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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 $n$?

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    $\begingroup$ I just want to register the opinion that the deleted answer by Lukas Heger is actually correct. $\endgroup$ Oct 6, 2023 at 15:53
  • $\begingroup$ @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. $\endgroup$ Oct 6, 2023 at 16:03
  • $\begingroup$ c.f. oeis.org/A000001 $\endgroup$
    – DanTheMan
    Oct 7, 2023 at 5:59

1 Answer 1

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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 $$c(n)=n! \sum_{l=1}^k \frac{1}{|\operatorname{Aut}(G_l)|}$$

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  • $\begingroup$ This can’t be right, take $n$ a prime number… $\endgroup$ Oct 6, 2023 at 15:35
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    $\begingroup$ @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. $\endgroup$ Oct 6, 2023 at 16:26
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    $\begingroup$ 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. $\endgroup$ Oct 6, 2023 at 17:58

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