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Topic: John Thompson's Lemma Revisited (Read 2199 times) |
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ecoist
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John Thompson's Lemma Revisited
« on: Aug 19th, 2006, 6:57pm » |
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Many years ago Zvonimir Janko reported to the group theory community that Thompson's Transfer Lemma is not a transfer lemma. He gave the proof I had found. That proof also works for the slight generalization:
Let G be a finite group of order greater than 2 whose sylow 2-subgroup S has the form S=CH, where C and H are subgroups of S with C cyclic of order greater than 1 and every G-conjugate of C has trivial intersection with H. Then G is not simple.
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| « Last Edit: Aug 19th, 2006, 8:20pm by ecoist » |
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ThudnBlunder
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Re: John Thompson's Lemma Revisited
« Reply #1 on: Aug 19th, 2006, 7:27pm » |
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Perhaps this should be in Putnam.
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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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ecoist
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Re: John Thompson's Lemma Revisited
« Reply #2 on: Aug 19th, 2006, 8:26pm » |
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Ok with me if the problem is moved to Putnam. Assumed that the Putnam thread was for more difficult problems.
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ThudnBlunder
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Re: John Thompson's Lemma Revisited
« Reply #3 on: Aug 19th, 2006, 10:12pm » |
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In my opinion, Easy problems ought to be easily understood by the not-so-mathematically-minded.
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| « Last Edit: May 15th, 2007, 3:22am by ThudnBlunder » |
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Michael Dagg
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Re: John Thompson's Lemma Revisited
« Reply #4 on: Aug 21st, 2006, 4:48pm » |
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Nice problem!
Mechanisms for the classification of finite simple groups attempted to try
to determine simple groups whose Sylow 2-subgroup was of this-or-that type.
Note that the dihedral case is just the kind of thing you have here, with, conjugation of C
nevertheless: |C|=2 and with H being the large cyclic subgroup of S.
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| « Last Edit: Aug 21st, 2006, 6:45pm by Michael Dagg » |
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Michael Dagg
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Michael Dagg
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Re: John Thompson's Lemma Revisited
« Reply #5 on: Nov 1st, 2006, 7:09pm » |
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You have to watch ecoist as he likes intersections within which and onto [which] are interesting!
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| « Last Edit: Nov 1st, 2006, 7:33pm by Michael Dagg » |
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Michael Dagg
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ecoist
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Re: John Thompson's Lemma Revisited
« Reply #6 on: May 14th, 2007, 5:29pm » |
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In response to M_D (and sneakily reminding all that no one has yet posted a solution), it is crucial that the conjugates of C intersect H trivially. For, if C is cyclic of order 4 and H is of order 2 not in the center of the Sylow 2-subgroup HC, it is not enough that H contain no conjugates of C. The simple group of order 168 is a counterexample.
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Eigenray
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Re: John Thompson's Lemma Revisited
« Reply #7 on: May 14th, 2007, 11:57pm » |
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I see. I was trying to generalize from the point of view of the transfer proof, which uses Ver:G  S/S' (|C|=2). But this is "not a transfer lemma": it generalizes the case |H|=1, which is a standard exercise.
Let C=<x> be cyclic of order m, and let n=[G:CH].
Let G =  giH be a decomposition into [G:H]=mn left cosets. Left multiplication gives an action  : G Smn. Under this action, xr has a fixed point iff it lies in some conjugate of H; by assumption, this happens iff m|r. It follows that x acts as a product of n disjoint m-cycles; since m is even and n is odd, this is an odd permutation. The kernel of sign  is then normal of index 2.
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