Vtyjkyuk

If we define $n^n^2$ distinct binary operations from $Stimes S to S$ [where $S$ is a set of cardinality $n$], how many of these operations make $S$ a group?
Can we further way generalise this result?

I don't think there's any closed formula for it.

To start with, we can take the number of groups of order $n$. However that is not yet the number you are after, since different operations can result in the same group, just with the elements relabelled. Let's call that number $g(n)$.

Now there are $n!$ ways to relabel the elements of a group, that is, apply a permutation on the elements on the set. So we get as upper bound of the number of group operations $n!cdot g(n)$.

But that's still not the solution, because relabelling can result in the same operation again. This is the case exactly if the permutation is an automorphism of the group. Of course if you do an automorphism followed by another permutation, you get the same operation as if you had just done the other permutation.

So ultimately for each group of order $n$, you need to divide $n!$ by the order of its automorphism group, and then add all those numbers together.

For example, for $n=4$ we find that there are exactly two groups of order 4, the cyclic group and the Klein four-group.

The automorphism group of the cyclic group has order 2 (the only automorphisms are the identity and the exchange of 1 and 3), while the automorphism group of the Klein four-group has order 6 (any permutation of the non-identity elements is an automorphism).

Therefore the number of group operations on $S=a,b,c,d$ is
$$frac4!2 + frac4!6 = 12 + 4 = 16$$

We can verify this number by explicitly enumerating the possibilities:

Basically, your choices are:

• First, choose which group you want, cyclic or Klein.




• Then, choose which element should be the neutral one. There are of course 4 options.




• If you chose the Klein four-group, then everything is now fixed (so you get a total of 4 options for that).




• If you chose the cyclic group, then you can choose which of the remaining elements is the unique element of order $2$; there are 3 choices. After that choice, again everything is fixed, therefore you have $4cdot 3 = 12$ total choices.

So in total you have $4+12=16$ possible group actions, as obtained by the formula before.

I don't think there's a way to calculate the number without explicitly identifying the groups and their automorphism groups.