a galois related problem
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From: hagman on 14 Feb 2010 17:42 On 13 Feb., 17:57, "victor_meldrew_...(a)yahoo.co.uk" <victor_meldrew_...(a)yahoo.co.uk> wrote: > On 13 Feb, 12:15, Timothy Murphy <gayle...(a)eircom.net> wrote: > > > Steve Dalton wrote: > > > A simpler argument for (odd) prime n: > > > > D_n in S_n is a subgroup generated by an n-cycle and a non-centralizing, > > > normalizing involution. Such an involution can fix at most one point. > > > What exactly do you mean by "D_n in S_n"? > > There is a natural way of embedding D_n in S_n, > > but it is not obvious that this is the way the Galois group in question > > is embedded in S_n. > > When n is composite, there may be other ways of embedding D_n > in S_n. For example it's possible to embed D_15 in S_8. However > for this Galois-theoretical application, one needs an emebedding > of D_n as a *transitive* subgroup of S_n, and I think that > this can only be done in the "standard" way. Probably a question with a simple counterexample: Is the following true in general: CONJECTURE: Let G be the Galois group of an irreducible rational P polynomial of degree n. Let A be the set of complex roots of P and let B be an n-element set such that G acts transitively on B. Then there exist a bijection f:A->B compatible with the actions, i.e. g(f(a)) = f(g(a)) for all g in G, a in A. ?
From: Chip Eastham on 14 Feb 2010 17:48 On Feb 14, 12:48 pm, Chip Eastham <hardm...(a)gmail.com> wrote: > On Feb 14, 9:12 am, Kent Holing <K...(a)statoil.com> wrote: > > > A follow-up question: > > Of all transitive subgroups of Sn and of order 2n; does it exist one such not isomorphic to Dn? > > If n is an odd prime, the only groups of order 2n are > cyclic or dihedral (can be deduced from Sylow theory). > Also if n is prime, Sym(n) does not contain a cyclic > group of order 2n (by disjoint cycle representation > the order of a permutation in Sym(n) is either n or > coprime to n). Thus for odd prime n we know there > are no subgroups of Sym(n) of order 2n not isomorphic > to Dih(n), even without invoking transitivity. [But a > posteriori these are transitive by virtue of having a > cycle of length n. Cf. Beachy/Blair, AAOL Lemma 8.6.4] > > http://www.math.niu.edu/~beachy/aaol/galois.html#compute > > Also: For n = 4 the only subgroups of Sym(4) of order > 8 are dihedral (again without invoking transitivity) > and these are the three Sylow 2-subgroups of Sym(4), > and all of them are transitive. I should correct my last statement: > For n = 6 the only subgroups of Sym(6) of order 12 > are dihedral (but not all of them are transitive). Not all subgroups of order 12 in Sym(6) are dihedral, e.g. there are copies of Alt(4) in there, but all the transitive subgroups of order 12 are dihedral. I was trying to work in the point we saw earlier in the discussion that not all the the copies of Dih(6) in Sym(6) are transitive. regards, chip
From: Steve Dalton on 14 Feb 2010 09:16 > On 13 Feb., 17:57, "victor_meldrew_...(a)yahoo.co.uk" > <victor_meldrew_...(a)yahoo.co.uk> wrote: > > On 13 Feb, 12:15, Timothy Murphy > <gayle...(a)eircom.net> wrote: > > > > > Steve Dalton wrote: > > > > A simpler argument for (odd) prime n: > > > > > > D_n in S_n is a subgroup generated by an > n-cycle and a non-centralizing, > > > > normalizing involution. Such an involution can > fix at most one point. > > > > > What exactly do you mean by "D_n in S_n"? > > > There is a natural way of embedding D_n in S_n, > > > but it is not obvious that this is the way the > Galois group in question > > > is embedded in S_n. > > > > When n is composite, there may be other ways of > embedding D_n > > in S_n. For example it's possible to embed D_15 in > S_8. However > > for this Galois-theoretical application, one needs > an emebedding > > of D_n as a *transitive* subgroup of S_n, and I > think that > > this can only be done in the "standard" way. > > Probably a question with a simple counterexample: > Is the following true in general: > > CONJECTURE: > Let G be the Galois group of an irreducible rational > P polynomial of > degree n. > Let A be the set of complex roots of P and let B be > an n-element set > such that > G acts transitively on B. > Then there exist a bijection f:A->B compatible with > the actions, i.e. > g(f(a)) = f(g(a)) for all g in G, a in A. > > ? Well there certainly exist isomorphic transitive subgroups of S_n which are not conjugate in S_n. This is essentially your question*. For example, G = <(1,4)(2,3,5,6), (1,2,3)(4,5,6)> and H = <(2,3,5,6), (1,2,3)(4,5,6)> are both transitive subgroups of S_6, and G ~ H ~ S_4, but they are not conjugate in S_6. * I am ignoring the fact that we are talking about Galois groups. I am uncertain if that makes a difference or not. In the above, I am "prescribing" the action of G and H by choosing embeddings into S_6, but perhaps every sextic polynomial with Galois group S_4 only has one type of action on the roots. I don't know. Steve
From: Derek Holt on 15 Feb 2010 08:44 On 15 Feb, 00:16, Steve Dalton <sdal...(a)math.sunysb.edu> wrote: > > On 13 Feb., 17:57, "victor_meldrew_...(a)yahoo.co.uk" > > <victor_meldrew_...(a)yahoo.co.uk> wrote: > > > On 13 Feb, 12:15, Timothy Murphy > > <gayle...(a)eircom.net> wrote: > > > > > Steve Dalton wrote: > > > > > A simpler argument for (odd) prime n: > > > > > > D_n in S_n is a subgroup generated by an > > n-cycle and a non-centralizing, > > > > > normalizing involution. Such an involution can > > fix at most one point. > > > > > What exactly do you mean by "D_n in S_n"? > > > > There is a natural way of embedding D_n in S_n, > > > > but it is not obvious that this is the way the > > Galois group in question > > > > is embedded in S_n. > > > > When n is composite, there may be other ways of > > embedding D_n > > > in S_n. For example it's possible to embed D_15 in > > S_8. However > > > for this Galois-theoretical application, one needs > > an emebedding > > > of D_n as a *transitive* subgroup of S_n, and I > > think that > > > this can only be done in the "standard" way. > > > Probably a question with a simple counterexample: > > Is the following true in general: > > > CONJECTURE: > > Let G be the Galois group of an irreducible rational > > P polynomial of > > degree n. > > Let A be the set of complex roots of P and let B be > > an n-element set > > such that > > G acts transitively on B. > > Then there exist a bijection f:A->B compatible with > > the actions, i.e. > > g(f(a)) = f(g(a)) for all g in G, a in A. > > > ? > > Well there certainly exist isomorphic transitive subgroups of S_n which are not conjugate in S_n. This is essentially your question*. > For example, > G = <(1,4)(2,3,5,6), (1,2,3)(4,5,6)> and > H = <(2,3,5,6), (1,2,3)(4,5,6)> > are both transitive subgroups of S_6, and G ~ H ~ S_4, but they are not conjugate in S_6. > > * I am ignoring the fact that we are talking about Galois groups. I am uncertain if that makes a difference or not. > In the above, I am "prescribing" the action of G and H by choosing embeddings into S_6, but perhaps every sextic polynomial with Galois group S_4 only has one type of action on the roots. I don't know. > If S_4 (or any finite group G) occurs as a Galois group over Q at all (and S_4 does of course), then any transitive permutation representation of G occurs as a Galois group over Q, so your two examples must both arise as Galois groups. To see this, suppose K is a Galois extension of Q with |K:Q|=|G| and Galois group G. The transitive permutations of G are equivalent to the coset actions on subgroups of G and are in 1-1 correspondence with the conjugacy classes of subgroups of G. Let H be any subgroup of G, and let L be the fixed field of that subgroup under that action of G on K, so |L:Q|=|G:H|. There exists a in L with Q(a)=L. Let f be the minimal polynomial of a over Q. Then the Galois group of f acting on the roots of f is permutation equivalent (and hence conjugate in S_|L:Q|) to the coset action of G on the cosets of H. Derek Holt.
From: Kent Holing on 16 Feb 2010 05:11
Can anybody give an example of equation of degree n with a Galois group of order 2n but not Dn? |