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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?

Thanks!

  • 5
    There are $49487365422$ of order 10242012-11-20
  • 0
    This was a result of Besche, Eick and O'Brien. Note that O'Brien $\Rightarrow$ computer usage. See [mathscinet](http://www.ams.org/mathscinet-getitem?mr=1935567). Basically, the paper is a survey discussing methods and algorithms used to construct (small) groups.2012-11-20
  • 6
    No doubt this about _isomorphism classes_ of groups.2012-11-20
  • 0
    Is there a rough estimate (say one significant digit) of the number of groups of order 2048?2012-11-20
  • 3
    @yatima2975: From [this article](http://www.math.auckland.ac.nz/~obrien/research/gnu.pdf) : "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."2012-11-21
  • 0
    Is there a chart (or database to build one) to visualize the distribution of group orders less than $n$ for individual $n$ up to ~2000?2012-11-22
  • 0
    @alancalvitti: Try this one http://oeis.org/A000001/b000001.txt2012-11-23

2 Answers 2

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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.

  • 0
    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.2012-11-20
  • 1
    (I wonder if there are more groups of order $3^{10}$ than of order $2^{10}$? Genericity proofs are...unsavoury...at least to my pallet...)2012-11-20
  • 0
    (...because a genericity result based on order would loose the fact that there are more groups of order $p^n$ than of $2>n$ purely because $2...which doesn't quite ring true with reality...for some interpretation of reality...)2012-11-20
  • 0
    There are probably more groups of order $3^{10}$ than $2^{10}$, but what the conjecture means is that if you denote by $\mu(n)$ the number of groups of order less than or equal to $n$ and by $\rho(n)$ the number of $2$-groups of order less than or equal to $n$, then $\lim_{n\to\infty}\rho(n)/\mu(n)=1$.2012-11-20
  • 6
    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.2012-11-20
  • 0
    @AlexanderGruber: I know this, I am just saying that it is somewhat unsavoury. For example, one can do what you are saying and perhaps prove that almost all groups are two-groups, but then one can use exponents of $p$-groups and perhaps prove that "almost no" p-groups are $2$-groups (assuming there are more groups of order $p^n$ than of order $2^n$, $p>2$). If you change your definition of genericity you change your result...2012-11-20
  • 5
    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 $2^4=16$.2012-11-20
  • 12
    Almost all groups are infinite.2012-11-20
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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 $1024$.