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        riddles >> putnam exam (pure math) >> Permutation groups where only 1 fixes two letters
        
(Message started by: ecoist on May 9^th, 2007, 6:31pm)

Title: Permutation groups where only 1 fixes two letters
Post by ecoist on May 9^th, 2007, 6:31pm

Let G be a finite group with a subgroup H such that H is its own normalizer in G and any two conjugates of H intersect trivially. Using character theory, it can be shown that the identity and all elements of G not in any conjugate of H form a subgroup N normal in G. Is there a proof of this that does not use character theory?

(Oops! Had left out "any conjugate of" in the original post.)

Title: Re: Permutation groups where only 1 fixes two lett
Post by Obob on May 9^th, 2007, 7:11pm

The subgroup N, shown to exist via character theory, is usually defined as the set of elements of G that act on the coset space G/H without fixed points. The theorem invoked to prove N is in fact a normal subgroup is called Frobenius' theorem, which states that for a transitive permutation group G on a set X, such that no element other than the identity has more than one fixed point, the set of fixed-point free elements together with the identity gives a normal subgroup of G. I know I have been told by Jon Alperin that no known proof of Frobenius' theorem without character theory is known.

Now suppose we have a proof that the Frobenius kernel N of any Frobenius group G is normal. (A Frobenius group is a group G with a subgroup H satisfying the hypotheses of the riddle; the Frobenius kernel is the set of fixed-point free elements of G for the action on G/H, together with the identity.) Let us show that this implies Frobenius' theorem is true.

We are given a group G acting transitively and faithfully on a set X. The G-set X is then isomorphic to the coset space G/H as a G-set, since any transitive G-set is isomorphic to such a G-set. Every element of G fixes at most one coset in G/H. Now the stabilizer of g_1 H is g_1 H g_1^{-1}, so if g_1 H g_1^{-1} intersects g_2 H g_2^{-1} nontrivially, there is some g fixing both g_1 H and g_2 H, whence g_1 H = g_2 H, and therefore g_1 H g_1^{-1} = g_2 H g_2^{-1}. Hence any two conjugates of H intersect trivially or are equal. If g is not in H, then the stabilizer of g H meets the stabilizer of H trivially, since any element of G fixing both is the identity. This implies H is self-normalizing. Thus G is a Frobenius group with Frobenius complement H, and N is a normal subgroup. Thus Frobenius' theorem holds.

In particular, a character-theory free proof of the statement about Frobenius groups implies a character-theory free proof of Frobenius' theorem.

Title: Re: Permutation groups where only 1 fixes two lett
Post by ecoist on May 9^th, 2007, 11:06pm

You are misstaken, Obob. The number of elements in the set N is |G|-(|H|-1)[G:H], where [G:H] is the index in G of H. This is because H has [G:H] conjugates and two distinct conjugates have only the identity in common. Hence |N|=[G:H]. The statement as given is equivalent to G being a permutation group in which only the identity fixes two letters.

If Alperin is right, you have answered my question, but I find it hard to believe that character theory is required for this result.

Title: Re: Permutation groups where only 1 fixes two lett
Post by Obob on May 10^th, 2007, 8:24am

Didn't you define N to be the complement of H, together with 1? I agree that, with the correct definition of N, |N|=[G:H], and in fact G is the semidirect product of H and N.

Title: Re: Permutation groups where only 1 fixes two lett
Post by Obob on May 10^th, 2007, 8:27am

I haven't had the time to read it, but this book review seems to support Alperin's position.

http://www.ams.org/bull/1999-36-04/S0273-0979-99-00789-2/S0273-0979-99-00789-2.pdf

Title: Re: Permutation groups where only 1 fixes two lett
Post by ecoist on May 10^th, 2007, 10:25am

Sorry, Obob. Just now saw the error in my post and corrected it.

I'd like to see those "partial proofs" in Huppert's book!

Title: Re: Permutation groups where only 1 fixes two lett
Post by Obob on May 10^th, 2007, 10:33am

Unfortunately Amazon doesn't have search inside this book for Huppert's book, and it appears to be checked out at my library.