[GAP Forum] Frobenius group

[Holt, Derek]

Dear Fatemeh, dear Forum,

I would suggest the following approach, which I think will work, although there might be other ways of doing it.

Consider each integer n from 1 to 1000 in turn. For each such n find all expressions of n as a product n = dh, where d,h > 1 and h divides d-1.

There could be more than such expression e.g. 42 = 21 x 2 = 7 x 6.

You can discard all cases such as 30 = 6 x 5, where d is twice an odd number because a Frobenius kernel cannot have twice odd order.

Now, if n has one or more expressions as dh then using the small groups library in GAP, consider each group G of order n in turn, and for each such G consider all possible factorisations n = dh.

First check that G has a normal nilpotent subgroup K of order d. If it
does not, then you proceed to the next group or the next factorisation.
To do that, you could find all Sylow p-subgroups for p dividing d, and check that they are all normal in G and, if so, take K to be the subgroup generated by these Sylow subgroups. K is a candidate for the Frobenius kernel of G.

Before going further you could use the fact that if h is even then K must be abelian and if h is divisible by 3 then K must have nilpotency class at most 2. If not, you can rule out this case.

Assuming K exists, it must have a unique conjugacy class of subgroups H of order h. There is a function in GAP to find complements, which you could use to find H, but in many cases h will be a prime power, in which case you can just take a Sylow subgroup. H is a candidate for the Frobenius complement. 

Finally, you need to check whether H really is a Frobenius complement.
To do that, you could compute the permutation action on the cosets of H and see if that is a Frobenius group. Another way, which might be quicker, is to find the subgroups P of H of prime order. If H is a Frobenius complement, then there is a single conjugacy class of such subgroups for each prime dividing |H|. If so, then check that C_K(P) = 1 for all such P. If so, then G = KH is a Frobenius group.

I would guess that n = 768 with k = 256, h = 3, might be the hardest case, because there are over a million groups of order 768, but it should still work if you are patient.

I hope this helps.
Derek Holt.

On Sat, Sep 10, 2016 at 12:49:58PM +0430, fatemeh moftakhar wrote:
> Dear forum
> I need the list of all Frobenius groups up to order 1000. Is there any
> command or package in GAP to do this?
>
> Best regards
> Fatemeh Moftakhar
>
> --
> Regards;
> Miss Fatemeh Moftakhar
> PhD Candidate,
> Department of Pure Mathematics,
> Faculty of Mathematical Sciences,
> University of Kashan, Kashan, Iran
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