The spectrum of group-based Latin squares

TL;DR

This paper explores the existence of group-based Latin squares, providing constructions for certain groups and characterizing when such Latin squares exist based on group properties.

Contribution

It introduces new constructions for Latin squares based on semi-direct product groups and characterizes their existence for various group orders.

Findings

Group-based Latin squares exist for orders 1, 2, and 4.

Existence of non-abelian groups determines Latin square existence for other orders.

Constructed Latin squares for many semi-direct product groups.

Abstract

We construct sequencings for many groups that are a semi-direct product of an odd-order abelian group and a cyclic group of odd prime order. It follows from these constructions that there is a group-based complete Latin square of order $n$ if and only if $n \in \{ 1,2,4\}$ or there is a non-abelian group of order $n$.

Equations

$$(0,n-1,1,n-2,2,n-3,\dots ,n/2)$$

$$G={\mathbb{Z}}_q{⋉}_{\alpha }A=\{(u,v):u\in {\mathbb{Z}}_q,v\in A\},(u,v)(x,y)=(u+x,{\alpha }^x(v)+y).$$

$$\begin{array}[]{l}(0,g_{1}),(0,g_{2}),\ldots,(0,g_{t}),\\ (1,h_{11}),(\lambda^{q-2},h_{21}),(\lambda^{q-3},h_{31}),\ldots,(\lambda,h_{q-1,1}),\\ (1,h_{12}),(\lambda^{q-2},h_{22}),(\lambda^{q-3},h_{32}),\ldots,(\lambda,h_{q-1,2}),\\ \hskip 28.45274pt\vdots\\ (1,h_{1,m-1}),(\lambda^{q-2},h_{2,m-1}),(\lambda^{q-3},h_{3,m-1}),\ldots,(\lambda,h_{q-1,m-1}),\\ (\lambda,0),(\lambda^{2}/(\lambda-1),0),(\lambda^{3}/(\lambda-1)^{2},0),\ldots,(\lambda^{q-2}/(\lambda-1)^{q-3},0),(\lambda-1,0),\\ (0,g_{t+1}),(0,g_{t+2}),\ldots,(0,g_{m}).\end{array}$$

$$\begin{array}[]{l}(0,g_{2}-g_{1}),(0,g_{3}-g_{2}),\ldots,(0,g_{t}-g_{t-1});(1,h_{11}-\alpha(a_{t})),\\ (\lambda^{q-2}-1,h_{21}-\alpha^{\lambda^{q-2}-1}(h_{11})),(\lambda^{q-3}-\lambda^{q-2},h_{31}-\alpha^{\lambda^{q-3}-\lambda^{q-2}}(h_{21})),\ldots,\\ \hskip 28.45274pt(\lambda-\lambda^{2},h_{q-1,1}-\alpha^{\lambda-\lambda^{2}}(h_{q-2,1}));(1-\lambda,h_{12}-\alpha^{1-\lambda}(h_{q-1,1})),\\ (\lambda^{q-2}-1,h_{22}-\alpha^{\lambda^{q-2}-1}(h_{12})),(\lambda^{q-3}-\lambda^{q-2},h_{32}-\alpha^{\lambda^{q-3}-\lambda^{q-2}}(h_{22})),\ldots,\\ \hskip 28.45274pt(\lambda-\lambda^{2},h_{q-1,2}-\alpha^{\lambda-\lambda^{2}}(h_{q-2,2}));(1-\lambda,h_{13}-\alpha^{1-\lambda}(h_{q-1,2})),\\ \hskip 28.45274pt\vdots\\ (\lambda^{q-2}-1,h_{2,m-1}-\alpha^{\lambda^{q-2}-1}(h_{1,m-1})),(\lambda^{q-3}-\lambda^{q-2},h_{3,m-1}-\alpha^{\lambda^{q-3}-\lambda^{q-2}}(h_{2,m-1})),\ldots,\\ \hskip 28.45274pt(\lambda-\lambda^{2},h_{q-1,m-1}-\alpha^{\lambda-\lambda^{2}}(h_{q-2,m-1}));(0,-h_{q-1,m-1}),\\ (\lambda^{2}/(\lambda-1)-\lambda,0),(\lambda^{3}/(\lambda-1)^{2}-\lambda^{2}/(\lambda-1),0),\ldots,\\ \hskip 28.45274pt(\lambda-1-\lambda^{q-2}/(\lambda-1)^{q-3},0);(-(\lambda-1),g_{t+1}),\\ (0,g_{t+2}-g_{t+1}),(0,g_{t+3}-g_{t+2}),\ldots,(0,g_{m}-g_{m-1}).\end{array}$$

$${\lambda }^2/(\lambda -1)-\lambda ,{\lambda }^3/(\lambda -1{)}^2-{\lambda }^2/(\lambda -1),\dots ,\lambda -1-{\lambda }^{q-2}/(\lambda -1{)}^{q-3}$$

$$\begin{array}[]{l}h_{11},h_{12},\ldots,h_{1,m-1}\\ h_{21},h_{22},\ldots,h_{2,m-1}\\ \hskip 28.45274pt\vdots\\ h_{q-1,1},h_{q-1,2},\ldots,h_{q-1,m-1}\\ g_{2}-g_{1},g_{3}-g_{2},\ldots,g_{t}-g_{t-1},-h_{q-1,m-1},g_{t+2}-g_{t+1},\ldots,g_{m}-g_{m-1}\\ h_{21}-\alpha^{\lambda^{q-2}-1}(h_{11}),h_{22}-\alpha^{\lambda^{q-2}-1}(h_{12}),\ldots,h_{2,m-1}-\alpha^{\lambda^{q-2}-1}(h_{1,m-1})\\ h_{31}-\alpha^{\lambda^{q-3}-\lambda^{q-2}}(h_{21}),h_{32}-\alpha^{\lambda^{q-3}-\lambda^{q-2}}(h_{22}),\ldots,h_{3,m-1}-\alpha^{\lambda^{q-3}-\lambda^{q-2}}(h_{2,m-1})\\ \hskip 28.45274pt\vdots\\ h_{q-1,1}-\alpha^{\lambda-\lambda^{2}}(h_{q-2,1}),h_{q-1,2}-\alpha^{\lambda-\lambda^{2}}(h_{q-2,2}),\ldots,h_{q-1,m-1}-\alpha^{\lambda-\lambda^{2}}(h_{q-2,m-1})\\ h_{12}-\alpha^{1-\lambda}(h_{q-1,1}),h_{13}-\alpha^{1-\lambda}(h_{q-1,2}),\ldots,h_{1,m-1}-\alpha^{1-\lambda}(h_{q-1,m-2}),g_{t+1}\end{array}$$

$$h_{11}-\alpha (g_t)=c_1-\alpha ({\alpha }^{q-1}(c_1))=0$$

$$\begin{array}[]{l}c_{1},c_{2},\ldots,c_{m-1}\\ \alpha^{\lambda-1}(c_{1}),\alpha^{\lambda-1}(c_{2}),\ldots,\alpha^{\lambda-1}(c_{m-1})\\ \alpha^{\lambda-1}(c_{m-1}),a_{2}-a_{1},a_{3}-a_{2},\ldots,a_{m}-a_{m-1}\\ -\alpha^{\lambda-1}(c_{1})-\alpha^{\lambda^{q-2}-1}(c_{1}),-\alpha^{\lambda-1}(c_{2})-\alpha^{\lambda^{q-2}-1}(c_{2}),\ldots,-\alpha^{\lambda-1}(c_{m-1})-\alpha^{\lambda^{q-2}-1}(c_{m-1})\\ c_{1}+\alpha^{\lambda^{q-3}-\lambda^{q-2}}(\alpha^{\lambda-1}(c_{1})),c_{2}+\alpha^{\lambda^{q-3}-\lambda^{q-2}}(\alpha^{\lambda-1}(c_{2})),\ldots,c_{m-1}+\alpha^{\lambda^{q-3}-\lambda^{q-2}}(\alpha^{\lambda-1}(c_{m-1}))\\ \hskip 28.45274pt\vdots\\ -\alpha^{\lambda-1}(c_{1})-\alpha^{\lambda-\lambda^{2}}(c_{1}),-\alpha^{\lambda-1}(c_{2})-\alpha^{\lambda-\lambda^{2}}(c_{2}),\ldots,-\alpha^{\lambda-1}(c_{m-1})-\alpha^{\lambda-\lambda^{2}}(c_{m-1})\\ c_{2}+c_{1},c_{3}+c_{2},\ldots,c_{m-1}+c_{m-2},a_{1}+\alpha^{q-1}(c_{1}).\par\end{array}$$

$$(g_1,g_2,\dots ,g_k,g_k+k,g_{k-1}+k,\dots ,g_1+k)$$

$$a_1,a_1,a_2,a_3,\dots a_{m-1}$$

$$0,0,a_2,a_3,\dots a_{m-1}.$$

$$\begin{array}[]{l}2k,1,2k,1,\ldots,2k,1,2\\ 2k-2,2,2k-2,2,\ldots,2k-2,2,4\\ \vdots\\ 1,2k,1,2k,\ldots,1,2k,2r-1\end{array}$$

$$A=B\times {\mathbb{Z}}_{r_1}\times {\mathbb{Z}}_{r_2}\times \cdots \times {\mathbb{Z}}_{r_k}$$

$$(1,1),(2,0),(2,1),(0,2),(2,2),(1,0),(1,2),(0,1)$$

$$(1,1),(2,1),(0,2),(1,2),(2,2),(0,0),(1,0),(2,0),(0,1).$$

$$2\ell ,2\ell -2,\dots ,2,4\ell ,4\ell -2,\dots ,2\ell +2,2\ell +1,2\ell +3,\dots ,4\ell -1,1,3,\dots ,2\ell -1$$

$$2\ell ,2\ell +1,\dots ,4\ell ,0,1,2,\dots ,2\ell -1.$$

$$2\ell +1,2\ell -1,\dots ,1,4\ell +1,4\ell -1,\dots ,2\ell +3,2\ell +2,2\ell +4,\dots ,4\ell +2,2,3,\dots ,2\ell$$

$$2\ell +1,2\ell +2,\dots ,4{\ell }_2,0,1,2,\dots ,2\ell .$$

$$(c_1,d_1),(c_2,d_1),\dots ,(c_{m-1},d_1);(c_1^{\prime },d_2),(c_2^{\prime },d_2),\dots ,(c_m^{\prime },d_2);\dots ;(c_1^{\prime },d_n),\dots ,(c_m^{\prime },d_n)$$

$$(c_1^{\prime },d_1),(c_2^{\prime },d_1),\dots ,(c_m^{\prime },d_1);(c_1^{\prime },d_2),(c_2^{\prime },d_2),\dots ,(c_m^{\prime },d_2);\dots ;(c_1^{\prime },d_n),\dots ,(c_m^{\prime },d_n)$$

$${\alpha }^{\lambda -1}(c_{m-1})\notin ⟨c_1+c_{m-1}-{\alpha }^{q-1}(c_1)⟩$$

$$\{c_1,c_{m-1}\}=\{(-2,0,\dots ,0),(1,0,\dots ,0)\}$$

Full text

Keywords: complete Latin square; directed terrace; Keedwell’s Conjecture; sequencing.

MSC2010: 05B15

1 Introduction

A Latin square of order $n$ is an $n\times n$ array of symbols from a set of size $n$ with each symbol appearing once in each row and once in each column. A Latin square is row-complete or Roman if each pair of distinct symbols appears in adjacent positions in a row once in each order. It is complete if both it and its transpose are row-complete.

Interest in complete and row-complete Latin squares was originally prompted by their usefulness in the design of experiments where neighboring treatments, whether in space or time, might interact. See, for example, [4].

The Cayley table of a finite group of order $n$ is a Latin square. The principal question for this work is to determine at what orders there is a group for which it is possible to permute the rows and columns of its Cayley table to give a complete Latin square. To this end, consider the following definition. Let $G$ be a group of order $n$ and ${\bf a}=(a_{1},a_{2},\ldots,a_{n})$ be an arrangement of the elements of $G$. Define ${\bf b}=(b_{1},b_{2},\ldots,b_{n-1})$ by $b_{i}=a_{i}^{-1}a_{i+1}$ for each $i$. If ${\bf b}$ includes each non-identity element of $G$ exactly once then ${\bf b}$ is a sequencing of $G$ and ${\bf a}$ is a directed terrace for $G$. Call a group that admits a sequencing sequenceable.

Theorem 1.1**.**

[11]* The rows and columns of the Cayley table of a group $G$ of order $n$ may be permuted to give a complete Latin square if and only if $G$ is sequenceable.*

Proof idea.

Let $(g_{1},\ldots,g_{n})$ and $(h_{1},\ldots,h_{n})$ be arrangements of the elements of $G$ and let the $(i,j)$-entry of a Cayley table of $G$ be given by $g_{i}h_{j}$. Then it is a complete Latin square if and only if $(g_{1}^{-1},g_{2}^{-1},\ldots,g_{n}^{-1})$ and $(h_{1},h_{2},\ldots,h_{n})$ are both directed terraces for $G$. ∎

Let $\mathbb{Z}_{n}=\{0,1,\ldots,n-1\}$ be the additively written cyclic group of order $n$. When $n$ is even,

[TABLE]

is a directed terrace for $\mathbb{Z}_{n}$. The first use of a “zig-zag” construction of this type is Walecki’s in 1892 [19]; see [1] for more of its history. The first use of it to control neighbor balance in Latin squares seems to be due to Williams in 1949 [28].

The systematic consideration of sequenceability for arbitrary groups was initiated by Gordon [11] where, as well as proving Theorem 1.1, it is shown that an abelian group is sequenceable if and only if it has exactly one involution. Hence if $n$ is even there is a group-based complete Latin square of order $n$. The result also implies that for odd orders we must turn our attention to non-abelian groups. Several families of groups of odd order are known to be sequenceable, including: a group of order $p^{m}$ for each odd prime $p$ and each $m\geq 3$ [25]; many groups of order $pq$ for distinct primes $p$ and $q$ [16, 25]; and a group of each order $3m$ where $m$ is powerful (i.e. for each prime $p$ dividing $m$, $p^{2}$ also divides $m$) [21].

We construct sequencings for some semi-direct products $\mathbb{Z}_{q}\ltimes A$ where $A$ is an abelian group of odd order and $q$ is an odd prime, including all possible such groups when $A$ is cyclic. These constructions allow us to determine the full spectrum of orders at which a group-based complete Latin square exists:

Theorem 1.2**.**

There is a group-based complete Latin square of order $n$ if and only if $n=1$, $n$ is even, or there exists a non-abelian group of order $n$. That is, if and only if $n=1$, $n$ is even, or $n$ has either a prime divisor $p$ with $p^{3}|n$ or a prime-power divisor $p^{k}$ such that $p^{k}\equiv 1\pmod{q}$ for some prime divisor $q$ of $n$.

This result also gives the spectrum for group-based row-complete Latin squares. However, whereas all known complete Latin squares are group-based, there are alternative methods known for constructing row-complete squares. Row-complete Latin squares are known to exist at orders 1 and 2 and at every composite order [13, 28]. They are known not to exist at order 3, 5 or 7. The question remains open at other odd primes.

On the question of which groups are sequenceable, the abelian case is settled as mentioned above, and the three non-abelian groups of orders 6 and 8 are not sequenceable. Keedwell’s Conjecture is that all other non-abelian groups are sequenceable. In addition to those already mentioned, groups known to satisfy Keedwell’s Conjecture include dihedral groups [15, 18], soluble groups with a single involution [3], and groups of order at most 255 [22]. See [20] for a survey of this and related problems.

In the next section we develop a framework for constructing sequencings of groups of the form $G=\mathbb{Z}_{q}\ltimes A$ for prime $q$ and odd-order abelian $A$. In Section 3 we show that the framework can be successfully completed for a variety of choices of $A$, whence Theorem 1.2.

2 The construction

Let $A$ be an abelian group of order $m$ with an automorphism $\alpha$ of prime order $q$. Let

[TABLE]

This is a group of order $n=mq$. Let $\lambda$ be a primitive root of $q$ such that $\lambda/(\lambda-1)$ is also a primitive root; Wang [25] shows that the existence of such a $\lambda$ follows from results of [6].

Here is the template we use for a proposed directed terrace of $G$:

[TABLE]

This generalizes the structures of directed terraces in [21] and [25] (each of which grew out of that of [16]).

The corresponding sequencing (terms immediately after a semi-colon correspond to quotients formed from terms on different rows):

[TABLE]

What requirements must we satsify to ensure these are a directed terrace and sequencing?

Observe that

[TABLE]

lists all of the elements $\mathbb{Z}_{q}\setminus\{0,1\}$. We therefore require that $h_{11}-\alpha(g_{t})=0$ so that every element of the form $(x,0)$, for $x\in\mathbb{Z}_{q}\setminus\{0\}$, appears in the sequencing.

We need that $g_{1},g_{2},\ldots g_{m}$ includes every element of $A$ exactly once and that each of the following sequences list all elements of $A\setminus\{0\}$:

[TABLE]

In order to meet these requirements, we introduce two auxilliary types of sequence. Let ${\bf a}=(a_{1},a_{2},\ldots,a_{m-1})$ be an arrangement of the non-identity elements of $A$ and define ${\bf b}=(b_{1},b_{2},\ldots,b_{m-1})$ by $b_{i}=a_{i+1}-a_{i}$, where the indices are calculated modulo $m-1$ (so $b_{m-1}=a_{1}-a_{m-1}$). If ${\bf b}$ also contains all of the non-identity elements of $A$, then ${\bf b}$ is a rotational sequencing or R-sequencing of $A$ and ${\bf a}$ is the associated directed rotational terrace or directed R-terrace.

Similarly, let ${\bf c}=(c_{1},c_{2},\ldots,c_{m-1})$ be an arrangement of the non-identity elements of $A$ and define ${\bf d}=(d_{1},d_{2},\ldots,d_{m-1})$ by $d_{i}=c_{i}+c_{i+1}$, where, again, the indices are calculated modulo $m-1$ (so $d_{m-1}=c_{m-1}+c_{1}$). If ${\bf d}$ also contains all of the non-identity elements of $A$, then ${\bf a}$ is a #-harmonious sequence and $A$ is #-harmonious.

Both of these objects are of interest in their own right. For example, the existence of either a rotational sequencing or a #-harmonious sequence for a group implies that that group has a complete mapping and hence its Cayley table has an othogonal mate. Rotational sequencings were introduced by Ringel [24] and studied extensively in [9]. They have various different but equivalent formulations in the literature. #-harmonious sequences were introduced and studied in [5] and were studied further and named in [26]. In each case, the existence question is completely settled in abelian groups (each of them may also be studied in non-abelian groups, which is not relevant here):

Theorem 2.1**.**

[2, 5, 9]* An abelian group has a rotational sequencing if and only if it does not have exactly one involution. An abelian group has a #-harmonious sequence if and only if it is not $\mathbb{Z}_{3}$ and does not have exactly one involution.*

We shall see some aspects of the known construction methods for rotational sequencings and #-harmonious sequences in the next section. Here we see how to combine them to make the task of completing the template successfully more tractable:

Theorem 2.2**.**

Let $A$ be an abelian group of odd order $m$ with an automorphism $\alpha$ of odd prime order $q$ and let $\lambda$ be a primitive root of $q$ such that $\lambda/(\lambda-1)$ is also a primitive root of $q$. Let ${\bf a}=(a_{1},a_{2},\ldots,a_{m-1})$ be a directed rotational terrace and ${\bf c}=(c_{1},c_{2},\ldots,c_{m-1})$ be a #-harmonious sequence for $A$. If $a_{1}=c_{1}+c_{m-1}-\alpha^{q-1}(c_{1})$ and $a_{m-1}=a_{1}-\alpha^{\lambda-1}(c_{m-1})$ then $G=\mathbb{Z}_{q}\ltimes_{\alpha}A$ is sequenceable.

Proof.

First, we use ${\bf a}$ and ${\bf c}$ to assign values to the $g_{i}$ and $h_{ij}$ in the template. We shall see that the difference/sum properties they have, combined with the extra conditions, allow us to satisfy all of the constraints.

Let $g_{1}=\alpha^{q-1}(c_{1})$ and let $g_{i}=a_{i-1}+\alpha^{q-1}(c_{1})$ for $i>1$. As ${\bf a}$ is a directed rotational terrace the sequence $g_{1},g_{2},\ldots g_{m}$ includes every element of $A$ once.

For odd $i$ set $h_{ij}=c_{j}$; for even $i$ set $h_{ij}=-\alpha^{\lambda-1}(c_{j})$. Set $t=1$. This gives

[TABLE]

as required. We now need to consider the sequences that are required to contain all of the non-zero elements of $A$. With the assignments of the $g_{i}$, $h_{ij}$ and $t$, these become:

[TABLE]

The first two each contain the non-zero elements of $A$ as ${\bf c}$ is a #-harmonious sequence. The third does because it comprises all of the rotational sequencing elements apart from $a_{1}-a_{m-1}$ and in its place we have $\alpha^{\lambda-1}(c_{m-1})=a_{1}-a_{m-1}$.

That each of the fourth through to the penultimate sequences have the required elements follows from the properties of the automorphism $\alpha$ and that $c_{1},c_{2},\ldots,c_{m-1}$ are distinct and non-zero.

Finally, the last one is satisfied because we have all the sums of the #-harmonious sequence ${\bf c}$, except $c_{m-1}+c_{1}$, and in its place we have $a_{1}+\alpha^{q-1}(c_{1})=c_{1}+c_{m-1}$. ∎

Theorem 2.2 is not the only way to successfully complete the template. For example, in [25] there is an alternative scheme for assigning the $h_{ij}$ elements when $A$ is cyclic of prime order and in [21] a different way of assigning the $g_{i}$ elements is used when $q=3$. These alternatives (and others) have potential for proving the sequenceability of more groups than we consider in the next section.

3 The directed terraces

The target for this section is the proof of Theorem 1.2. First we consider some structural properties of groups. The following is well-known; see, for example, [23].

Lemma 3.1**.**

Let $n=p_{1}^{a_{1}}\cdots p_{t}^{a_{t}}$, where the $p_{i}$ are distinct odd primes. There is a nonabelian group of order $n$ if and only if one of the following conditions applies:

$a_{i}\geq 3$* for some $i$,* 2. 2.

$n$* is cube-free and $p_{i}\equiv 1\pmod{p_{j}}$ for some $i,j$,* 3. 3.

$n$* is cube-free and $p_{i}^{2}\equiv 1\pmod{p_{j}}$ for some $i,j$*

We are able to meet all of these orders using groups of the form $\mathbb{Z}_{q}\ltimes A$ for prime $q$ and abelian $A$. When $A\cong\mathbb{Z}_{m}$ there is the necessary automorphism of order $q$ when either $q^{2}|m$ or $p|m$ for some prime $p\equiv 1\pmod{q}$. These groups cover the first two cases of Lemma 3.1. For the third case we use groups of the form $A\cong\mathbb{Z}_{p}^{2}\times B$ where $p^{2}\equiv 1\pmod{q}$ and $p\nmid|B|$. As it is little extra work, when $3\nmid|B|$ we consider groups of the form $A\cong\mathbb{Z}_{p}^{k}\times B$ for any $k\geq 2$.

We shall need an explicit construction for rotational sequencings for cyclic groups. Define a graceful permuation of length $k$ to be an arrangement $(g_{1},g_{2},\ldots,g_{k})$ of the integers $\{1,2,\ldots,k\}$ such that the absolute differences ${\bf h}=(h_{1},h_{2},\ldots,h_{k-1})$ given by $h_{i}=|g_{i+1}-g_{i}|$ are distinct. This is equivalent to a graceful labeling of a path; graceful labelings of graphs are well-studied, see [10] for details.

For example, $(1,k,2,k-1,\ldots,\lfloor k/2\rfloor+1)$ is a graceful permutation of length $k$ known as the Walecki Construction, see [1].

Here is the connection between graceful permutations and rotational sequencings:

Lemma 3.2**.**

[9]* If $(g_{1},g_{2},\ldots,g_{k})$ is a graceful permutation then*

[TABLE]

is a directed rotational terrace for $\mathbb{Z}_{2k+1}$.

We also need the following fact:

Lemma 3.3**.**

[8, 12]* There is a graceful permutation $(g_{1},g_{2},\ldots,g_{k})$ with $g_{1}=x$ for each $1\leq x\leq k$.*

To extend rotational sequencings to non-cyclic groups we need a strengthening of the definition. Let ${\bf a}=(a_{1},a_{2},\ldots,a_{m-1})$ be a directed R-terrace. If $a_{i}=a_{i-1}+a_{i+1}$ for some $i$ then ${\bf a}$ is a directed R∗-terrace and its associated rotational sequencing is an R∗-sequencing. If $i=1$ then both the directed R∗-terrace and the R∗-sequencing are standard. Any directed R∗-terrace may be made standard by re-indexing.

Theorem 3.4**.**

[9]* Suppose $3\nmid 2k+1$ and $A$ is an abelian group of odd order $m$. If $A$ is R∗-sequenceable then so is $A\times\mathbb{Z}_{2k+1}$.*

Proof Construction.

Let $(a_{1},a_{2},\ldots,a_{m-1})$ be a standard directed R∗-terrace for $A$.

Exactly as in [9] we list the sequences of first and second coordinates of a standard directed R∗-terrace for $A\times\mathbb{Z}_{2k+1}$ separately.

The first coordinates are given by $a_{1},a_{2},\ldots a_{m-1}$, followed by $k$ copies of the length-$m$ sequence

[TABLE]

and then $k$ copies of the length-$m$ sequence

[TABLE]

The second coordinates are given by $m-2$ 0s, followed by

[TABLE]

(where each line has $m$ elements starting with $(m-1)/2$ pairs of the form $x,-x$) and then one final 0. ∎

The requirement that $m$ is odd in Theorem 3.4 is unnecessary, but we do not require the construction for the even case. We particularly need:

Corollary 3.5**.**

[9]* Let $A$ and $B$ be abelian groups of odd order. Suppose that $A$ has a standard directed R∗-terrace $(a_{1},\ldots,a_{m-1})$ and $3\nmid|B|$. Then for any $i$ with $1\leq i\leq m-3$, the group $A\times B$ has a standard directed R∗-terrace. *

Proof.

We may write $B$ as the direct product of cyclic groups of odd order. The result follows from repeated applications of the construction in the proof of Theorem 3.4. ∎

Call the process of Theorem 3.4 and Corollary 3.5 the FGM Construction.

To construct #-harmonious sequences we need the closely-related notion of a harmonious sequence. Let ${\bf c}=(c_{1},c_{2},\ldots,c_{m})$ be an arrangement of the elements of an abelian group $A$ and define ${\bf d}=(d_{1},d_{2},\ldots,d_{m})$ by $d_{i}=c_{i}+c_{i+1}$ where the indices are calculated modulo $m$ (so $d_{m}=c_{m}+c_{1}$). If ${\bf d}$ is also an arrangement of the elements of $A$ then ${\bf c}$ is a harmonious sequence and $A$ is harmonious.

Lemma 3.6**.**

[5]* Every odd-order abelian group except $\mathbb{Z}_{3}$ is #-harmonious.*

Proof Construction.

Let $A$ be an abelian group of odd order other than $\mathbb{Z}_{3}$ and write

[TABLE]

where $B$ is either $\mathbb{Z}_{3}^{2}$ or cyclic of order greater than 3. The construction is by induction.

We in fact show that these groups are harmoniously matched, meaning that they have a harmonious sequence and a #-harmonious sequence that start with the same element as each other and end with the same element as each other.

To see that $\mathbb{Z}_{3}^{2}$ is harmoniously matched, consider the #-harmonious sequence

[TABLE]

and the harmonious sequence

[TABLE]

For $r=4\ell+1$ the #-harmonious sequence

[TABLE]

is matched with the harmonious sequence

[TABLE]

For $r=4\ell+3$ with $\ell>0$ the #-harmonious sequence

[TABLE]

is matched with the harmonious sequence

[TABLE]

Note also that $(0,1,2)$ is a harmonious sequence for $\mathbb{Z}_{3}$.

Now, let $C$ and $D$ be abelian groups of odd order with $(c_{1},c_{2},\ldots,c_{m-1})$ and $(c^{\prime}_{1},c^{\prime}_{2},\ldots,c^{\prime}_{m})$ a matched #-harmonious and harmonious sequence for $C$ and $(d_{1},d_{2},\ldots,d_{n})$ a harmonious sequence for $D$, indexed so that $d_{1}=0$. Then

[TABLE]

is a #-harmonious sequence for $C\times D$ and

[TABLE]

is a harmonious sequence. These may be re-indexed to be matched (for example, by taking endpoints $(c^{\prime}_{1},d_{2})$ and $(c^{\prime}_{2},d_{2})$). ∎

Call a #-harmonious sequence constructed via the proof of Lemma 3.6 a BGJH #-harmonious sequence.

We can now move to the main results. First, we consider the case when $A$ is cyclic.

Theorem 3.7**.**

Let $q$ be an odd prime. Let $A=\mathbb{Z}_{m}$, where $m$ is odd and either $q^{2}|m$ or $p|m$ for some prime $p\equiv 1\pmod{q}$ and let $\alpha$ be an automorphism of $\mathbb{Z}_{m}$ of order $q$. Then $G=\mathbb{Z}_{q}\ltimes_{\alpha}\mathbb{Z}_{m}$ is sequenceable.

Proof.

We aim to satisfy the conditions of Theorem 2.2. The arithmetical conditions on $m$ ensure that we are considering exactly those $m$ for which there is a value $r$ such that multiplication by $r$ is an automorphism of order $q$ of $\mathbb{Z}_{m}$. Let $\alpha$ be multiplication by $r$ and let $\lambda$ be a primitive root of $q$ such that $\lambda/(\lambda-1)$ is also a primitive root.

Let $m=2k+1$. The BGHJ #-harmonious sequence for $\mathbb{Z}_{m}$ has 1 and $-2$ as adjacent elements. As $\gcd(k+1,m)=1$, multiplication by $r^{1-\lambda}(k+1)$ is an automorphism of $\mathbb{Z}_{m}$. Apply this automorphism to the BGHJ #-harmonious sequence to get one with $r^{1-\lambda}(k+1)$ and $-2r^{1-\lambda}(k+1)$ adjacent. Index this #-harmonious sequence so that $c_{1}=-2r^{1-\lambda}(k+1)$ and $c_{m-1}=r^{1-\lambda}(k+1)$.

Suppose that $c_{1}+r^{1-\lambda}(k+1)-r^{q-1}c_{1}\neq 0$. Let $g_{1}$ be $c_{1}+r^{1-\lambda}(k+1)-r^{q-1}c_{1}$ or $c_{1}+r^{1-\lambda}(k+1)-r^{q-1}c_{1}-k$, whichever, when considered as an integer, is in the range $1\leq g_{1}\leq k$. Let ${\bf g}$ be a graceful permutation with first element $g_{1}$ (which exists by Lemma 3.3), and let ${\bf a}=(a_{1},a_{2},\ldots,a_{m-1})$ be either the directed rotational terrace constructed from ${\bf g}$ via Lemma 3.2 or its reverse, so that $a_{1}=c_{1}+r^{1-\lambda}(k+1)-r^{q-1}c_{1}$.

Now, $a_{1}=c_{1}+c_{m-1}-\alpha^{q-1}(c_{1})$ and, as $a_{1}-a_{m-1}=k+1$, we have $a_{m-1}=a_{1}-\alpha^{\lambda-1}(c_{m-1})$, and the conditions of Theorem 2.2 are satisfied.

If $c_{1}+r^{1-\lambda}(k+1)-r^{q-1}c_{1}=0$, then reversing the initial roles of 1 and $-2$ and running the same process gives alternative values for the variables such that $c_{1}+r^{1-\lambda}(k+1)-r^{q-1}c_{1}\neq 0$ and the argument goes through as before. ∎

Next we move to non-cyclic $A$. We shall need certain facts about the automorphism groups of abelian groups. A comprehensive description of their structure can be found in [14].

Given two elements $g$ and $h$ of an abelian group, say that they are independent if $\langle g\rangle\cap\langle h\rangle=\{0\}$. Let $(g_{1},h_{1})$ and $(g_{2},h_{2})$ be pairs of independent elements in an abelian group $A$. If all four elements have the same prime order, then there is an automorphism of $A$ that maps $g_{1}$ to $g_{2}$ and $h_{1}$ to $h_{2}$.

We shall be considering groups of the form $A=\mathbb{Z}_{p}^{k}\times B$ where $p$ is prime, $k\geq 2$ and $p\nmid|B|$. If $p^{k}\equiv 1\pmod{q}$, for some prime $q$ then $A$ has an automorphism of order $q$ with $\alpha\mathbin{\upharpoonright}\mathbb{Z}_{p}^{k}$ also of order $q$. Further, we restrict our attention to $\alpha$ of this form such that $\alpha\mathbin{\upharpoonright}\mathbb{Z}_{p}^{k}$ (which is a subgroup of $\mathop{\rm GL}\nolimits(k,p)$) is not diagonalisable, of which there is always at least one.

Suppose $\alpha$ and $\beta$ are two automorphisms of $A$. If $\alpha$ and $\beta$ are conjugate then $\mathbb{Z}_{q}\ltimes_{\alpha}A\cong\mathbb{Z}_{q}\ltimes_{\beta}A$. We may therefore assume that $\alpha\mathbin{\upharpoonright}\mathbb{Z}_{p}^{k}$, which we think of as a matrix in $\mathop{\rm GL}\nolimits(k,p)$ multiplying elements of $A$ (considered as column vectors) from the left, is in rational canonical form. That is, it is a block-diagonal matrix where each block has 1s on the subdiagonal and 0s everywhere else except for the last column, see [7, Chapter 12]. In fact, as $\alpha$ and $\beta$ are conjugate if and only if $\alpha^{c}$ and $\beta^{c}$ are conjugate for positive $c$, we shall usually assume that a particular power of $\alpha\mathbin{\upharpoonright}\mathbb{Z}_{p}^{k}$ is in rational canonical form.

The first non-cyclic case we consider is when $3\nmid|A|$:

Theorem 3.8**.**

Let $p$ and $q$ be odd primes with $p\neq 3$ and $p^{k}\equiv 1\pmod{q}$. Let $B$ be an abelian group of odd order with $3\nmid|B|$. Let $\alpha$ be an automorphism of $\mathbb{Z}_{p}^{k}\times B$ of order $q$ such that $\alpha\mathbin{\upharpoonright}\mathbb{Z}_{p}^{k}$ is of order $q$ and not diagonalisable. Then $G=\mathbb{Z}_{q}\ltimes_{\alpha}(\mathbb{Z}_{p}^{k}\times B)$ is sequenceable.

Proof.

The non-diagonalisable condition implies that $k\geq 2$. As noted before the statement of the theorem, we may assume that some non-trivial power of $\alpha\mathbin{\upharpoonright}\mathbb{Z}_{p}^{k}$ is in rational canonical form with the additional condition that the $(1,1)$-entry is 0. Let $\lambda$ be a primitive root of $q$ such that $\lambda/(\lambda-1)$ is also a primitive root and assume that $\alpha^{\lambda-1}\mathbin{\upharpoonright}\mathbb{Z}_{p}^{k}$ is in this form.

To apply Theorem 2.2, we require a directed R-terrace $(a_{1},\ldots a_{m-1})$ and a #-harmonious sequence $(c_{1},\ldots,c_{m-1})$ such that $a_{1}=c_{1}+c_{m-1}-\alpha^{q-1}(c_{1})$ and $a_{m-1}=a_{1}-\alpha^{\lambda-1}(c_{m-1})$. It is sufficient to find a directed R-terrace with $a_{1}$ and $a_{m-1}$ a pair of independent elements of prime order and a #-harmonious sequence with $c_{1}+c_{m-1}-\alpha^{q-1}(c_{1})$ and $c_{1}+c_{m-1}-\alpha^{q-1}(c_{1})-\alpha^{\lambda-1}(c_{m-1})$ also a pair of independent elements of prime order.

The group $\mathbb{Z}_{p}^{k}$ has a directed R∗-terrace [9]. Any arrangement of the elements of $\mathbb{Z}_{p}^{k}\setminus\{0\}$ unavoidably has many pairs of adjacent independent elements and hence any directed R∗-terrace does too. Using this in the FGM construction gives a directed R∗-terrace with a pair of independent elements of order $p$ which, by re-indexing, we may set to be $a_{1}$ and $a_{m-1}$.

As noted in the proof of Theorem 3.7, the BGHJ #-harmonious sequence for $\mathbb{Z}_{p}$ has $-2$ and $1$ as adjacent elements. Further, using the BGHJ construction, we find that $\mathbb{Z}_{p}^{k}\times B$ has a #-harmonious sequence with $(-2,0,\ldots,0)$ and $(1,0,\ldots,0)$ adjacent. Our target is to show that the condition

[TABLE]

holds, which implies the result we require. We may re-index the #-harmonious sequence so that

[TABLE]

and there are two ways to do so.

We have that $\alpha^{\lambda-1}((1,0,\ldots,0))=(0,1,0,\ldots,0)$. For the condition to fail in the case $c_{m-1}=(1,0,\ldots,0)$ we must have $\alpha^{q-1}((-2,0,\ldots,0))=(1,x,0\ldots,0)$ for some $x\in\mathbb{Z}_{p}$. For the condition to fail in the case $c_{m-1}=(-2,0,\ldots,0)$ we must have $\alpha^{q-1}((1,0,\ldots,0))=(1,y,0\ldots,0)$ for some $y\in\mathbb{Z}_{p}$. But as $\alpha^{q-1}((-2,0,\ldots,0))=-2\alpha^{q-1}((1,0,\ldots,0))$, at least one of the two potential allocations of $c_{1}$ and $c_{m-1}$ does not violate the condition. ∎

Lastly we cover the cases where $3\mid|A|$ but $27\nmid|A|$.

Theorem 3.9**.**

Let $p$ and $q$ be odd primes with $p\neq 3$ and $p^{2}\equiv 1\pmod{q}$. Let $B$ be an abelian group of odd order with $3\nmid|B|$. If $\alpha$ is an automorphism of $\mathbb{Z}_{p}^{2}\times\mathbb{Z}_{3}\times B$ of order $q$ such that $\alpha\mathbin{\upharpoonright}\mathbb{Z}_{p}^{2}$ is of order $q$ and not diagonalisable, then $G_{1}=\mathbb{Z}_{q}\ltimes_{\alpha}(\mathbb{Z}_{p}^{2}\times\mathbb{Z}_{3}\times B)$ is sequenceable. If $\alpha$ is an automorphism of $\mathbb{Z}_{p}^{2}\times\mathbb{Z}_{9}\times B$ of order $q$ such that $\alpha\mathbin{\upharpoonright}\mathbb{Z}_{p}^{2}$ is of order $q$ and not diagonalisable, then $G_{2}=\mathbb{Z}_{q}\ltimes_{\alpha}(\mathbb{Z}_{p}^{2}\times\mathbb{Z}_{9}\times B)$ is sequenceable.

Proof.

The structure of the proof is much the same as for Theorem 3.8 in that we are looking for a directed R-terrace $(a_{1},\ldots,a_{m-1})$ and a #-harmonious sequence $(c_{1},\ldots,c_{m-1})$ for $\mathbb{Z}_{p}^{2}\times\mathbb{Z}_{3}\times B$ (or $\mathbb{Z}_{p}^{2}\times\mathbb{Z}_{9}\times B$) such that $a_{1}$ and $a_{m-1}$ are a pair of independent elements of prime order $p$ and $c_{1}+c_{m-1}-\alpha^{q-1}(c_{1})$ and $c_{1}+c_{m-1}-\alpha^{q-1}(c_{1})-\alpha^{\lambda-1}(c_{m-1})$ are also a pair of independent elements of order $p$.

The argument for the existence of the latter of these is identical to that of Theorem 3.8. The difficulty arises in the former case as the FGM construction for R∗-terraces does not directly apply. In order to use the FGM construction we need to move the $\mathbb{Z}_{3}$ or $\mathbb{Z}_{9}$ factor into the base case and find directed R∗-terraces for $\mathbb{Z}_{p}^{2}\times\mathbb{Z}_{3}$ and $\mathbb{Z}_{p}^{2}\times\mathbb{Z}_{9}$ that have the independent pairs of adjacent elements we need.

First consider $\mathbb{Z}_{p}^{2}\times\mathbb{Z}_{3}\equiv\mathbb{Z}_{3p}\times\mathbb{Z}_{p}$. The directed R-terrace $(a_{1},\ldots a_{3p-1})$ for $\mathbb{Z}_{3p}$ constructed from the Walecki Construction using Lemma 3.2 is in fact a directed R∗-terrace with $a_{2p}$ the sum of its neighbors [9]. It also has $a_{4}=3(p-1)/2$ and $a_{5}=3$, which are elements of order $p$. Now, applying the construction of Theorem 3.4 we find a directed R∗-terrace of $\mathbb{Z}_{3p}\times\mathbb{Z}_{p}$ with $(3(p-1)/2,1)$ and $(3,-1)$ adjacent, which are independent elements of order $p$. From here we can re-index and follow the method of Theorem 3.8 to get the sequencing for $G_{1}$.

Now consider $\mathbb{Z}_{p}^{2}\times\mathbb{Z}_{9}\equiv\mathbb{Z}_{9p}\times\mathbb{Z}_{p}$. For this case we need part of the more complex constructions of [2]. However, as the aspects of that construction that we require are clearly stated within the paper, we do not recapitulate the full construction. They use a device that they call “the gadget” [2, Definition 3.5] which when used on a directed R∗-terrace of a group $G$ and a group $H$ that has orthomorphisms with an additional property, it produces a directed R∗-terrace for $G\times H$. Further, if $a_{i}=a_{i-1}+a_{i+1}$ gives the R∗ property in the directed R∗-terrace of $G$ then we have $(a_{i},0)=(a_{i-1},0)+(a_{i+1},0)$ doing the same for the resulting directed R∗-terrace of $G\times H$ [2, Lemma 3.7]. It is allowable to take $G=\mathbb{Z}_{p}$ and $H=\mathbb{Z}_{9}$ in this construction [2, Corollary 3.8], which gives a standard directed R∗-terrace $(a_{1},\ldots,a_{9p-1})$ for $\mathbb{Z}_{9p}$ with $a_{1}$, $a_{2}$ and $a_{9p-1}$ all of order $p$.

We may now use the FGM Construction to give a directed R∗-terrace for $\mathbb{Z}_{9p}\times\mathbb{Z}_{p}$. This gives us adjacent elements of the form $(a_{1},1)$ and $(a_{1},-1)$, which are independent elements of order $p$. Re-index and follow the method of Theorem 3.8 to get a sequencing for $G_{2}$. ∎

Taken together with Lemma 3.1, Theorems 3.7, 3.8 and 3.9 prove Theorem 1.2. While they also make some progress towards Keedwell’s Conjecture, a full proof does not appear to be on the horizon. Progress on the spectrum of not-necessarily-group-based complete Latin squares will require a totally different approach.

Acknowledgements

The authors gratefully acknowledge the support of a Marlboro College Faculty Professional Development Grant.