---
tag: 4RF117CH63
kind: example
name: Steiner quasigroup
msc: [20N05, 05B07, 08A99]
instance_of: [XAC78GN2YJ]
uses: [7BPSC4T4M3, XAC78GN2YJ, 9FS1JFZSX4, CDVRMVED2V]
sources:
  - url: https://en.wikipedia.org/wiki/Steiner_system
---

## Example

Let $S$ be a set with a Steiner triple system on it, that is, a [[Steiner system|7BPSC4T4M3]] $S(2, 3, n)$.
Its blocks are $3$-element subsets of $S$, and every two distinct points lie in exactly one block.
The *Steiner quasigroup* of this system is $S$ with the binary operation $\cdot$ given by:

1. $x \cdot x = x$ for every $x \in S$;
2. $x \cdot y = z$ for $x \ne y$, where $\set{x, y, z}$ is the unique block that contains $x$ and $y$.

This operation is [[commutative|XAC78GN2YJ]] and [[idempotent|9FS1JFZSX4]], and it satisfies $x \cdot (x \cdot y) = y$ for all $x, y$.
For $x \ne y$ the last law holds because $x \cdot y = z$ and $x \cdot z = y$ come from the same block.

Check that $S$ is a quasigroup, that is, every equation $a \cdot y = b$ and $y \cdot a = b$ has exactly one solution $y$.
The law $x \cdot (x \cdot y) = y$ says that the map $y \mapsto a \cdot y$ is its own inverse, so it is a bijection.
Hence the equation $a \cdot y = b$ has the unique solution $y = a \cdot b$.
By commutativity the equation $y \cdot a = b$ has the same unique solution.

## Remarks

The smallest case is $S = \set{0, 1, 2}$ with the single block $\set{0, 1, 2}$.
There the operation is $x \cdot y = -x - y \bmod 3$.
For instance $1 \cdot 2 = -3 \equiv 0$, which is the third point of the block, and $1 \cdot 1 = -2 \equiv 1$.
Conversely, a commutative idempotent quasigroup that satisfies $x \cdot (x \cdot y) = y$ comes from a Steiner triple system, whose blocks are the sets $\set{x, y, x \cdot y}$ with $x \ne y$.
The multiplication table of a Steiner quasigroup is a symmetric [[Latin square|CDVRMVED2V]] with entry $x$ in row $x$ and column $x$.

## Explanations

### L1 human

A Steiner triple system groups points into triangles so that any two points share exactly one triangle.
The Steiner quasigroup turns this into a multiplication: the product of two points is the third corner of their triangle.
The product of a point with itself is the point.
Multiplying twice by the same point brings you back, which makes every equation $a \cdot y = b$ solvable in exactly one way.
On the three points $0, 1, 2$ the product of two different points is the remaining one.