2-group ordering sequences

Creighton Dement crowdog at crowdog.de
Sun Apr 3 00:47:43 CEST 2005


Sorry, a correction: 

The sentence  "The 2nd direction involves recording
what the greatest possible n is corresponding to each m(G)", below,
should read 
"...what the smallest possible n is corresponding to each m(G)"

There is also "third" direction which would generate integer sequences I
forgot to mention: to each m(G) such that for all x_1, ..., x_m in G,
(x_1,...,x_m)^n
orders itself to (1,...,1) for some n, the (strictly increasing)
sequence all such n.

Sincerely, 
Creighton 

- - - - - - - - - - - - - - - - - - - - - - - - - - - - -

> Date: Sun,  3 Apr 2005 00:28:24 +0200
> Subject: 2-group ordering sequences
> From: "Creighton Dement" <crowdog at crowdog.de>
> To: seqfan at ext.jussieu.fr

> Dear Seqfans,
> 
> I just had a little idea concerning what might return some nice
> sequences concerning 2-groups, etc. (apparently, every once in a blue
> moon, I am still able to think of something which is only indirectly
> connected with the floretions).
> 
> Let G be a 2-group (or other group which fits similar requirements -
> surely there are others) and let ^ be the operation described at
> http://www.crowdog.de/16401.html and let 1 be the unit.
> Note: that page is was written well over a year ago and, if I remember
> correctly, I still needed to go over it again to correct some mistakes
> (I think these had mostly to do with notation, but who knows...).
> 
> Then there are at least two separate directions which lead to integer
> sequences. One direction is finding m(G) (a natural number which
> depends on G) such that for all x_1, ..., x_m in G, (x_1,...,x_m)^n
> orders itself to (1,...,1) for some n.   I presume this number will be
> the same for a large number of groups- it may even provide some sort
> of
> classification scheme for them. The 2nd direction involves recording
> what the greatest possible n is corresponding to each m(G). It would
> be great if  "theory alone" led to these numbers (I doubt that will be
> easy)- until then, an exhaustive search should work to calculate at
> least the first few terms.
> 
> Sincerely,
> Creighton
> 
> 
> 
> 
> 
> 








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