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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?


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There are $49487365422$ of order 1024 – PAD Nov 20 '12 at 15:32
This was a result of Besche, Eick and O'Brien. Note that O'Brien $\Rightarrow$ computer usage. See mathscinet. Basically, the paper is a survey discussing methods and algorithms used to construct (small) groups. – user1729 Nov 20 '12 at 15:34
No doubt this about isomorphism classes of groups. – Marc van Leeuwen Nov 20 '12 at 17:28
Is there a rough estimate (say one significant digit) of the number of groups of order 2048? – yatima2975 Nov 20 '12 at 23:06
@yatima2975: From this article : "gnu(2048) is still not precisely known, but it strictly exceeds 1774274116992170, which is the exact number of groups of order 2048 that have exponent-2 class 2, and can confidently be expected to agree with that number in its first 3 digits." – Mikko Korhonen Nov 21 '12 at 10:59
up vote 48 down vote accepted

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.

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A while ago I tried to find a reference for this "almost all..." result. I think it is just a folklore statement, with the paper which is the subject of this thread proffered as evidence. – user1729 Nov 20 '12 at 15:42
(I wonder if there are more groups of order $3^{10}$ than of order $2^{10}$? Genericity proofs least to my pallet...) – user1729 Nov 20 '12 at 15:44
Of course it's possible to deduce the number of isomorphism classes of groups of (finite) order $n$: write down all possible $n$ by $n$ multiplication tables, check which satisfy the group axioms, check every bijection between each pair to see if it's a group isomorphism. Since everything is finite, this can all be computed in finite time. The hard part is finding ways to do it in a sane amount of time. – Chris Eagle Nov 20 '12 at 15:55
According to the list linked in the answer, there are 504 groups of order $3^6=729$ and 267 groups of order $2^6=64$. There are 15 groups of order $5^4=625$ and also of order $3^4=81$ and 14 groups of order $2^4=16$. – Mark Bennet Nov 20 '12 at 16:40
Almost all groups are infinite. – Marc van Leeuwen Nov 20 '12 at 17:29

This is true. The amount of groups of order at most 2000 (up to isomorphism) was calculated precisely for the first time in 2001 by Besche, Eick and O'Brien. Here is the announcement of their result:

We announce the construction up to isomorphism of the $49 910 529 484$ groups of order at most $2000$.

In table 1 the number of groups of order $1024$ is given, it is $49 487 365 422$. Hence ~99.2% of all groups of order at most $2000$ have order $1024$.

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