37
$\begingroup$

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 $\{n: f(n) = 1\}$?

$\endgroup$
7
  • 2
    $\begingroup$ (In case people want some data and known results, oeis.org/A000001) $\endgroup$ Nov 13, 2013 at 4:32
  • 2
    $\begingroup$ Presumably the density of $n$ with $f(n)=1$ is zero, because there are lots of semidirect products. $\endgroup$
    – Lucia
    Nov 13, 2013 at 4:40
  • 3
    $\begingroup$ 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. $\endgroup$ Nov 13, 2013 at 6:06
  • 4
    $\begingroup$ I'd rather have thought this question would be a good candidate for migration to Math.SE ... . $\endgroup$
    – Stefan Kohl
    Nov 13, 2013 at 14:04
  • 2
    $\begingroup$ @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. $\endgroup$ Nov 13, 2013 at 17:47

4 Answers 4

55
$\begingroup$

$f(n)=1$ 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).

$\endgroup$
33
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Thanks — that's an interesting paper. To clarify, does $\gamma$ denote the Euler–Mascheroni constant? $\endgroup$ Nov 13, 2013 at 5:08
  • 1
    $\begingroup$ @Daniel, Yes. It arises (essentially) through the use of Merten's formula. $\endgroup$
    – Mark Lewko
    Nov 13, 2013 at 5:18
18
$\begingroup$

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.

$\endgroup$
9
$\begingroup$

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 $\log \log n$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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