Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $f(n)$ denote the number of (isomorphism classes of) groups of order $n$. A couple easy facts:

  1. If $n$ is not squarefree, then there are multiple abelian groups of order $n$.
  2. If $n \geq 4$ is even, then the dihedral group of order $n$ is non-cyclic.

Thus, if $f(n) = 1$, then $n$ is a squarefree odd number (assuming $n \geq 3$). But the converse is false, since $f(21) = 2$.

Is there a good characterization of $n$ such that $f(n) = 1$? Also, what's the asymptotic density of $\{n: f(n) = 1\}$?

share|improve this question
(In case people want some data and known results, oeis.org/A000001) –  Andres Caicedo Nov 13 '13 at 4:32
Presumably the density of $n$ with $f(n)=1$ is zero, because there are lots of semidirect products. –  Lucia Nov 13 '13 at 4:40
Since we always have the cyclic group of order $n$, then $f(n)=1$ if and only if $n$ is a cyclic number. The cyclic numbers are well known: they are the square-free integers $n=p_1\cdots p_r$, where $p_1\lt p_2\lt\cdots\lt p_r$ in which $p_i$ does not divide any of $p_j-1$ for all $j\neq i$. See e.g. Pete Clark's answer here and references cited there. –  Arturo Magidin Nov 13 '13 at 6:06
I'd rather have thought this question would be a good candidate for migration to Math.SE ... . –  Stefan Kohl Nov 13 '13 at 14:04
@StefanKohl: The first part (values of $n$ for which $f(n)=1$), definitely yes; the second part (asymptotic density of cyclic numbers) seems a better fit for MO, though. –  Arturo Magidin Nov 13 '13 at 17:47

4 Answers 4

up vote 43 down vote accepted

$f(n)=1$ if and only if $\gcd(n,\phi(n))=1$, where $\phi$ is the Euler phi-function. These $n$ are tabulated at http://oeis.org/A003277

The result is found in Tibor Szele, Über die endichen Ordnungszahlen, zu denen nur eine Gruppe gehört, Comment. Math. Helv. 20 (1947) 265–267, MR0021934 (9,131b).

share|improve this answer

Let $G(x)$ denote the number of $n \leq x$ such that there is exactly $1$ isomorphism class of groups of order $n$. Then: $$G(x) \sim e^{-\gamma}\frac{x}{\log\log\log(x)} $$ where $\gamma$ is Euler's constant. This is a result of Erdos, Murty and Murty. Their paper also contains other interesting results on the distribution of values of the group order function.

share|improve this answer
Thanks — that's an interesting paper. To clarify, does $\gamma$ denote the Euler–Mascheroni constant? –  Daniel Hast Nov 13 '13 at 5:08
@Daniel, Yes. It arises (essentially) through the use of Merten's formula. –  Mark Lewko Nov 13 '13 at 5:18

The original question has already been answered, so I thought I would provide a slightly more general version.

The short paper http://www.math.ku.dk/~olsson/manus/three-group-numbers.pdf describes those orders for which there are precisely 1, 2 or 3 groups of the given order.

share|improve this answer

If $p$ is the smallest prime dividing $n$, then if $n=pm$ and $p|\phi(m)$ then there exists a semidirect product of the cyclic group of order $p$ and the cyclic group of order $m$. So $f(n)$ is not $1$ for such $n$. Now given a prime $p$, most values of $m$ that are odd and coprime to $p$ will have $\phi(m)$ being a multiple of $p$ (all we need is some prime factor of $m$ to be $1\pmod p$, and usually $m$ will have some such factors). Since most numbers won't be coprime to all small primes, this will give a proof that the density of numbers with $f(n)=1$ is zero.

Note: Mark Lewko posted the interesting reference to Erdos, Murty & Murty while I was writing the answer above. Comparing our answers, one can see that the numbers with $f(n)=1$ are closely related to the numbers $n$ having no prime factor below $\log \log n$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.