Count of Commutative Binary Operations with Fixed Identity

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

Let $$x \in S$$.

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


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

Proof
This follows by the arguments of Count of Binary Operations with Fixed Identity and Count of Commutative Binary Operations on a Set.

From Count of Binary Operations on a Set, there are $$n^{\left({n^2}\right)}$$ 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 $$\left({S - \left\{{x}\right\}}\right) \times \left({S - \left\{{x}\right\}}\right) \to S$$.

So the question boils down to establishing how many different unordered pairs there are in $$\left({S - \left\{{x}\right\}}\right)$$.

That is, how many doubleton subsets there are in $$\left({S - \left\{{x}\right\}}\right)$$.

From Cardinality of Set of Subsets, this is given by:
 * $$\binom {n - 1} 2 = \frac {\left({n - 1}\right) \left({n - 2}\right)} 2$$

To that set of doubleton subsets, we also need to add those ordered pairs where $$x = y$$. There are clearly $$n - 1$$ of these.

So the total number of pairs in question is $$\frac {\left({n - 1}\right) \left({n - 2}\right)} 2 + n - 1 = \frac {n \left({n-1}\right)} 2$$.

The result follows from Cardinality of Set of All Mappings.

Comment
The number grows rapidly with $$n$$:

$$\begin{array} {c|cr} n & \frac {n \left({n-1}\right)}2 & n^{\frac {n \left({n-1}\right)}2}\\ \hline 1 & 0 & 1 \\ 2 & 1 & 2 \\ 3 & 3 & 27 \\ 4 & 6 & 4 \ 096 \\ 5 & 10 & 9 \ 765 \ 625 \\ \end{array}$$