Constructing a Steiner Triple System

Question

How can I construct a Steiner Triple System with $v=9$?

Is it possible to do it by using other Steiner Triple Systems ($v=1$, $v=3$, $v=7$)?

I found the answer on internet, but I can't find good materials with examples on how to get to that answer.

$v = 9$

$S = \{1,2,\ldots,9\}$

$T = \{123, 147, 159, 168, 456, 258, 267, 249, 789, 369, 348, 357\}$ (how can I get to this?)

Thank you in advance!

Answer 1

The affine plane of order $3$ is such a system.

This Steiner system is perhaps easier to understand if you are familiar with $\mathbb{Z}/3\mathbb{Z}$, which is the field with three elements.

Let $S=\left\{(x,y)\,|\,x,y\in\mathbb{Z}/3\mathbb{Z}\right\}$. Elements of $S$ are points in a plane that has $9$ elements. The $x$ and $y$ coordinates of each point are elements of $\mathbb{Z}/3\mathbb{Z}$.

Let $T$ be the set of lines connecting these points. A line will be a set either of the form

$$\left\{(x,y)\in S\,|\,y=mx+b\right\}$$

for some $m,b\in\mathbb{Z}/3\mathbb{Z}$, or a set of the form

$$\left\{(x,y)\in S\,|\,x=a\right\}$$

for some $a\in\mathbb{Z}/3\mathbb{Z}$.

Our set of vertices, $S$, is a plane with $9$ elements. Each triple in our set $T$ of triples is a line in this plane. We end up with a Steiner Triple System because we can show that in this plane, every pair of points determines a unique line.

Answer 2

One way to construct Steiner triple systems is using symmetric idempotent latin squares, i.e. latin squares in which $L_{i,i} = i$ and $L_{i,j} = L_{j,i}$ for all $i,j \in \{1,2,\ldots,|L|\}$. This is commonly known as The Bose Construction. Following Linder and Rodgers book, we proceed as follows

Let $L$ be a symmetric idempotent latin square of order $2n+1$. We can then define an STS(6n+3) by setting $V = \{0,1,\ldots,2n\} \times \{0,1,2\}$ and setting $B = B_1 \cup B_2$, where $B_1 = \{(x,0),(x,1),(x,2)\,:\, 0 \leq x \leq 2n\}$ and $B_2 = \{(x,i),(y,i),(L_{x,y},i+1 \pmod{3}) \,:\, 0 \leq x < y \leq 2n \,,\, 0 \leq i \leq 2\}$.

For example, letting $L = \begin{bmatrix} 0 & 2 & 1 \\2 & 1 & 0 \\1&0&2 \end{bmatrix}$, you can construct an STS(9) which is equivalent (in a strong sense) to the STS(9) you have given.

For a nice write up on these idea, see these slides compiled by Lucia Moura.