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        riddles >> putnam exam (pure math) >> John Thompson's Lemma Revisited
        
(Message started by: ecoist on Aug 19^th, 2006, 6:57pm)

Title: John Thompson's Lemma Revisited
Post by ecoist on Aug 19^th, 2006, 6:57pm

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.

Title: Re: John Thompson's Lemma Revisited
Post by THUDandBLUNDER on Aug 19^th, 2006, 7:27pm

Perhaps this should be in Putnam.

Title: Re: John Thompson's Lemma Revisited
Post by ecoist on Aug 19^th, 2006, 8:26pm

Ok with me if the problem is moved to Putnam. Assumed that the Putnam thread was for more difficult problems.

Title: Re: John Thompson's Lemma Revisited
Post by THUDandBLUNDER on Aug 19^th, 2006, 10:12pm

In my opinion, Easy problems ought to be easily understood by the not-so-mathematically-minded.

Title: Re: John Thompson's Lemma Revisited
Post by Michael_Dagg on Aug 21^st, 2006, 4:48pm

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.

Title: Re: John Thompson's Lemma Revisited
Post by Michael_Dagg on Nov 1^st, 2006, 7:09pm

You have to watch ecoist as he likes intersections within which and onto [which] are interesting!

Title: Re: John Thompson's Lemma Revisited
Post by ecoist on May 14^th, 2007, 5:29pm

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.

Title: Re: John Thompson's Lemma Revisited
Post by Eigenray on May 14^th, 2007, 11:57pm

I see. I was trying to generalize from the point of view of the transfer proof, which uses Ver:G http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif 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 = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/coprod.gifg_iH be a decomposition into [G:H]=mn left cosets. Left multiplication gives an action http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif: G http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif S_mn. Under this action, x^r 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 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/circ.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif is then normal of index 2.