# Number of groups of a given order

In general, for what $n$ do there exist two groups of order $n$? How about three groups of order $n$?

I know that if $n$ is prime, there only exists one group of order $n$, by Lagrange's Theorem, but how do you classify all other such $n$ that have $2, 3, 4, ...$ groups?

This question came to me during a group theory class, when we were making a table of groups of order $n$. For instance, all groups of order $4$ are isomorphic to either $C_4$ or $C_2\times C_2$.

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As far as I know, there aren't many nice results, unless you make strong assumptions about the prime factorisation of $n$. See this page. –  Joshua Pepper Mar 8 '14 at 17:57
I think my instructor told me there was a way to classify all such $n$ with 2 groups! He didn't show us, though. –  Mathemusician Mar 8 '14 at 18:02
This is a very difficult question. See en.wikipedia.org/wiki/Classification_of_finite_simple_groups –  R.. Mar 9 '14 at 6:18

The numbers for which there are precisely $1$, $2$ and $3$ groups are classified in the short paper http://www.math.ku.dk/~olsson/manus/three-group-numbers.pdf By Jørn Børling Olsson.

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A partial result: there exists a unique group of order $n$ (i.e. the cyclic group) if and only if $(n,\phi(n)) = 1$, where $\phi$ is the Euler phi function and $(a,b) = \operatorname{gcd}(a,b)$. This is certainly satisfied when $n$ is a prime, but when $n = 15$ we have $\phi(n) = 8$, and since $(15,8) = 1$ there is a unique group of order $15$, even though $15$ is not prime. A proof can be found here.

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Or equivalently, if $n$ is a product of distinct primes $p_i$ such that no $p_i$ divides any $p_j-1$; –  Derek Holt Mar 8 '14 at 18:06
Great reference! Thanks! –  Mathemusician Mar 8 '14 at 18:15

Here are two possibilities for $n$ for which there are precisely two groups of order $n$.

1. $n=p_1p_2 \cdots p_n$ for some $n \ge 2$ and distinct primes $p_i$, such that there is exactly one pair $(i,j)$ with $p_i$ divides $p_j-1$.

2. $n=p_1^2p_2\cdots p_n$ for some $n \ge 1$ and distinct primes $p_i$, where there are no $(i,j)$ with $p_i$ divides $p_j-1$, and also no $p_i$ divides $p_1+1$.

Are there any further possible $n$?

Contrary to what I wrote earlier, it is possible to have precisely three groups of order $n$. The values of $n$ with this property and $n \le 2000$ are:

 75, 363, 609, 867, 1183, 1265, 1275, 1491, 1587, 1725, 1805


I'll leave it to someone else to describe the complete set of such $n$.

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Can you explain how you came up with the second one? –  Mathemusician Mar 8 '14 at 18:12
Well there are two groups of order $p_1^2$, $C_{p_1^2}$ and $C_{p_1} \times C_{p_1}$. The divisibility condition ensures that any group of that order is the direct product of its Sylow subgroups. –  Derek Holt Mar 8 '14 at 18:24
I believe it is still open whether the "number of groups of order $n$" function is surjective. But it seems likely to be true, and it has been verified that the function takes all possible values $\leq 10000000$. See Blackburn et. al, Enumeration of Finite Groups, 21.6, pg. 238. Also page 8 of this article by Conway, Dietrich and O'Brien. –  Mikko Korhonen Mar 9 '14 at 9:50

Take a look at this wikipedia page on $p$-groups: http://en.wikipedia.org/wiki/P-group

The results in the subsection Among groups under the section Prevalence are astonishing.

The basic corollary is: Your question is very hard in general.

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Exactly 2 groups. There is a paper, which claims to classify "Orders for which there exist exactly two groups". This link contains the text of the paper in text (!) format. I didn't find a pdf.

Disclamer: I didn't check if the proofs in the paper are correct. I also don't know, if the paper was published in any peer-reviewed journal (probably it wasn't).

Exactly 3 groups. Proposition 2 of the paper above says, that if $n$ is not cube-free, then there are at least 5 groups of the order $n$. So you are only interested in cube-free numbers. Then you at least know all numbers $n$ with exactly 3 groups for $n<50000$ from here. Based on this one will know the probable answer and then one can try to prove it.

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