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Using the Charaterization of finite simple groups theorem we can answer the question. There is another way to solve this problem? Using the Charaterization theorem we have to follow a path with a lot of calculation and verify is some order is in the form of n! and i'm not sure this is simple even for a computer. This problem can be challenging and we can exclude things. If the group is abelian we know it is not simple. If it is not abelian we can try to prove that we can find a subgroup of index $n$. In that case we could Use the fact that if $G$ is simple and $H$ is a subgroup of index n with $n\geq 3$, $|G|$ divides $\frac{n!} 2$. So since $n$! cannot divide $\frac{n!}{2}$ and we can find this subgroup we have done. Is this a path?

gamer rossi
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    You do need $n \geq 3$ since $\mathbb{Z}/2\mathbb{Z}$ is simple. – Tob Ernack Oct 21 '24 at 21:33
  • Yes, infact the strategy works for $n\geq 3$ and for n=2 or n=1 is pretty obvious since 1!=1 and 2!=2 and 2 Is a prime. – gamer rossi Oct 21 '24 at 21:38
  • Why would a subgroup of index $n$ imply that $|G|$ divided $\frac {n!}2$? Simple groups tend to have lots and lots of subgroups...just not normal ones. – lulu Oct 21 '24 at 23:50
  • It seems like this has been asked before and it seems from the comments like this is believable but not necessarily true. I certainly don't see an easy proof or counterexample, but maybe someone who knows more will be able to help. – Chris Grossack Oct 22 '24 at 01:00
  • @lulu https://groupprops.subwiki.org/wiki/Order_of_simple_non-abelian_group_divides_half_the_factorial_of_index_of_proper_subgroup – gamer rossi Oct 22 '24 at 07:20
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    It might well be true, but you would have to use the Classification of Finite Simple Groups to do it. A sketch would be: obviously $G$ cannot be alternating, and one can check the sporadics by hand. If $G$ has Lie type in characteristic $p$, then $p=2$ or $p>2$. If $p>2$ then $|G|$ has a large power of $p$ compared with other powers, which factorials shouldn't have (as 2 should be the most common prime). For $p=2$, I guess one should proceed by finding gaps in the primes dividing $|G|$, which would probably work for $p$ odd as well. You could probably use the power of $3$ as well. – David A. Craven Oct 22 '24 at 20:13

2 Answers2

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I believe it is true that there is no simple group of order $n!$, but as noted in the comments there is no hope of proving this without CFSG.

For a proof using CFSG, you can first rule out alternating groups (obvious) and sporadic groups (just check).

For groups of Lie type, you can rule out order $n!$ by adapting the argument in the following paper of E. Artin:

Artin, Emil. The orders of the classical simple groups. Comm. Pure Appl. Math. 8 (1955), 455-472. https://doi.org/10.1002/cpa.3160080403

In this paper Artin determines the coincidences among the orders of finite simple groups known at the time. You can use the exact same argument as in Section 3 to prove the result.

Here is an outline for the proof:

Let $N$ be the order of a finite simple group of Lie type in characteristic $p$. First Artin proves the following lemma.

Lemma: Let $M$ be the largest power of $p$ dividing $N$. Then $N \leq M^3$ for type $L_2(q)$, and $N \leq M^{8/3}$ for other types.

Now suppose $N = m!$. Then the largest power of $p$ dividing $N$ is $$\sum_{i \geq 1} \lfloor m/p^i \rfloor < \frac{m}{p-1}.$$

So the Lemma gives $N < p^{3m/(p-1)}$ for type $L_2(q)$, and $N < p^{8m /(3(p-1))}$ for other types.

On the other hand $N = m! > m^m/e^m$, so taking $m$th roots gives $$m < \begin{cases} e \cdot p^{\frac{3}{p-1}}, & \text{type } L_2(q). \\ e \cdot p^{\frac{8/3}{p-1}}, & \text{other types.}\end{cases}$$

This already gives you an upper bound $m \leq 21$ and you can handle the rest with some case-by-case analysis, exactly as in Artin's paper.

testaccount
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This is indeed a path for small $n$. But it will not work in general. For $n=5$ it is done exactly like this, see here:

A group of order $120$ cannot be simple.

For $6!=720$ it it already much harder, as the answer by Derek Holt and the other answers show, see here:

No simple group of order 720.

For $n=7$, there is only one perfect group of order $7!=5040$, and it is not simple.

For $n=8$, i.e., for groups of order $8!=40320$ see here:

The structure of a finite group of order 40320

None of them is simple.

It looks like that one has to use more than the standard tools of a group theory course for the general case.

Dietrich Burde
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