This question may have more of a vague, less objective answer than usual for this site, so I apologise if it difficult to answer definitively.

Below, $p$, $q$ and $r$ are distinct primes.

  • A group of order $p$ is simple (these groups are precisely the prime-order cyclic groups.)

  • A group of order $p^n$ is not simple, for $n \geq 2$. (Here is a proof from this site.)

  • A group of order $pq$ is not simple. (Proof.)

  • A group of order $p^2q$ is not simple. (Proof.)

  • A group of order $p^2q^2$ is not simple. (Proof.)

  • A group of order $pqr$ is not simple. (Proof.)

It is not always possible to classify the simplicity this way - consider $p^2qr$. $60=2^2 \cdot 3 \cdot 5$. $A_5$ is simple, $\mathbb{Z}_{60}$ is not.

Is there a more general statement that can classify whether a group is either a) definitely simple, b) definitely not simple, c) could be either simple or not simple, in terms of the decomposition of the prime factors of the order, or certain results that rule out many more cases?

  • 2
    $\begingroup$ By a well-known result of Burnside, groups of order $p^aq^b$ are not simple for primes $p,q$ and $a,b \ge 0$. A simple group order must be divisible by $8$ or by $12$. $\endgroup$
    – Derek Holt
    Commented May 31, 2023 at 18:05
  • $\begingroup$ @DerekHolt Thank you, this is exactly what I am looking for. If anyone is aware of any other conditions, please do continue to list them. $\endgroup$
    – Robin
    Commented May 31, 2023 at 18:15

1 Answer 1


Probably the most important result here is the Feit-Thompson Theorem (or Odd Order Theorem):

Theorem (Feit & Thompson, 1963). Every finite group of odd order is solvable.

Corollary. Any non-abelian finite simple group has even order.

That is, if $G$ is simple but non-cyclic, then $2$ is necessarily a prime factor of $|G|$.

This result played a significant role in the Classification of finite simple groups. Its proof is notoriously long.

  • 1
    $\begingroup$ Thanks for such a quick answer, this is certainly what I am looking for, ruling out a lot of cases. Alongside the comment by @Derek Holt this helps substantially. I'll accept when I can. $\endgroup$
    – Robin
    Commented May 31, 2023 at 18:08
  • $\begingroup$ Further, if someone is able to provide even more results other than Burnside or Feit-Thompson, I would really appreciate them on this question as answers/comments. Thank you both! $\endgroup$
    – Robin
    Commented May 31, 2023 at 18:11
  • $\begingroup$ @Robin I would hold off accepting this for now - I suspect a more can be said, possibly by Derek Holt himself! $\endgroup$
    – user1729
    Commented May 31, 2023 at 18:15
  • 1
    $\begingroup$ Alright - if there is another answer posted, I will consider changing the accept to that if it has more to say, but thank you nonetheless. $\endgroup$
    – Robin
    Commented May 31, 2023 at 18:16

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