Possible group operations on a finite set

Question

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 $n$?

Answer 1

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

Answer 2

$\DeclareMathOperator{\Sym}{Sym}$ This is just a more detailed version of @Lukas Heger 's answer.

WLOG, let $X=\{1,2,\ldots, n\}$ and $G_1$, $G_2$, $\ldots$, $G_k$ be a complete list of groups of order $n$ up to isomorphism.

Let $X'$ be the set of binary operations $\circ\colon\, X\times X\to X$ such that $(X,\circ)$ is a group. We are to determine $|X'|$.

Let $\Sym(X)=S_n$ act on $X'$ from the left as follows: $\begin{array}{lccll}\cdot\,\,\colon& X'\times S_n &\longrightarrow& X'\\ &(\circ,\sigma)&\longmapsto& \circ_\sigma\colon& X\times X&\to&X\\ &&&&(x,y)&\mapsto&\sigma(\sigma^{-1}(x)\circ \sigma^{-1}(y))\end{array}\tag*{}$

It’s easy to check that we have $\circ_{\sigma \tau} = (\circ_{\tau})_\sigma$ (it may seem weird but it's a left group action with the group element written on the right.)

I didn't invent this action out of nowhere. There's a thing in the literature called structure “transport of structure using bijection”.

If $(G,\circ)$ is a group and $H$ is a set such that $\sigma:G\to H$ is a bijection, then a natural way to transport the structure of $G$ to $H$ would be to define $\cdot\,\colon \, H\times H\to H$ by $x\cdot y = \sigma(\sigma^{-1}(x)\circ \sigma^{-1}(y))$. Now it’s easy to see that $(H,\cdot)$ is a group and $\sigma:(G,\circ)\to (H,\cdot)$ is an isomorphism.

Surprisingly, this is so canonical when $G$ and $H$ are different… However, when $G=H$, it’s not immediately apparent…

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With the above group action in mind, we see that orbit of a group operation $\circ$ is given by: $S_n\cdot \circ = \{\circ_\sigma \mid \sigma\in S_n\}\tag*{}$

We have $(X,\circ)\cong (X,\circ_\sigma)$ for any $\sigma\in S_n$ with $\sigma:(X,\circ)\to (X,\circ_\sigma)$ giving the isomorphism. On the other hand, for $*\in X'$ s.t. $(X,*)\cong (X, \circ)$, there is an isomorphism $\varphi:(X,\circ)\to (X,*)$ and hence, $*=\circ_\varphi$.

Thus, the orbit of $\circ\in X'$ is the set of all those $*\in X'$ for which $(X,\circ)\cong (X,*)$.

The orbits of $X'$ can be indexed by the groups we have been given… $\mathcal O_i = \{\circ\in X'\mid G_i\cong(X,\circ) \}\quad\text{for $i=1,2,\ldots, k$}\tag*{}$

For a given $\circ\in X'$, there are some permutations $\varphi\in S_n$ s.t. $\circ=\circ_\varphi$. These are basically automorphisms of the group $(X,\circ)$. Thus, $|\text{stab}(\circ)|=|\text{Aut}(G_i)|$ where $i\in\{1,\ldots,k\}$ is such that $G_i\cong (X,\circ)$.

Using orbit stabilizer theorem, we have: $\displaystyle |\mathcal O_i| = \frac{n!}{|\text{Aut}(G_i)|}\tag*{}$ Now since $X'=\bigsqcup_{i=1}^n \mathcal O_i$, we have: $\displaystyle |X'| = \sum_{i=1}^k |\mathcal O_i| = \sum_{i=1}^k\frac{n!}{|\text{Aut}(G_i)|} \quad\blacksquare\tag*{}$