Count of Commutative Binary Operations with Identity

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

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


 * $N = n^{\frac {n \left({n-1}\right)}2 + 1}$

Proof
From Count of Commutative Binary Operations with Fixed Identity, there are $n^{\frac {n \left({n-1}\right)}2}$ 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 $n \times n^{\frac {n \left({n-1}\right)}2}$ different algebraic structures with an identity.

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

Hence the result.

Comment
The number grows rapidly with $n$:

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