For which $n$ is there a unique perfect group of order $n$? Are there infinitely many such $n$?

Some guesses for infinite sequences of such $n$: $|\mathrm{PSL}(2,p)|$, $|\mathrm{SL}(2,p)|$, $|A_m|$, $|S_m|$.

If you replace "perfect" with "simple", a complete understanding follows from the Classification. Apart from a few small coincidences and one well-understood infinite family, there are no coincidences. See this previous question for instance. But the picture for perfect groups is less clear to me.

A complete list of the orders of perfect groups up to $10^6$ is contained in Section 5.4 of the book by Holt and Plesken (Holt, Derek F.; Plesken, Wilhelm, Perfect groups. (1989). ZBL0691.20001.). There seem to be many $n$ for which there is a unique perfect group of order $n$, but there are also many $n$ (e.g. $2^{14} \cdot 60$) for which the number of perfect groups is very large. There are precisely five $n < 10^6$ for which there are both simple and non-simple perfect $G$ of order $n$:

  • 20160: $A_8$, $L_3(4)$, $2.(A_5 \times L_3(2))$
  • 181440: $A_9$, $A_6 \times L_2(8)$, $L_3(2) \times 3.A_6$
  • 262080: $L_2(64)$, $2^2.(A_5 \times L_2(13))$
  • 443520: $M_{22}$, $2^2.(L_3(2)\times L_2(11))$
  • 604800: $J_2$, $A_5^2 \times L_3(2)$, $2^2.(A_5 \times A_7)$
  • $\begingroup$ $|S_m|$ being the order of the Schur cover of $A_m$, obviously. $\endgroup$ Commented Jan 14, 2020 at 15:56
  • 3
    $\begingroup$ Is it obvious ( or at least known) when we can have $|S|^{a} = |T|^{b}$ for non-Abelian simple groups $S$ and $T$ and positive integers $a,b$? $\endgroup$ Commented Jan 14, 2020 at 16:04
  • 2
    $\begingroup$ @GeoffRobinson I don't know about that, but if you look at just $\mathrm{PSL}_2(q)$ for small $q$ ($q \leq 17$ is enough) then eventually you have more groups than prime divisors, so you will have many coincidences between direct products of these groups. I'm more interested in the other direction: can you be sure of some $n$, say $n = p(p-1)(p+1)$, that there is only one perfect group of order $n$? $\endgroup$ Commented Jan 14, 2020 at 16:36
  • 2
    $\begingroup$ @GeoffRobinson Your question about $|S|^a=|T|^b$ is known -- it only happens for $a=b$. Reference: W. Kimmerle et al., Proc. London Math. Soc. (3) 60 (1990), no. 1, 89–122. $\endgroup$ Commented Jan 14, 2020 at 22:02
  • $\begingroup$ @RichardLyons :Thanks for that reference Richard. $\endgroup$ Commented Jan 14, 2020 at 22:04

1 Answer 1


For the case $n = p(p-1)(p+1)$ for $p >3$ a prime, a theorem of M. Herzog (to be found in the article "Finite groups with a large cyclic Sylow subgroup" (MSN) in the conference proceedings "Finite Simple Groups" (Oxford, editors M. Powell and G. Higman, published 1971 by Academic Press (MSN))) answers your question postively. The Theorem of Herzog uses a 1958 Theorem of Brauer and Reynolds ("On a problem of E. Artin", Annals of Mathematics, 68 (MSN)) which probably suffices to cover the cases you are interested in.

Note that if $n = p(p-1)(p+1)$ with $p >3$ a prime, and $G$ is a perfect group of order $n$, then $G$ is not $p$-solvable (for if $G$ is $p$-solvable of order $n$, then $G$ either has a factor group of order $p$ or a non-trivial cyclic factor group of order dividing $p-1$)-(later edit: more generally, if $G$ is any $p$-solvable group with a Sylow $p$-subgroup of order $p$, then $G$ is not perfect).

The Theorem of Herzog/Brauer–Reynolds proves that a finite group $G$ of order divisible by $p$, but of order less than $p^{3}$ (for a prime $p >3$) which is not $p$-solvable is one of the following groups: $\operatorname{PSL}(2,p)$, $\operatorname{PSL}(2,p-1)$ (for $p >5$ a Fermat prime), $\operatorname{SL}(2,p)$, $\operatorname{PGL}(2,p)$, $\operatorname{PSL}(2,p) \times \mathbb{Z}/2\mathbb{Z}$.

The only perfect group of order $n$ on the list is ${\rm SL}(2.p)$ and the only perfect group of order $\frac{n}{2}$ is ${\rm PSL}(2,p)$, showing that ${\rm SL}(2,p)$ and ${\rm PSL}(2,p)$ ( for $p >3$ a prime) are the unique perfect groups of their orders.

  • $\begingroup$ Nice, thank you! $\endgroup$ Commented Jan 14, 2020 at 19:17
  • $\begingroup$ Details of second paragraph (please correct me if you had something simpler in mind): If $G$ is $p$-solvable and has order $n$ then some quotient $G/N$ has a normal subgroup of order $p$, and this extension is split by Schur--Zassenhaus. Your two cases are whether this split product is direct or not. $\endgroup$ Commented Jan 16, 2020 at 11:25
  • $\begingroup$ I think that works, but the argument I had in mind was the following: G/Op′(G) has a self centralizing normal Sylow p-subgroup of order p (eg by Hall-Higman centralizer Lemma), so is isomorphic o a subgroup of order divisible by $p$ of a Frobenius group of order p(p−1), so is itself either a Frobenius group or cyclic of order $p$. If cyclic of order $p$, then G has a factor group of order p, while if the Frobenius complement is non-trivial, it is a cyclic subgroup of order dividing p−1 which is a homomorphic image of G. $\endgroup$ Commented Jan 18, 2020 at 9:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.