Count of Binary Operations on Set

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Theorem

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

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

$N = n^{\paren {n^2} }$


Sequence of Values of $N$

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

$\begin{array} {c|rr} n & n^2 & N = n^{\paren {n^2} } \\ \hline 1 & 1 & 1 \\ 2 & 4 & 16 \\ 3 & 9 & 19 \ 683 \\ 4 & 16 & 4 \ 294 \ 967 \ 296 \\ \end{array}$

There are still only $4$ elements in a set, and already there are over $4$ thousand million different possible algebraic structures.


Proof

A binary operation on $S$ is by definition a mapping from the cartesian product $S \times S$ to the set $S$.

Thus we are looking to evaluate:

$N = \card {\set {f: S \times S \to S} }$

The domain of $f$ has $n^2$ elements, from Cardinality of Cartesian Product of Finite Sets of Finite Sets.

The result follows from Cardinality of Set of All Mappings.

$\blacksquare$


Examples

Order $2$ Structure

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

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

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

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

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

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


Also see


Sources

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