Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Related to A000679 (Number of groups of order $2^n$), how many non-Abelian groups of order $2^n$ are there?

share|improve this question
I don't think you'll be able to do any better than subtracting the number of abelian groups from the number of groups, since the latter is complicated and the former is simple. As a hint for what the former looks like: en.wikipedia.org/wiki/… –  Qiaochu Yuan Jul 2 '10 at 23:48
@Qiaochu Yuan: Ah, okay. That would work. And if I can apply the fundamental theorem correctly, the number of Abelian $p$-groups of order $p^n$ is the partition function of $n$. Thank you for the idea. –  HYYY Jul 3 '10 at 0:23
add comment

4 Answers

up vote 8 down vote accepted

It is true that there is no known formula for the number of isomorphism classes of groups of order $n$, but there is a very nice asymptotic formula for $p$-groups.

In particular, the number of isomorphism classes of groups of order $p^n$ grows as $p^{\frac{2}{27}n^3 + O(n^{8/3})}$. This function grows very rapidly, and there is a folklore conjecture that "almost all groups are $2$-groups."


share|improve this answer
For more about results related to the statement that Matt gave see Bjorn Poonen's paper arxiv.org/abs/math/0608491 –  Noah Snyder Jul 3 '10 at 3:00
add comment

It is as Qiaochu Yuan says, and worse: the number of abelian groups of order 2^n are not complicated. They are direct products of cyclic groups of order 2^m. The number of non-abelian groups is unknown. We don't actually have a formula. We just have exhaustive analyses for a few small values of n.

share|improve this answer
add comment

Marshall Hall and James Senior published a book The Groups of order 2n (n <= 6) in the 1960's. It's in the common room in the math department here at UCSB. Using rather obscure notation it arranges the conjugacy classes into lattices.

Online you can find a list of all groups of order 64.

share|improve this answer
That'd be a pretty handy book to have if you're an algebraicist,John. –  Andrew L Jul 3 '10 at 2:43
The number of groups of order $2^n$ for $0\leq n\leq 10$ is given by 1, 1, 2, 5, 14, 51, 267, 2328, 56092, 10494213, 49487365422. See oeis.org/A000679. –  Richard Stanley Jul 3 '10 at 2:59
add comment

I don't think there's a general method of approaching this.There are several methods of obtaining all the Abelian groups of order n for specific n-such as the finite groups of prime order are all cyclic and therefore Abelian,etc. Unless n is very large,proceeding in this manner will usually get the number of groups down to a managable size and what remains should be non-Abelian.

I think that's the best you can do unless you want to add the condition that the remaining groups are simple-in which case,a Sylow analysis would be appropriate and would considerably simplify things.

There are a few general results on non-Abelian groups-like a finite group G is nonAbelian if there are 2 elements in G whose commutator is nontrivial i.e. not the identity. But I don't know if you can use these kinds of results to get the kind of general formula you want-I don't think you can,although I could be wrong on this.

The best discussions I know of such matters can be found in Herstien's Topics in Algebra,2nd edition and I.Martin Issacs' Finite Group Theory.

share|improve this answer
For $p$ a prime, the only simple groups of order $p^n$ for some $n$ are the cyclic groups of order $p$. This follows easily from the fact that groups of order $p^n$ are nilpotent. –  Andy Putman Jul 3 '10 at 2:49
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.