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I'm looking for a finite group $G$ which has the following properties (simultaneously):

a) $G$ has a $2$-subgroup $P$ such that $N_G(P)/P \cong A_6$, the alternaing group acting on $6$ points

b) $|G| > |N|$, where $N=N_G(P)$ from a)

c) the order of $G$ is not too large (roughly $1\ 000 < |G| < 30\ 000$).

I was thinking about the outer automorphism group of $A_6$, but it did not lead anywhere.

Thank you.

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1 Answer 1

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The smallest possiility for $N_G(P)$ is the group ${\rm SL}(2,9)$ (with $|P|=2$) of order $720$, so I think asking for $|G| < 30000$ might be optimistic.

There is an example of order $58320$ with structure $3^4:{\rm SL}(2,9)$. So it has a normal elementary abelian subgroup of order $3^4$ with complement ${\rm SL}(2,9)$.

You can access it in GAP with $\mathtt{PerfectGroup}(58320,2)$.

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  • $\begingroup$ Thank you. I am confused: isn't $|N_G(P)|=|G|$ in your example with the perfect group? Or maybe I did a miscalculation or understood something wrong. $\endgroup$
    – Stein Chen
    Commented Feb 26, 2022 at 16:11
  • $\begingroup$ No, as I said, $N_G(P)$ is a subgroup of order $720$ isomorphic to ${\rm SL}(2,9)$. It is a complement to the normal subgroup of $N$ of order $81$. Note that $P$ has order $2$ and the nontrivial element of $P$ inverts all elements of $N$. $\endgroup$
    – Derek Holt
    Commented Feb 26, 2022 at 18:03
  • $\begingroup$ Ok. I tried the following in GAP and it did not work. May I kindly ask, if you have an idea about where to start looking for the mistake? R:=PerfectGroup(58320,1); iso:=IsomorphismPermGroup(R); G:=Image(iso); LoadPackage("PERMUT"); p:=2; Syl:=SylowSubgroup(G,p); ccsSyl:=ConjugacyClassesSubgroups(Syl); H:=Representative(ccsSyl[2]); Order(H); N:=Normaliser(G,H); Order(N); and GAP claims that the order of $H$ is $2$, that all $2$-subgroups of $G$ which have order $2$ are conjugated in $G$ (I did not include the code here), and that the order of $G$ is equal to the order of $N$. $\endgroup$
    – Stein Chen
    Commented Feb 27, 2022 at 16:18
  • $\begingroup$ Oh dear I am sorry! I use both Magma and GAP and I assumed that they would use the same numbering for the perfect groups, but I was wrong. The group I am describing is $\mathtt{PerfectGroup}(58320,2)$ in GAP. I have corrected my answer. Try your calculation again with that change. $\endgroup$
    – Derek Holt
    Commented Feb 27, 2022 at 17:01
  • $\begingroup$ Thank you very much! Now everything works. $\endgroup$
    – Stein Chen
    Commented Feb 27, 2022 at 17:11

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