Talk:Frobenius group

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[edit] Examples

How exactly (i.e. which kind of set and action) is the Frobenius structure realized in the case of 2 × 2 upper triangular matrices in the case of an arbitrary ring R (the third example)? One might be tempted to represent this group by affine transformations a*x+b, where a is invertible in the ring. But the fixed point equation in this case ax+b=x is equivalent to (a-1)x= -b which could have more than 1 solutions if (a-1) is not invertible...(which might happen). --212.18.24.11 16:59, 15 Apr 2005 (UTC)

This was an error and has been corrected. —The preceding unsigned comment was added by R.e.b. (talk• contribs) 20:08, 15 April 2005 (UTC)

Upper triangular matrices are Frobenius if thecharacteristic is 2. Examples:

 TL(2,q) for q=3,5,7,9,11,13,17,19,23,25,27,...
 have complement subgroups which are their own
 normalizers but all contain the non trivial
 center of TL(2,q), and so are not Frobenius.
   for q=4,8,16,32, ... this center is trivial.

--213.224.110.147 07:25, 4 July 2007

I think the matrix example should just be removed. The previous comment points out that if the group has a nontrivial center, then it is not a Frobenius group. In fact, for all q, TL(2,q) has a center of size q-1, since [z,0;0,z] commutes with [a,b;0,c]. While the 2x2 upper triangular matrices modulo their center do form a Frobenius group, they just form the same group mentioned right before them, AGL(1,q). JackSchmidt 16:28, 4 July 2007 (UTC)

The action of AGL(1,q) as a transformation group is not on GF(q) but on the right cosets of the complement subgroup, x -> ax, a<>0 by right multiplication. 213.224.110.136 07:31, 5 July 2007

The action of AGL(1,q) is on GF(q), which, by the orbit-stabilizer theorem, is in natural correspondence with the right cosets Hb={ x -> ax+b : a in GF(q), a <> 0 } of the complement subgroup H={ x -> ax : a <> 0 }. Your edit on the main article is good, but so was the original text. JackSchmidt 12:20, 5 July 2007 (UTC)

I have objections against the term 'natural action on GF(q)' in the old text. There is no direct (geometrical) way to define this group as a permutation group on GF(q). You have to pass via the cosets and a bijection of these cosets onto GF(q) and there are q! of these. Which is the natural one? That depends on one's taste. You give (x -> ax+b) -> b. Why not (x -> ax+b) -> b+1? In any case you have to pass by the cosets of H, so why not stop there?

By the way there is a natural way to define these Frobenius groups by their action on GF(q) as follows: They are 1-point stabilizers (fixing the point at infinity) of the image PSL(2,q) of SL(2,q) in PGL(2,q) acting on the projective line PG(1,q) with q+1 points by the projective transformations

  x -> (ax+b)/(cx+d) a,b,c,d in GF(q) and
         ad-bc a square <> 0 in GF(q)

Here the complements(consisting of the elements fixing the point at infinity and one other point in PG(1,q)) are cyclic of order e(q-1) with e=1 for q even, 1/2 for q odd; and the kernel(consisting of the unity and the elements fixing the point at infinity and regular on the other points) is elementary abelian of order q and disjoint from its conjugates in PSL(2,q). 213.224.108.123 06:23, 6 July 2007

I think the action is natural, but only because of the specific way the group elements themselves are defined. The way the group is presented is explicitly as permutations on GF(q). Each element is explicitly a bijective function from GF(q) to GF(q) given by the rule x -> a*x + b. In the PGL(2,q) terms, c=0,d=1.

In other words, I think we agree. When the elements are functions, the action is natural. When the elements are not, then there is choice.

The stabilizer of a point in PGL(2,q) is isomorphic to AGL(1,q), and the stabilizer of a point in PSL(2,q) is isomorphic to "ASL(1,q)" which is only half as big when q is odd. Both AGL(1,q) and ASL(1,q) are usually Frobenius groups. For q=2, AGL(1,q) = ASL(1,q) = { x -> x, x->x+1 } is regular. For q=3, ASL(1,q) = { x->x, x->x+1, x->x+2 } is regular. For q>2, AGL(1,q) is a Frobenius group, and for q>3, ASL(1,q) is a Frobenius group, and for q>3 odd, ASL(1,q) is even a Frobenius group which is distinct from AGL(1,q).

Should we add ASL(1,q) to the examples?

As a reference, we can use Dixon&Mortimer's text on Permutation Groups, section 2.8. It has the nice description you gave of AGL <= PGL, for instance. In fact, later in section 3.4, it generalizes both AGL and ASL as example 3.4.1:

"Let U denote a [nonidentity] subgroup of the group of units of a field F. Then the set G consisting of all permutations of F of the form x -> ax + b with a in U and b in F is a Frobenius group where the point stabilizer of 0 [in F] is { x -> ax : a in U } =~ U. The elements which fix no points, together with the identity, are the translations { x -> x+b : b in F }. These translations constitute a normal subgroup K =~ (F,+) of G."

JackSchmidt 14:03, 6 July 2007 (UTC)

I begin to understand your insistence in calling x-> ax+b natural actions on GF(q). Frobenius groups are special cases of transitive permutation groups. Transitive permutation representations are always generated by the action of the group on the cosets of a subgroup, but only rarely such a permutation representation is a Frobenius group. Such cases are the AGL(1,q) groups acting on the cosets of a (Frobenius) complement. But this action is not permutation equivalent with the natural action of these groups on GF(q) as affine groups if q is not a prime. The latter are not even permutation representations if q is not a prime (try q=4 x->2x: 0 and 2 are both transformed in 0). So with this natural action on GF(q) the AGL(1,q) groups are not Frobenius groups if q is not a prime. The natural action of these groups considered as 1-point stabilizers in PSL(2,q) acting on the projective line PG(1,q) is a permutation representation of the AGL(1,q) that is permutation equivalent with their action on the cosets of H. So these actions may be called natural actions of Frobenius groups on GF(q), not the actions as affine transformations. 213.224.111.158 09:08, 7 July 2007

In GF(q) with q=4, 0=2 are the same element, and x->2x is the same as x->0, which is not an element of AGL(1,q). GF(4) is the field with 4 elements, not Z/4Z.

The action of AGL(1,q) as a subgroup of PGL(2,q) acting in your geometrical way, and the action of AGL(1,q) are identical. I am confused that you seem to say they are different. In PGL(2,q), we have the functions on PG(1,q) = GF(q) union {oo} which take x in PG(1,q) to (a*x+b)/(c*x+d), for arbitrary elements a,b,c,d in GF(q) with ad-bc nonzero. These functions take oo to "a/c" where this is in GF(q) if c is nonzero, and is oo if c is zero. In other words, the stabilizer of the point oo is precisely those functions which have c=0, and so WLOG d=1. In other words, the stabilizer of the point oo is precisely those functions x -> a*x + b.

Here is a tiny example (too tiny for AGL(1,q) be a Frobenius group) to show what I mean: PGL(2,2) has six elements, namely { x -> x, x -> x+1, x -> 1/x, x -> 1/(x+1), x -> x/(x+1), x -> (x+1)/x }. These six elements are functions on PG(1,q) = { 0,1,oo }. As permutations in cycle notation with trivial cycles shown, the six elements are { (0)(1)(oo), (0,1)(oo), (0,oo)(1), (0,1,oo), (0)(1,oo), (0,oo,1) }. The stabilizer of oo is then just the first and second elements, since all other elements move oo. AGL(1,2) = { x -> x, x -> x + 1 } = { (0)(1), (0,1) }.

JackSchmidt 19:14, 7 July 2007 (UTC)

You are right that my argument was completely wrong. AGL(1,q) acting naturally on GF(q) as affine transformations define Frobenius groups. This action is permutation equivalent with the action on the cosets of a complement. Thanks for the lesson. 213.224.111.158 11:10, 8 July 2007

I think your comments about PGL could be a good addition to the article. Also, as you pointed out, another Frobenius group is { x -> a*x + b | a,b in GF(q), a is a nonzero square }.

I think PGL(2,q) is a fairly important group for lots of reasons, and the fact that its point stabilizer has such a nice structure is certainly interesting.

One idea would be to bring in the material from chapter XI on Zassenhaus groups from Huppert and Blackburn's text, Finite Groups III. If a permutation group has the property that for every two triples of distinct points there is one and only one group element taking one to the other, then the group is called sharply 3-transitive, and the group must be PGL(2,q) or this slightly weird subgroup of PGammaL(2,q) called M(q) in the book and PSL(2,q).2_3 in the ATLAS, at least when q=p^2.

I don't know exactly the best way to bring these in, but surely PGL(2,q) and PSL(2,q) are important enough to be mentioned somewhere as naturally containing one of the main examples of Frobenius groups.

JackSchmidt 14:41, 9 July 2007 (UTC)

[edit] Anon question

Did Zassenheus really prove that if the Frobenius complement contains a generalized quaternion subgroup that the stucture is like the one given? I have found this result if one assumes that the Frobenius complement is non-solvable in which case the sylow 2-subgroups would be generalized quarternion but the statement seems to apply that for a solvable Frobenius group, the complement is metacyclic (as all Sylow groups are cyclic). Is this true?

67.184.176.21 07:17, 8 March 2006 (UTC)

So based on the this link http://verdi.algebra.unilinz.ac.at/Projects/FWF/P15691/antrag/node4.html, it appears that there is a mistake as suspected. They have given an example at the very bottom of the page of a Frobenius group with abelian Frobenius kernel and quaternion Frobenius complement so Zassenheus didn't show what is claimed on the page.

67.184.176.21 17:35, 8 March 2006 (UTC)

I'm not sure if I understand the statement in the examples of Frobenius groups with regards to using a non-trivial subgroup of the Frobenius kernel to construct another Frobenius group. Do you mean replacing the Frobenius kernel of an existing Frobebius group with any subgroup and keeping the complement the same? I know that if the order of the kernel is k and that of the compleement is h that h divides k-1, but this would mean you could take a cyclic subgroup of the kernel of prime order and then h would have to divide p-1 for every prime dividing the order of the kernel. Is this true?

67.184.176.21 18:54, 9 March 2006 (UTC)

These were errors and have been fixed. R.e.b. 21:05, 9 March 2006 (UTC)

[edit] Alternate definitions

Frobenius groups have a number of nice alternative definitions. It might good to add a few alternate definitions. However, before I did, I wanted to see if someone knew what happened here:

One particularly simple definition was reverted in the edit [1] by R.E.B. as being unintuitive and incorrect. While I will certainly agree the non-definition phrasing was awkward and the revert was needed, the definition was quite correct and is proposition 8.5.5 of Robinson's text on group theory and *definition* 7.1 in Isaac's text on character theory. The definition is intuitively interesting as it says that not only is the subgroup self-normalizing, but it is self-normalizing in a very strong way, H^x n H = 1 for x not in H.

A few other nice definitions are given in Isaacs's text on character theory. I think the definitions given in problem 7.1 are nice in that they show that in a semidirect product, being a Frobenius group is equivalent to either the kernel being a strong sort of self-centralizing, or the complement being a strong sort of self-centralizing.