# How to find / construct a finite group with special properties?

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.

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 $$\mathtt{PerfectGroup}(58320,2)$$.
• 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. Commented Feb 26, 2022 at 16:11
• 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 $N$. Commented Feb 26, 2022 at 18:03
• 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 $N$. Commented Feb 27, 2022 at 16:18
• 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. Commented Feb 27, 2022 at 17:01