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](https://math.stackexchange.com/questions/413021/are-there-any-distinct-finite-simple-groups-with-the-same-order) 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](https://zbmath.org/?q=an:0691.20001).). 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: $J\_2$, $A\_5^2 \times L\_3(2)$, $2^2.(A\_5 \times A\_7)$



When is there a unique perfect group of order $n$?
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](https://mathscinet.ams.org/mathscinet-getitem?mr=338163)) in the conference proceedings "Finite Simple Groups" (Oxford, editors M. Powell and G. Higman, published 1971 by Academic Press ([MSN](https://mathscinet.ams.org/mathscinet-getitem?mr=327886)))) answers your question postively. The Theorem of Herzog uses a 1958 Theorem of Brauer and Reynolds ("[On a problem of E. Artin](https://www.jstor.org/stable/1970164)", Annals of Mathematics, 68 ([MSN](https://mathscinet.ams.org/mathscinet-getitem?mr=100635))) 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.


