When is there a unique perfect group of order $n$?

Question

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: $J_2$, $A_5^2 \times L_3(2)$, $2^2.(A_5 \times A_7)$

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.