Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We know that there a isomorphism class of symmetric group structure of order $n!$.

Can we conclude any other information about the isomorphism classes of groups of order $n!$ ?

share|cite|improve this question
For n = 1, 2, ..., 6, the number of groups of order n! is 1, 1, 2, 15, 47, 840. The 7th term seems to be unknown. You may also like this: . I couldn't find anything more general. – Billy Jul 16 '13 at 2:07
@Billy ,Do you mean , you don't know it but others do ? or Do you mean you and mathematicians don't know ? – Maths Lover Jul 16 '13 at 2:18
I mean I computed the first few, inputted it into the OEIS, and it told me that the 7th term "seems to be unknown". How out-of-date that is I don't know, and it doesn't provide its sources, so nor do I. :) – Billy Jul 16 '13 at 2:19
I somewhat doubt mathematicians have classified groups of order $7!$, simply because that is a very specific big number, but on the other hand via prime factorization and enumeration of smaller $p$-groups it might be possible to do so (perhaps using a computer and good algorithms). I don't believe there is any nice characterization or counting of groups of order $n$ in general and I doubt narrowing ourselves to groups whose orders are factorials would much mitigate the craziness inherent in the growth and behavior of isoclasses. – anon Jul 16 '13 at 2:23
@PeteL.Clark What I'm more concerned about is if (informally speaking) the sequence is meaningful specifically in relation to possible group structures (simply noting that $|S_n|=n!$ seems a bit superficial to me) that might potentially make the problems of characterizing / counting / asymptotically analyzing any easier or the answers any more elegant. Compare to, for example, restricting ourselves to squarefree numbers, where we can actually provide an exact formula for the number of groups of given order. Picking just any sequence shouldn't be expected to change the game so to speak. – anon Jul 16 '13 at 2:48

For $n \in \mathbb{Z}^+$, let $g(n)$ be the number of isomorphism classes of groups of order $n$. I think that most sufficiently inquisitive students of group theory have asked themselves what can be said about this function. For instance:

1) For which $n$ is $g(n) = 1$?

This is equivalent to: for which $n$ is every group of order $n$ cyclic. This is known: it is necessary and sufficient that $\operatorname{gcd}(n,\varphi(n)) = 1$. I believe the result goes back to Burnside; a nice treatment can be found here.

2) What are upper and lower bounds on $g(n)$?

By 1), $g(n) = 1$ infinitely often. It is certainly also at least two infinitely often: in fact for any prime number $p$, $g(p^2) = 2$. So there is no hope for a precise asymptotic formula for $g(n)$. The sequence is discussed here: by its label, I guess it is a rather important integer sequence!

There are certainly some interesting results here: for instance, for $n \in \mathbb{Z}^+$, let $m(n)$ be the largest power to which any prime divides $n$ (so e.g. $m(n) = 1 \iff n$ is squarefree). Then:

Theorem (Pyber): $g(n) \leq n^{2/27 m(n)^2 + O(m(n)^{3/2})}$.

It is known that $g$ grows very rapidly along prime powers. In fact, if you fix an exponent $a$ and ask about $g(p^a)$ for a variable prime number $p$, you get some very interesting questions, which are for instance related to elliptic curve theory (!!). For a nice discussion, see e.g. here. In particular that reference gives:

Theorem (Newman-Seeley): $p^{2/27 n^3 - 6n^2} \leq g(p^n) \leq p^{2/27 n^3 + O(n^{5/2})}$.

Okay, you asked about $g(n!)$. The answer is that it will be huge. This follows from the following (rather crude: surely someone else can do better) multiplicative property of $g(n)$:

$g(mn) \geq \max g(m), g(n)$.

Indeed this follows from:

Theorem Krull-Remak-Schmidt: if $G \times H \cong G \times K$, then $H \cong K$.

Thus $g(n!)$ is at least as large as $g(2^{\sum_{n=1}^{\infty} \lfloor \frac{n}{2^i} \rfloor})$, since $2^{\sum_{n=1}^{\infty} \lfloor \frac{n}{2^i} \rfloor}$ is the largest power of $2$ dividing $n!$.

I'm a little too pressed right now to write out the details, but if you put that together with the lower bound in the Newman-Seeley Theorem, you should get a lower bound on $g(n!)$ which is both explicit and huge.

share|cite|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.