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Can you say how to find number of non-abelian groups of order n?

Suppose n is 24 ,then from structure theorem of finite abelian group we know that there are 3 abelian groups.But what can you say about the number of non-abelian groups of order 24?

The following link is a list of number of groups of order n: http://oeis.org/wiki/Number_of_groups_of_order_n .But here also they did not mention anything how to find number of non-abelian groups of order n.

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Do you really want to know about methods for general $n$, or just for $n=24$? – Chris Eagle Feb 13 at 11:18
I want a general result – math89 Feb 13 at 11:56

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Can be pretty messy to do this, but using semidirect products you can get quite some answers.

In you example, $\,24=2^3\cdot 3\,$ and from Sylow Theorems and some element counting either the Sylow $\,2-$ subgroup or the $\,3-$subgroup must be normal. From here you obtaing some action by automorphisms of the other Sylow subgroup (i.e., the one that's not necessarily normal) on the normal one (i.e., a homomorphism to the normal one's automorphisms group), and from here we can construct semidirect products.

Of course, the above cannot properly be developed in this site and by this means, so you better grab some good group theory books and read it there.

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Can you refer some book(/article) where I can find it.I have Herstein book but it is not there. – math89 Feb 13 at 11:59
(1) Dummit & Foote's "abstract Algebra", (2) Milne's "Group Theory", which you can download in jmilne.org/math/CourseNotes/gt.html . It is an excellent introduction to this. (3) Rotman's "An Introduction to the Theory of Groups" , (4) The books in group theory by Rose, Robinson, Kargapolov-Merzlakov, etc., etc.... – DonAntonio Feb 13 at 13:14
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The point is, the numbers grow very fast, particularly for prime powers. Look at this list for groups of order $2^{k}$, for $k \le 10$. (I will try and provide a better reference later.)

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Actually the numbers are not following some order,they are increasing rapidly. – math89 Feb 13 at 12:01
@user62142, there is some kind of pattern, as the Higman-Sims estimate for the number of (isomorphism classes of) $p$-groups of order $p^n$ is $ \exp\left(\log(p) \left(\frac{2}{27} + O(n^{-1/3})\right) n^3\right), $ see: plms.oxfordjournals.org/content/s3-15/1/151.full.pdf – Andreas Caranti Feb 13 at 12:08
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The short answer is that there is no formula for the number of non-abelian groups of order $n$, nor is there an algorithm for computing the number of such groups of order $n$.

Note that we do not really have a formula for the number of abelian groups. We can express the number in terms of the number of partitions of an integer; this is something we know a lot about and there are algorithms for computing it for a given value.

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