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

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

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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

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

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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

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.

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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.

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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
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