MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A Steiner triple system is a decomposition of $K_n$ into $K_3$, such as $S=\{013,026,045,124,156,235,346\}$. Steiner triple systems give rise to a Steiner Latin squares, such as $L$ below.

\[L=\left(\begin{matrix} 0 & 3 & 6 & 1 & \bf{5} & 4 & 2 \\\\ 3 & 1 & 4 & 0 & 2 & \bf{6} & 5 \\\\ 6 & 4 & 2 & 5 & 1 & 3 & \bf{0} \\\\ \bf{1} & 0 & 5 & 3 & 6 & 2 & 4 \\\\ 5 & \bf{2} & 1 & 6 & 4 & 0 & 3 \\\\ 4 & 6 & \bf{3} & 2 & 0 & 5 & 1 \\\\ 2 & 5 & 0 & \bf{4} & 3 & 1 & 6 \end{matrix}\right)\]

We define $L=(l_{ij})$ by $l_{ii}=i$ for all $i$ and $l_{ij}=k$ whenever $ijk$ is a triangle in $S$.

Note: Typically, Steiner Latin squares are viewed in an algebraic context and referred to as "Steiner quasigroups" -- Steiner quasigroups correspond to isomorphism classes of Steiner Latin squares, whereas for this question, I'm interested in the "labelled" case.

In some instances, such as $L$ above, the Latin square obtained is a diagonally-cyclic Latin square. That is, $L$ satisfies the identity $l_{(i+1)(j+1)}=l_{ij}+1 \pmod n$, for all $i,j \in \mathbb{Z}_n$, where the indices are taken modulo $n$ also. I've highlighted (in bold) an orbit of an entry of $L$ under this symmetry.

Which Steiner triple systems give rise to diagonally-cyclic Steiner Latin squares?

The above question was my original question, however, as Douglas Zare points out, these are precisely the Steiner triple systems that admit the automorphism $(0,1,\ldots,n-1)$.

Proof: If $L$ is a Steiner Latin square derived from $S$ then $(i,j,k)$ is an entry in $L$ if and only if it $ijk$ is an element of $S$. For $L$ to be diagonally-cyclic, if $(i,j,k)$ is an entry of $L$ then so is $(i+1,j+1,k+1) \pmod n$. Therefore $(i+1)(j+1)(k+1)$ is also an element of $S$. The converse is true by definition.

Since that was such an easy task, lets look at a (hopefully) more interesting question.

Let $L=(l_{ij})$ be a diagonally-cyclic Steiner Latin square. Let $\sigma$ be the permutation defined by $\sigma(j)=l_{0j}$ (that is, the first row of $L$). Then $\sigma$ is an orthomorphism of $\mathbb{Z}_n$. That is, $\sigma$ is a permutation of $\mathbb{Z}_n$ and the map defined by $i \mapsto \sigma(i)-i \pmod n$ is also a permutation of $\mathbb{Z}_n$.

In the above example $\sigma=(0)(13)(26)(45)$.

Which orthomorphisms of $\mathbb{Z}_n$ arise from diagonally-cyclic Steiner Latin squares?

share|cite|improve this question

The triple 0ab is equal to 0ba. If 0ab occurs in triple system, then in the corresponding Steiner Latin square the (0,a) entry is b and the (0,b) entry is a. The permutation corresponding the 0 row must have the transposition $(a\ b)$. In fact, row $j$ written a permutation must be a product of disjoint transpositions (a.k.a an involution) that fixes only $j$.

If a Latin square admits an orthomorphism $\sigma$ in one row, then it admits an orthomorphism in every row. This follows from every row being a permutation of the first row and so the differences $\sigma(i)-i$ are all permuted, too.

Let $\rho$ be the permutation $(1\ 2\ \dots\ n)$. In a diagonally-cyclic Steiner Latin square, let the permutation $\pi_1$ correspond to row $1$ and then row j corresponds to the permutation be $\rho^j(\pi_1)$. If $\pi_1$ is an orthomorphism corresponding to the first row of a diagonally-cyclic Latin square, then $\rho^j(\pi_1)$ is required to be a orthomorphism, too.

If $\pi_1$ is an involutary orthomorphism corresponding to the first row of a diagonally-cyclic Steiner Latin square, then $\rho^j(\pi_1)$ must be an involutary orthomorphism with a single fixed point. So these are the necessary conditions for an orthomorphism to arise from a diagonally-cyclic Steiner Latin square. The question now becomes "Are they sufficient?". I suspect the answer is no. You need to find some starter cycles to produce a triple system. Maybe looking at the Latin subsquares will help.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.