7

How can I prove that there are no perfect groups of order $3024$?

My attempt is the following:

Each non-trivial finite perfect group admits a non-abelian simple quotient.

This holds because if the group $G$ is simple, we have done, otherwise has to admit a normal subgroup $K$. However $G/K$ is a non-trivial perfect group and or it’s simple or admits a normal subgroup $H$. Then $K_1:=\pi^{-1}(H)$ is a normal subgroup of $G$ containing $K$, where $\pi \colon G\to G/K$ is the projection map. Moreover the order of $K_1$ is strictly greater than the order of $K$. Thus in a finite number of steps we get a normal subgroup of $G$ such that the quotient has to be simple and not-abelian.

The only not-abelian simple groups dividing $3024$ are $PSL(2,\mathbb{F}_7)$ and $PSL(2, \mathbb{F}_8)$.

In the second case we have that the group $G$ of order $3024$ is an extension of $PSL(2, \mathbb{F}_8)$ by a kernel $K$ of order $6=3\cdot 2$. The kernel has to admit a $3-$ Sylow normal subgroup $S$, that is characteristic in $K$, and so normal in $G$. So $G/S$ is a perfect group of order $3024/3=1008$ but there are not perfect groups of order $1008$.

The second case is quite difficult. Suppose that the group $G$ of order $3024$ is an extension of $PSL(2, \mathbb{F}_7)$ by a kernel $K$ of order $18=3^2\cdot 2$. The kernel has to admit a $3-$ Sylow normal subgroup $S$, that is characteristic in $K$, and so normal in $G$. So $G/S$ is a perfect group of order $3024/9=336$. The only perfect group of order $336$ is $SL(2, \mathbb{F}_7)$. Thus seems that there is not a contradiction.

However I’m sure that there not exist perfect groups of order $3024$, so someone of you does know some hint that can help me to solve this problem?

Shaun
  • 47,747
Federico Fallucca
  • 10,123
  • 4
    There is indeed no perfect group of order $3024$, see OEIS. – Dietrich Burde Apr 04 '22 at 16:38
  • 3
    See also the answer to this question. – Dietrich Burde Apr 04 '22 at 16:43
  • @DietrichBurde Can you explain better how can I use that question? – Federico Fallucca Apr 04 '22 at 17:21
  • 1
    Considering the case of $K\lhd G$, $G/K\simeq PSL(2,\Bbb{F}_7)$. If $P$ is one of the $8$ Sylow $7$-subgroups of $G/K$, then its preimage $H\le G$ has order $2\cdot3^2\cdot7$ and hence a single Sylow $7$-subgroup $P'$. It follows that $G$ also has $8$ Sylow $7$-subgroups. Furthermore, both $K$ and $P'$ are normal in $H$ and intersect trivially so must centralizer each other. It follows that the centralizer of $K$ projects surjectively onto all of $G/K$, as the Sylow $7$-subgroups of the latter generate it all. – Jyrki Lahtonen Apr 04 '22 at 20:19
  • 1
    (cont'd) I don't know what all of that says about $G$ for I am mostly ignorant about group theory. Burnside transfer and all that is a bit beoynd me. It does feel like the observation in the previous comment may allow you to use a known result. – Jyrki Lahtonen Apr 04 '22 at 20:28
  • 4
    How do you know that there is no perfect group of order $1008$? And how do you know that ${\rm SL}(2,7)$ is the only perfect group of order $336$? Assuming those two facts, you can proceed by showing that $S \le Z(G)$, and hence $G$ has a quotient of order $1008$. – Derek Holt Apr 05 '22 at 07:25
  • @DerekHolt I know that one because MAGMA said me . And I think that MAGMA works using maybe your book https://mathscinet.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=TI&pg6=PC&pg7=ALLF&pg8=ET&review_format=html&s4=Holt&s5=Perfect%20Groups&s6=&s7=&s8=All&sort=Newest&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq&r=1&mx-pid=1025760 – Federico Fallucca Apr 05 '22 at 10:53
  • @DerekHolt why I proved what I want if will I show that $S\leq Z(G)$ ? In that case why $G$ has a quotient of order $1008$. I’m sorry but I don’t know group theory very well – Federico Fallucca Apr 05 '22 at 10:57
  • 2
    You didn't answer my question. How do you know that there is no perfect group of order $1008$? It seems to me that you need to prove that as well. Proving that $S \le Z(G)$ requires some work. Once you have proved that, it follows immediately that $G$ has a normal subgroup of order $3$ contained in $S$, so it has a quotient of order $1008$. – Derek Holt Apr 05 '22 at 11:18
  • @DerekHolt I answered, I asked to MAGMA to tell me what are the perfect groups of order $1008$ and it told me there weren’t. I’m sure that in the book that I mentioned there is the reason why. If you tell me that it’s not so difficult to prove it, I can try to do it by my hands – Federico Fallucca Apr 05 '22 at 11:51
  • 1
    The quotient $G/C_G(K)$ embeds into $\operatorname{Aut}(K)$ which is solvable if $|K| = 6$ or $|K| = 18$. So if $G$ is perfect it follows that $K \leq Z(G)$. Thus $G$ is a perfect central extension of $PSL(2,7)$ or $PSL(2,8)$. Looking at the Schur multipliers, you know that for $q \neq 4,9$ the only nontrivial perfect central extension of $PSL(2,q)$ is $SL(2,q)$, so you have a contradiction. – spin Apr 05 '22 at 14:33
  • 4
    Magma can also tell you immediately that there is no perfect group of order $3024$ (typing $\mathtt{NumberOfGroups(PerfectGroupDatabase(),3024);}$ gives $0$). That just means that somebody else has proved it (in fact perfect groups have now been classified by Alexander Hulpke up to order two million). – Derek Holt Apr 05 '22 at 17:00
  • 2
    I think the discussion is caused in part by the fact that it is unclear what results you assume at the start. The simple groups of order dividing 3024? Their irreducible modules in characterstic $p$? The perfect groups of order $\le 3023$? Also note that "classification" results often don't have a nice proof one would want to follow by hand, but the results only follow from trying out all possibilities (and ultimately finding that nothing works out). – ahulpke Apr 05 '22 at 23:27
  • @DerekHolt Yeah someone else was proved that there are no perfect groups of order 1008 and that the only perfect group of order 336 is SL(2,7). Thus I’m assuming this two results and I would to prove that there is no a perfect group of order 3024 – Federico Fallucca Apr 05 '22 at 23:31

0 Answers0