversal.

We have established Conjecture 7 for those non-Abelian groups whose Sylow 2-subgroups are nontrivial and cyclic (these groups do not possess transversals, so a near transversal must be non-extendable). Perhaps our techniques can be used to find maximal independent sets of size $n-1$ in latin square graphs based upon non-Abelian groups with non-cyclic or trivial Sylow 2-subgroups. As far as we know, Conjecture 7 has not been attacked directly. However, many partial results are known due to its connection to sequenceable groups (see e.g. [11]).