In Algebra: Chapter 0, the author made a remark (footnote on page 82), saying that more than 99% of groups of order less than 2000 are of order 1024.

Is this for real? How can one deduce this result? Is there a nice way or do we just check all finite groups up to isomorphism?

Thanks!

**Answer**

Here is a list of the number of groups of order n for n=1,…,2015. If you add up the number of groups of order other than 1024, you get 423,164,062. There are 49,487,365,422 groups of order 1024, so you can see the assertion is true. (In fact the percentage is about 99.15%.)

As far as I know there is no reasonable way to deduce *a priori* the number of isomorphism classes of groups of a given order, though I believe that combinatorial group theory has some methods for specific cases. A general rule of thumb is that there are a *ton* of 2-groups, and in fact I have heard it said that “almost all finite groups are 2-groups” (though I cannot cite a reference for this statement).

EDIT: As pointed out in the comments, “almost all finite groups are 2-groups” is still a conjecture. There is an asymptotic bound on the number of p-groups of order pn, however. Denoting by μ(p,n) the number of groups of order pn, μ(p,n)=p(227+O(n−1/3))n3, which is proven here. This colossal growth along with the results of Besche, Eick & O’Brien seem to be what primarily motivated the conjecture.

**Attribution***Source : Link , Question Author : Hui Yu , Answer Author : Alexander Gruber*