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Topic: Permutation groups where only 1 fixes two letters (Read 808 times) |
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ecoist
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Permutation groups where only 1 fixes two letters
« on: May 9th, 2007, 6:31pm » |
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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.)
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| « Last Edit: May 10th, 2007, 10:21am by ecoist » |
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Obob
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Re: Permutation groups where only 1 fixes two lett
« Reply #1 on: May 9th, 2007, 7:11pm » |
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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.
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| « Last Edit: May 10th, 2007, 10:37am by Obob » |
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ecoist
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Re: Permutation groups where only 1 fixes two lett
« Reply #2 on: May 9th, 2007, 11:06pm » |
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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.
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Obob
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Re: Permutation groups where only 1 fixes two lett
« Reply #3 on: May 10th, 2007, 8:24am » |
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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.
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ecoist
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Re: Permutation groups where only 1 fixes two lett
« Reply #5 on: May 10th, 2007, 10:25am » |
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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!
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Obob
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Re: Permutation groups where only 1 fixes two lett
« Reply #6 on: May 10th, 2007, 10:33am » |
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Unfortunately Amazon doesn't have search inside this book for Huppert's book, and it appears to be checked out at my library.
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