If $G$ is a finite group whose order is divisible by a prime $p$ and $p^r$ is the maximal power of $p$ that divides it, the Sylow theorems tell us that the number $n_p$ of Sylow $p$-subgroups of $G$ is congruent to $1$ modulo $p$ and a divisor of $\lvert G\rvert/p^r$, and that if $n_p=1$ then $G$ is not simple. With this information alone we can decide that some numbers are not the order of a finite simple group.

How big is the set of numbers that these two facts exclude as candidates to be the order of a simple group?

By big here one can mean, say, the density in $\mathbb{N}$. One can also weigh numbers by the number of actual groups of each order, so as to turn the question into

What proportion of finite groups are known to be non-simple by using only these two Sylow theorems on the number of they Sylow subgroups?

Since most groups are apparently $p$-groups, the answer to this is probably one.

  • $\begingroup$ Probably the (conjectural) answer to the second question will be: proportion 1, just because among groups, those that are 2-groups are preponderant, and are covered by this result as non-(simple non-abelian). The first question is possibly more interesting and it would be useful to define more formally this set of integers without reference to groups. $\endgroup$
    – YCor
    Sep 11, 2021 at 22:43
  • $\begingroup$ Yup, by zero I meant one :-) $\endgroup$ Sep 11, 2021 at 22:46
  • $\begingroup$ It seems that Will Sawin used one group-free definition of this set: the set $W$ of $n$ such that for every prime divisor $p$ of $n$, there exists $a\ge 1$ such that $1+ap$ divides $n$, and shows $W$ has natural density $0$. (I don't know if there other orders that can be easily ruled out using Sylow's theorems.) $\endgroup$
    – YCor
    Sep 12, 2021 at 5:33
  • 1
    $\begingroup$ I checked a little more: many numbers $n$ are in $W$ but can be ruled out using Sylow's theorems too, e.g., $n=12$. Also $n=4.3^m$ for $m\ge 2$ is ruled out because $n_3=4$ means a nontrivial homomorphism $G\to S_4$, which can't be injective; similarly $n=3.2^m$ is ruled out getting a homomorphism $G\to S_3$. $n=30$, $n=105$ (which are in $W$) are ruled out just counting elements of Sylows, there are too many. $\endgroup$
    – YCor
    Sep 12, 2021 at 7:43
  • $\begingroup$ @YCor, indeed, the Sylow theorems allow us to rule out many other orders, but my question was really only about those orders that are ruled out by the two facts I mentioned, the complement of W. These are the orders that get excluded with least effort, if you want. $\endgroup$ Sep 13, 2021 at 4:37

1 Answer 1


Such numbers have density $1$. In fact, for a density $1$ set of numbers it suffices to apply Sylow's theorem to the largest prime factor.

In other words, for a density $1$ set of numbers $n$, with $p$ the largest prime factor of $n$, there is no number congruent to $1$ mod $p$ dividing $n$ other than $1$.

Proof: We start from a classical fact of analytic number theory, that for each $\epsilon>0$ there is $\delta>0$ such that a density $1-\epsilon$ fraction of numbers have a prime factor of size $>n^{\delta}$. So it suffices to restrict attention to numbers with a prime factor of size $n^{\delta}$.

The number of numbers between $X$ and $2X$ which have a large prime factor do not satisfy the condition is at most the number of triples $p,a,b$ where $p> X^{\delta}$ is prime, $a>0$, and $$X< p (1+pa) b< 2X$$

Indeed, take $p$ to be the largest prime factor, $1+pa$ a nontrivial divisor congruent to $1$ mod $p$, and $b$ to be whatever factor remains.

Now note that for each $p,a$, the number of possible values of $b$ is at most $\frac{2 X}{ p (1+pa) }\leq \frac{2X}{p^2 a}$. Summing over $a$ from $1$ to $X$, the number of such triples with fixed $p$ is $O ( \frac{X }{p^2} \log X)$ by the harmonic series. Summing over $p$ from $X^\delta$ to $\infty$, the number of such triples is $O( \frac{X \log X}{ X^{\delta}} ) = o(X)$ so the set of numbers not satisfying the condition has density zero, as desired.


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