Welcome to MathOverflow

Question

3

1

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

flag	asked Jul 2 2010 at 23:41 HYYY 955●4●10

4

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 2010 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 2010 at 0:23

Matthew Kahle · Accepted Answer · 2010-07-03 01:41:44Z UTC

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."

http://en.wikipedia.org/wiki/P-group

John Mangual · Answer 2 · 2010-07-03 02:32:28Z UTC

2

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.

answered Jul 3 2010 at 2:32

John Mangual
6,410●4●43

	That'd be a pretty handy book to have if you're an algebraicist,John. – Andrew L Jul 3 2010 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 research.att.com/~njas/sequences/A000679. – Richard Stanley Jul 3 2010 at 2:59

Daniel Mehkeri · Answer 3 · 2010-07-03 00:21:33Z UTC

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.

Andrew L · Answer 4 · 2010-07-03 01:12:04Z UTC

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.

Number of non-Abelian groups of order $2^n$

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

4 Answers

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.