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Topic: John Thompson's Lemma Revisited (Read 2189 times) 

ecoist
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John Thompson's Lemma Revisited
« on: Aug 19^{th}, 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 2subgroup S has the form S=CH, where C and H are subgroups of S with C cyclic of order greater than 1 and every Gconjugate of C has trivial intersection with H. Then G is not simple.

« Last Edit: Aug 19^{th}, 2006, 8:20pm by ecoist » 
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ThudnBlunder
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Re: John Thompson's Lemma Revisited
« Reply #1 on: Aug 19^{th}, 2006, 7:27pm » 
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Perhaps this should be in Putnam.


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ecoist
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Re: John Thompson's Lemma Revisited
« Reply #2 on: Aug 19^{th}, 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 19^{th}, 2006, 10:12pm » 
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In my opinion, Easy problems ought to be easily understood by the notsomathematicallyminded.

« Last Edit: May 15^{th}, 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 21^{st}, 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 2subgroup was of thisorthat 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.

« Last Edit: Aug 21^{st}, 2006, 6:45pm by Michael Dagg » 
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Michael Dagg
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Re: John Thompson's Lemma Revisited
« Reply #5 on: Nov 1^{st}, 2006, 7:09pm » 
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You have to watch ecoist as he likes intersections within which and onto [which] are interesting!

« Last Edit: Nov 1^{st}, 2006, 7:33pm by Michael Dagg » 
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ecoist
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Re: John Thompson's Lemma Revisited
« Reply #6 on: May 14^{th}, 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 2subgroup 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 14^{th}, 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 = g_{i}H be a decomposition into [G:H]=mn left cosets. Left multiplication gives an action : G S_{mn}. Under this action, x^{r} has a fixed point iff it lies in some conjugate of H; by assumption, this happens iff mr. It follows that x acts as a product of n disjoint mcycles; 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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