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Does there exist an answer to the next question:

How many groups of order $n$ ($|G|=n$, $n\in \mathbb{N}$) are exist up to isomorphism?

(The groups are not necessarily Abelian)

I am curious about this question.

If the answer ($\forall n\in\mathbb{N}$) does not exist,does the answer exist for $n=$? $6$? $10$?
(The question about $n=1,2,3,4$ (and primes or p*q) is easy because the groups are abelian).

Or Shahar
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  • If $n=\text{prime}$, then $G$ is cyclic. If $n=2 \times \text{ (prime)}$, then it can be either cyclic $C_{n}$ or isomorphic to $D_n$ (dihedral group). – Anurag A Aug 06 '20 at 06:28
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    For every $n$, the question definitely has an answer. For $n$ prime, there is only one such group. For $n = 6$, there are only 2 such groups. See http://oeis.org/wiki/Number_of_groups_of_order_n – Doctor Who Aug 06 '20 at 06:29
  • @DoctorWho when you say "definitely has an answer" you got to be careful. It doesn't have a closed form for every $n$, instead only asymptotic expression. – Anurag A Aug 06 '20 at 06:32
  • Thanks for answering! – Or Shahar Aug 06 '20 at 06:33
  • @AnuragA I didn't say there was a closed form formula. I stated that given a fixed $n$, there is an answer to the question "How many groups of order $n$ are there up to isomorphism?" – Doctor Who Aug 06 '20 at 06:35
  • @AnuragA there is no asymptotic expression. There are arbitrarily large $n$ for which the number of groups is exactly one, but the number over all $n$ is unbounded. – Gerry Myerson Aug 06 '20 at 06:35
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    There is a tabulation, and many links, at http://oeis.org/A000001 – Gerry Myerson Aug 06 '20 at 06:36
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    This is a very difficult question. There is an exact formula for squarefree $n$ and small prime powers, and an asymptotic formula for general prime powers. – anon Aug 06 '20 at 06:37
  • @GerryMyerson You are right. I should have been more careful in writing. What I meant was that there is at best an asymptotic bounds for infinitely many $n$'s. In fact, in my first comment (see above) itself, I had covered the case that for some groups with certain order, the no. of isomorphic groups are either $1$ or $2$. – Anurag A Aug 06 '20 at 06:39
  • @runway44 the number for squarefree $n$ depends on the residue classes of factors of $n$ modulo other factors of $n$. I'm not sure I'd say there's an exact formula. – Gerry Myerson Aug 06 '20 at 06:39
  • @runway44 it's a nice formula, for sure. – Gerry Myerson Aug 06 '20 at 06:53
  • And of course someone should mention the well-known open problem of whether there exists an $n$ such that there are exacly $n$ isomorphism classes of groups of order $n$. – Derek Holt Aug 06 '20 at 08:06

1 Answers1

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Let

$$g(n)=\text{number of isomorphism classes of groups of order }n$$

Then we can rephrase your question as:

Do we know the value of $g(n)$ for any $n\in\mathbb{N}$?

There are algorithms to calculate $g(n)$. The simpliest is as follows: for a given set $G$ of size $n$ we can simply go through all possible functions $G\times G\to G$ and verify how many of them produce groups (by checking axioms) and then filter out non-isomorphic (we can again go through all possible homomorphisms). The complexity of this is huge though, since there are $n^{n^2}$ potential candidates for a group operation. There are other algorithms, however all of them have complexity beyond being practically useful.

Unfortunately we don't know a closed formula for $g(n)$. We do know a bit, for special $n$. If $p$ is prime then

$$g(pq)=1\text{ if }q\text{ is another prime such that }p<q\text{ and }p\text{ does not divide }q-1$$ $$g(p)=1$$ $$g(p^2)=2$$ $$g(p^3)=5$$

$g(p^m)$ gets eventually very complicated, and soon (for $m\geq 5$ I think?) turns out to be too hard to solve precisely at the moment. Some approximations exist though, e.g. Higman-Sims asymptotic formula:

$$g(p^m)\sim p^{\frac{2}{27}m^3+O(m^{8/3})}$$

This gets even harder for naturals which are not prime powers.

If the answer ($\forall n\in\mathbb{N}$) does not exist,does the answer exist for $n=5$? $6$? $7$?

$g(5)=g(7)=1$ since these are primes. $g(6)=2$ (namely $\mathbb{Z}_6$ and $S_3$) but it is not trivial (although not hard as well).

freakish
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  • I have +1 the answer, but I think that an algorithm to calculate $g(n)$ is a formula for $g(n)$ - not a very illuminating one, but nevertheless a formula. – ancient mathematician Aug 06 '20 at 08:22
  • @ancientmathematician yes, but it is not a closed formula in the following sense: https://en.wikipedia.org/wiki/Closed-form_expression which is a very desired situation. – freakish Aug 06 '20 at 08:29
  • I'm not disagreeing with you about the desirability of a "nice" formula, but if a primitive recursive function is not a closed form I don't see what qualifies! – ancient mathematician Aug 06 '20 at 14:55
  • @ancientmathematician every formula in the post above is closed, e.g. $g(p)=1$. This also includes polynomials, rational functions, roots, exponentials, logarithms, their compositions and so on. I encourage you to read the linked article. Recursion is almost always worse than closed formula, since the latter is typically faster to compute. – freakish Aug 07 '20 at 17:05