# Possible group operations on a finite set

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 $$n$$?

• I just want to register the opinion that the deleted answer by Lukas Heger is actually correct. Oct 6, 2023 at 15:53
• @AlexKruckman Good thing, they undeleted it. I was going through it; I need some time to understand it as I am not yet familiar with actions... However, I realize that the relation between $\circ_1$ and $\circ_2$ goes deeper. Oct 6, 2023 at 16:03
• c.f. oeis.org/A000001 Oct 7, 2023 at 5:59

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 $$n!/|\operatorname{Aut}(G)|$$
Let $$G_1, G_2, \ldots G_k$$ be a complete list of pairwise nonisomorphic groups of order $$n$$, then $$c(n)=n! \sum_{l=1}^k \frac{1}{|\operatorname{Aut}(G_l)|}$$
• This can’t be right, take $n$ a prime number… Oct 6, 2023 at 15:35
• @NickyHekster If $n = p$ the formula gives $p(p-2)!$ since $\mathrm{Aut}(C_p) \cong C_{p-1}$. What is the problem? There are $p$ ways to choose an identity element. Let $g$ be any other element. Then $g$ has to generate the group and we just need to choose the names of the elements $g^2, g^3, \dots, g^{p-1}$, which can be done in $(p-2)!$ ways. Oct 6, 2023 at 16:26
• I read it more carefully, I misunderstood the value $c(n)$ as being the number of non-isomorphic groups of order $n$. But that is not the case. +1 from me. Oct 6, 2023 at 17:58