Talk:Frobenius group
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[edit] Examples
How exactly (i.e. which kind of set and action) is the Frobenius structure realized in the case of 2 × 2 upper triangular matrices in the case of an arbitrary ring R (the third example)? One might be tempted to represent this group by affine transformations a*x+b, where a is invertible in the ring. But the fixed point equation in this case ax+b=x is equivalent to (a-1)x= -b which could have more than 1 solutions if (a-1) is not invertible...(which might happen). --212.18.24.11 16:59, 15 Apr 2005 (UTC)
This was an error and has been corrected. —The preceding unsigned comment was added by R.e.b. (talk • contribs) 20:08, 15 April 2005 (UTC)
[edit] Anon question
Did Zassenheus really prove that if the Frobenius complement contains a generalized quaternion subgroup that the stucture is like the one given? I have found this result if one assumes that the Frobenius complement is non-solvable in which case the sylow 2-subgroups would be generalized quarternion but the statement seems to apply that for a solvable Frobenius group, the complement is metacyclic (as all Sylow groups are cyclic). Is this true?
67.184.176.21 07:17, 8 March 2006 (UTC)
So based on the this link http://verdi.algebra.unilinz.ac.at/Projects/FWF/P15691/antrag/node4.html, it appears that there is a mistake as suspected. They have given an example at the very bottom of the page of a Frobenius group with abelian Frobenius kernel and quaternion Frobenius complement so Zassenheus didn't show what is claimed on the page.
67.184.176.21 17:35, 8 March 2006 (UTC)
I'm not sure if I understand the statement in the examples of Frobenius groups with regards to using a non-trivial subgroup of the Frobenius kernel to construct another Frobenius group. Do you mean replacing the Frobenius kernel of an existing Frobebius group with any subgroup and keeping the complement the same? I know that if the order of the kernel is k and that of the compleement is h that h divides k-1, but this would mean you could take a cyclic subgroup of the kernel of prime order and then h would have to divide p-1 for every prime dividing the order of the kernel. Is this true?
67.184.176.21 18:54, 9 March 2006 (UTC)
These were errors and have been fixed. R.e.b. 21:05, 9 March 2006 (UTC)
