Count of Commutative Binary Operations with Identity

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Theorem

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

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

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


Sequence of Values of $N$

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

$\begin{array} {c|cr} n & \dfrac {n \paren {n - 1} } 2 + 1 & n^{\frac {n \paren {n - 1} } 2 + 1} \\ \hline 1 & 1 & 1 \\ 2 & 2 & 4 \\ 3 & 4 & 81 \\ 4 & 7 & 16 \ 384 \\ 5 & 11 & 48 \ 828 \ 125 \\ \end{array}$


Proof

From Count of Commutative Binary Operations with Fixed Identity, there are $n^{\frac {n \paren {n - 1} } 2}$ such binary operations for each individual element of $S$.

As Identity is Unique, if $x$ is the identity, no other element can also be an identity.

As there are $n$ different ways of choosing such an identity, there are $n \times n^{\frac {n \paren {n - 1} } 2}$ different algebraic structures with an identity.

These are guaranteed not to overlap by the uniqueness of the identity.

Hence the result.

$\blacksquare$


Sources

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