# Count of Commutative Binary Operations on Set

Jump to navigation Jump to search

## Theorem

Let $S$ be a set whose cardinality is $n$.

The number $N$ of possible different commutative binary operations that can be applied to $S$ is given by:

$N = n^{\frac {n \paren {n + 1} } 2}$

### Sequence of Values of $N$

The sequence of $N$ for each $n$ begins:

$\begin{array} {c|cr} n & \dfrac {n \paren {n + 1} }2 & n^{\frac {n \paren {n + 1} } 2} \\ \hline 1 & 1 & 1 \\ 2 & 3 & 8 \\ 3 & 6 & 729 \\ 4 & 10 & 1 \ 048 \ 576 \\ \end{array}$

and so on.

## Proof

Let $\struct {S, \circ}$ be a magma.

From Cardinality of Cartesian Product, there are $n^2$ elements in $S \times S$.

The binary operations $\circ$ is commutative if and only if:

$\forall x, y \in S: x \circ y = y \circ x$

Thus for every pair of elements $\tuple {x, y} \in S \times S$, it is required that $\tuple {y, x} \in S \times S$.

So the question boils down to establishing how many different unordered pairs there are in $S$.

That is, how many doubleton subsets there are in $S$.

From Cardinality of Set of Subsets, this is given by:

$\dbinom n 2 = \dfrac {n \paren {n - 1} } 2$

To that set of doubleton subsets, we also need to add those ordered pairs where $x = y$.

There are clearly $n$ of these.

So the total number of pairs in question is:

$\dfrac {n \paren {n - 1} } 2 + n = \dfrac {n \paren {n + 1} } 2$

The result follows from Cardinality of Set of All Mappings.

$\blacksquare$

## Examples

### Order $2$ Structure

The Cayley tables for the complete set of commutative magmas of order $2$ are listed below.

The underlying set in all cases is $\set {a, b}$.

$\begin{array}{r|rr} & a & b \\ \hline a & a & a \\ b & a & a \\ \end{array} \qquad \begin{array}{r|rr} & a & b \\ \hline a & a & a \\ b & a & b \\ \end{array}$
$\begin{array}{r|rr} & a & b \\ \hline a & a & b \\ b & b & a \\ \end{array} \qquad \begin{array}{r|rr} & a & b \\ \hline a & a & b \\ b & b & b \\ \end{array}$
$\begin{array}{r|rr} & a & b \\ \hline a & b & a \\ b & a & a \\ \end{array} \qquad \begin{array}{r|rr} & a & b \\ \hline a & b & a \\ b & a & b \\ \end{array}$
$\begin{array}{r|rr} & a & b \\ \hline a & b & b \\ b & b & a \\ \end{array} \qquad \begin{array}{r|rr} & a & b \\ \hline a & b & b \\ b & b & b \\ \end{array}$