Count of Binary Operations with Fixed Identity

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Theorem

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

Let $x \in S$.

The number $N$ of possible different binary operations such that $x$ is an identity element that can be applied to $S$ is given by:

$N = n^{\paren {\paren {n - 1}^2} }$


Sequence of Values of $N$

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

$\begin {array} {c|cr} n & \paren {n - 1}^2 & n^{\paren {\paren {n - 1}^2} } \\ \hline 1 & 0 & 1 \\ 2 & 1 & 2 \\ 3 & 4 & 81 \\ 4 & 9 & 262 \, 144 \\ \end{array}$


Proof

Let $S$ be a set such that $\card S = n$.

Let $x \in S$ be an identity element.

From Count of Binary Operations on Set, there are $n^{\paren {n^2} }$ binary operations in total.

We also know that $a \in S \implies a \circ x = a = x \circ a$, so all operations on $x$ are already specified.


It remains to count all possible combinations of the remaining $n - 1$ elements.

This is effectively counting the mappings $\paren {S \setminus \set x} \times \paren {S \setminus \set x} \to S$.

From Count of Binary Operations on Set, this is $n^{\paren {\paren {n - 1}^2} }$ structures with $x$ as the identity element.

$\blacksquare$


Sources

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