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!
Here is a list of the number of groups of order $n$ for $n=1,\ldots,2015$. 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 $p^n$, however. Denoting by $\mu(p,n)$ the number of groups of order $p^n$, $$\mu(p,n)=p^{\left(\frac{2}{27}+O(n^{-1/3})\right)n^3},$$ which is proven here. This colossal growth along with the results of Besche, Eick & O'Brien seem to be what primarily motivated the conjecture.