On Sunday, April 27, 2014 3:32:46 PM UTC-7, Dima Pasechnik wrote: > > In the first case one gets a transitive group on 8 points, and in the > second case on 4 points. > > Is this a bug? > The really (only?) useful answer is to get the Galois group as a permutation group on the roots of the original polynomial, i.e., a group acting on 4 points. Imagine computing the Galois group of a degree 8 polynomial and finding it's Sym(8) (the generic case and quite easily proved). Would you rather get Sym(8) acting on 8 points or a simply transitive permutation group on 40320 elements?
Note that if you have a permutation group (on any set!) then getting its simply transitive representation is quite easy: that's just the group acting on itself (as a set) by left (or right) composition. It's silly to do that explicitly. Try your example with sage: K.<a> = NumberField(x^8 + x - 1) to see how silly computing G1 is. Computing G2 is quite speedy. -- You received this message because you are subscribed to the Google Groups "sage-nt" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To post to this group, send an email to [email protected]. Visit this group at http://groups.google.com/group/sage-nt. For more options, visit https://groups.google.com/d/optout.
