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We say that an positive integer $n$ is a

  • simple number if there exist a non abelian simple group of order $n$. Denote by $\mathfrak{s}$ this set.
  • prime-power number if it is of the form $n=p^a$, with $p$ prime. Denote by $\mathfrak{pp}$ this set.
  • Sylow congruent number if it is a non prime-power number for which the congruence conditions of Sylow's theorem do not force at least one Sylow subgroup to be normal. In other words we can have $n_p\neq 1$ for all primes $p$ dividing $n$. Denote by $\mathfrak{sc}$ this set.

Also denote by $\mathfrak{p}$ the set of prime numbers.

Even though we know of the finite simple groups (I don't but some do), it seems unrealistic to expect an clean characterization of $\mathfrak{s}$.

For a set of positive integers $\mathfrak{t}$ we define $\pi_{\mathfrak{t}}$ the counting function of that set, i.e. $\pi_{\mathfrak{t}}(x)$ gives the number of elements in $\mathfrak{t}$ less than or equal to $x$, for any real number $x$.

It is well-known that $\pi_{\mathfrak{p}}(x)\sim\frac{x}{\ln x}$ (Prime number theorem).

Do we know the asymptotic laws of distribution for the other sets defined above, i.e. asymptotic behaviors of $\pi_{\mathfrak{pp}}(x)$, $\pi_{\mathfrak{s}}(x)$ and $\pi_{\mathfrak{sc}}(x)$, or $\pi_{\mathbb{N}\setminus \mathfrak{pp}}(x)$, $\pi_{\mathbb{N}\setminus \mathfrak{s}}(x)$ and $\pi_{\mathbb{N}\setminus \mathfrak{sc}}(x)$?

I suspect that we have $\pi_{\mathbb{N}\setminus \mathfrak{sc}}(x)\sim x$, or $\pi_{\mathbb{N}\setminus \mathfrak{sc}}(x)\sim k x$ for some $0<k<1$ (based on some computation). At least if we restrict to odd numbers.

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I have not thought much about it, but I strongly suspect that the simple numbers are dominated by the orders $p(p^2-1)/2$ of the groups ${\rm PSL}(2,p)$ for odd primes $p$. –  Derek Holt Oct 20 '11 at 15:50
    
@Nathan: is 12 in SC? –  Jack Schmidt Oct 20 '11 at 15:57
    
@Jack: Yes 12 is in SC based only on the congruence criterion of Sylow Theorem, so are 24, 30, 36, 48, 56,... –  Nathan Portland Oct 20 '11 at 16:30
    
@Derek: you are right, very strongly dominated by these numbers. There are 493 non abelian simple groups of order $<10^{10}$, of which 428 are of the type PSL($2,p$) for odd primes $p$. How does one prove it though? –  Nathan Portland Oct 20 '11 at 18:04
    
@Nathan: I show how to prove a similar domination for pp. s is the same idea, just the sum is over more indices. All but finitely many simple groups are in families of the form X(r,p,k) and their orders are well approximated by a prime power (usually some large power of p^k), and so the only ones that don't wash out are the ones where r=k=1. –  Jack Schmidt Oct 20 '11 at 18:20

1 Answer 1

up vote 2 down vote accepted

Here is a simpler problem that should get you going: Instead of counting prime powers, let's just count prime squareds.

$$\pi_{\mathfrak{p}^2}(x) = \#\{ p^2 : p^2 < x, p \in \mathfrak{p} \} = \#\{ p^2 : p < \sqrt{x}, p \in \mathfrak{p} \} = \pi_{\mathfrak{p}}(\sqrt{x})$$

You can estimate this using the good ole PNT:

$$\pi_{\mathfrak{p}^2}(x) \sim \frac{2\sqrt{x}}{\ln(x)}$$


Now several of your sets are basically counting prime powers. $\mathfrak{pp}$ for instance has a nice expression as:

$$\pi_{\mathfrak{pp}}(x) = \sum_{k=1}^{\log_2(x)} \pi_{\mathfrak{p}^k}(x) = \sum_{k=1}^{\log_2(x)} \pi_{\mathfrak{p}}(x^{(1/k)})$$

The asymptotics are the same:

$$\pi_{\mathfrak{pp}}(x) \sim \frac{x}{\ln(x)}$$

Similarly, I suspect there is no trouble with:

$$\pi_{\mathfrak{s}}(x) \sim \frac{3\sqrt[3]{x}}{\ln(x)}$$


Calculus fun:

$$\lim_{x\to\infty} \frac{ x/\ln(x) } { \sum_{k=1}^{\log_2(x)} kx^{(1/k)}/\ln(x) } = \lim_{x\to\infty} \frac{x}{\sum_{k=1}^{\log_2(x)} kx^{(1/k)}}$$

Clearly the operand of the limit is less than 1 for all $x > 1$, so we squeeze from below:

$$\begin{align} 1 &\geq \lim_{x\to\infty} \frac{x}{\sum_{k=1}^{\log_2(x)} k x^{(1/k)}} \\ \\ &\geq \lim_{x\to\infty} \frac{x}{x + \sqrt{x} \sum_{k=2}^{\log_2(x)} k} \\ \\ &\geq \lim_{x\to\infty} \frac{x}{x+\sqrt{x} (\log_2(x))^2} = 1 \end{align}$$ In the second line, I use that $\sqrt{x} \geq x^{(1/k)}$ for $k \geq 2$, and then in the third line I overestimate an arithmetic series. In the last line, we divide top and bottom by x and the second term of the denominator goes to 0.

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Thanks a lot Jack, I am skeptical about $\pi_{\mathfrak{s}}(x) \sim \frac{\sqrt{x}}{2\ln(x)}$, because $\pi_{\mathfrak{s}}(10^{10})= 493$, which is about 5 times smaller than the right side of $\sim$. I know it grows slowly, so it may be correct after all... –  Nathan Portland Oct 20 '11 at 18:46
    
@Nathan: Oh sorry, PSL(2,p) has order about p^3 not p^2. I also had the k in the wrong place. All fixed now. The calculus is less fun (arithmetic series instead of harmonic), but no big change. –  Jack Schmidt Oct 20 '11 at 19:14
    
Thanks again Jack. –  Nathan Portland Oct 20 '11 at 20:36

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