Suppose $X=\{x_1, x_2, \ldots, x_n\}$ is a finite set of $n$ elements.
I learned that there are $n^{n^2}$ binary operations $*:X\times X \to X$ and $n^{n(n+1)/2}$ of them are commutative.
I was wondering how many of them are associative. I came across this post; there's a very complicated formula in terms of $n$.
As it was discussed there, a semi-group being a set with associative binary operation, its easy to see that "number of associative binary operations on $X$" is equal to "the number of distinct non-isomorphic semi-groups of order $n$".
How many binary operations $*$ are there on $X$ such that $(X,*)$ is a group? Let's denote this by $c(n)$.
I looked online to find if there's an explicit formula for $c(n)$. I couldn't find anything.
$c(n)$ is not equal to the number of distinct non-isomorphic groups of order $n$ because it is possible that there are group operations $\circ_1\neq \circ_2$ for which $(X,\circ_1) \cong (X,\circ_2)$.
However, intuitively, I feel that $\circ_1$ and $\circ_2$ are basically somehow rearrangements of each other. I apologise, I am unable to explain.
I kind of believe that $\frac{c(n)}{n!}$ is equal to number of distinct non-isomorphic groups of order $n$. Is this correct? If not, is there any way to avoid counting operations like $\circ_1$ and $\circ_2$ more than once and hence relate $c(n)$ to the number of distinct non-isomorphic groups of order $n$?
The group $S_X\cong S_n$ acts on the set of group operations on the set $X$ by conjugation. The set of orbits may be identified with the set of isomorphism classes of groups of order $n$.
Now let $G$ a group of order $n$. By the orbit-stabilizer theorem, we get that the orbit of a group operation on $X$ isomorphic to $G$ is of size $n!/|\operatorname{Aut}(G)|$
Combining these two observations, we get the following result:
Let $G_1, G_2, \ldots G_k$ be a complete list of pairwise nonisomorphic groups of order $n$, then $$c(n)=n! \sum_{l=1}^k \frac{1}{|\operatorname{Aut}(G_l)|}$$