a galois related problem
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From: Chip Eastham on 13 Feb 2010 02:22 On Feb 12, 7:25 pm, Timothy Murphy <gayle...(a)eircom.net> wrote: > Chip Eastham wrote: > > Let's review the argument already advance. > > > The Galois group is transitive on the roots of an irreducible > > polynomial (you asked for coefficients to be rational, and for > > reasons we need not go into here, they irreducible polynomial > > may without loss of generality be taken to be monic with integer > > coefficients). > > > Now complex automorphism will be an element of this Galois group, > > and if there are 3 real roots, complex automorphism "fixes" just > > those three roots (and transposes the other two pairs). > > > Now consider the symmetries D7 of the regular septagon. If any > > three distinct vertices are fixed, so are the rest of them > > (e.g. there can be no rotation or reflection). So the case of > > exactly three real roots cannot arise if the Galois group is D7. > > The action of D_n on a regular n-gon gives an embedding of D_n in S_n. > But does every embedding of D_n in S_n arise in this way? Hi, Timothy: Well, let's try to argue this for D_7 to start with. D_7 contains a unique cyclic subgroup of order 7. Indeed the 14 elements of D_7 consist of the C_7 subgroup together with 7 elements of order 2 (in the context of symmetries of a regular septagon, or septagon if you prefer, the reflections thru a line from each vertex to the midpoint of the opposing edge). Fixing any element of order 7 imposes some order on the "base points" of S_7 where D_7 is embedded, i.e. the representation of such generator as WLOG (1 2 3 4 5 6 7), and we need to consider how the elements of order 2 would be similarly represented. Any element in S_7 of order 2 must be a product of disjoint transpositions. It cannot be a single transposition, as then together with our cycle of order 7 we would generate all of S_7 (not merely a copy of D_7). If the elements of order 2 consisted only of two disjoint transpositions, then (since the rotation of order 7 above is an even permutation) all of D_7 would be contained in the alternating group A_7. But this is not the case; A_7 does not contain a copy of D_7: http://for.mat.bham.ac.uk/atlas/html/A7.html#maxs http://www.projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.rmi/1228834296 Therefore at least one of the elements of order 2 must consist of 3 disjoint transpositions, and carrying out conjugations by the rotations yields that all elements of order 2 are of this form. More work is needed to check that the details of the 3 disjoint transpositions are as we expect from the permutation of a regular heptagon, but we have already done enough to rule out three points being fixed by any group element other than the identity. regards, chip
From: Steve Dalton on 12 Feb 2010 17:50 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. Steve
From: Timothy Murphy on 13 Feb 2010 07:15 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. Nb I think the result stated is probably true, but I'm not sure if the proof is valid, for the above reason. -- Timothy Murphy e-mail: gayleard /at/ eircom.net tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland
From: Steve Dalton on 12 Feb 2010 22:26 > 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. > > Nb I think the result stated is probably true, > but I'm not sure if the proof is valid, for the above > reason. > > > -- > Timothy Murphy > e-mail: gayleard /at/ eircom.net > tel: +353-86-2336090, +353-1-2842366 > s-mail: School of Mathematics, Trinity College, > Dublin 2, Ireland If you embed D_n in S_n, no matter how you do it, it must contain an n-cycle and a non-centralizing, normalizing involution (assuming n is odd prime). But then the involution fixes exactly one point, call it "1". Then the n-cycle corresponds to rotation of a regular n-gon, and the involution corresponds to reflection in a line passing through vertex "1". So it is acting on the n points as a dihedral group. So if you have an irreducible septic equation, and the Galois group is D_7, that D_7 is acting on the 7 roots exactly as a dihedral group would act on them; this is what Chip Eastham also showed. For contrast, consider < (123)(45678), (23)(48)(57) >, which is a copy of D_15 in S_15 which does *not* act as a dihedral group on the 15 points. Steve
From: Timothy Murphy on 13 Feb 2010 08:30
Chip Eastham wrote: >> > The Galois group is transitive on the roots of an irreducible >> > polynomial (you asked for coefficients to be rational, and for >> > reasons we need not go into here, they irreducible polynomial >> > may without loss of generality be taken to be monic with integer >> > coefficients). >> >> > Now complex automorphism will be an element of this Galois group, >> > and if there are 3 real roots, complex automorphism "fixes" just >> > those three roots (and transposes the other two pairs). >> >> > Now consider the symmetries D7 of the regular septagon. If any >> > three distinct vertices are fixed, so are the rest of them >> > (e.g. there can be no rotation or reflection). So the case of >> > exactly three real roots cannot arise if the Galois group is D7. >> >> The action of D_n on a regular n-gon gives an embedding of D_n in S_n. >> But does every embedding of D_n in S_n arise in this way? > Well, let's try to argue this for D_7 to start with. > > D_7 contains a unique cyclic subgroup of order 7. > Indeed the 14 elements of D_7 consist of the C_7 > subgroup together with 7 elements of order 2 (in > the context of symmetries of a regular septagon, > or septagon if you prefer, the reflections thru > a line from each vertex to the midpoint of the > opposing edge). > > Fixing any element of order 7 imposes some order > on the "base points" of S_7 where D_7 is embedded, > i.e. the representation of such generator as WLOG > (1 2 3 4 5 6 7), and we need to consider how the > elements of order 2 would be similarly represented. > > Any element in S_7 of order 2 must be a product of > disjoint transpositions. It cannot be a single > transposition, as then together with our cycle of > order 7 we would generate all of S_7 (not merely > a copy of D_7). > > If the elements of order 2 consisted only of two > disjoint transpositions, then (since the rotation > of order 7 above is an even permutation) all of > D_7 would be contained in the alternating group > A_7. But this is not the case; A_7 does not > contain a copy of D_7: > > http://for.mat.bham.ac.uk/atlas/html/A7.html#maxs > > http://www.projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.rmi/1228834296 > > Therefore at least one of the elements of order > 2 must consist of 3 disjoint transpositions, and > carrying out conjugations by the rotations yields > that all elements of order 2 are of this form. > > More work is needed to check that the details of > the 3 disjoint transpositions are as we expect > from the permutation of a regular heptagon, but > we have already done enough to rule out three > points being fixed by any group element other > than the identity. That seems reasonable to me. Note that I was not arguing that the assertion about polynomials with galois group D_7 was incorrect; I was merely suggesting that the argument was more complicated than that given. Actually, I would have thought that if p is an odd prime one could see more easily that the natural embedding of D_p in S_p is the only one (up to conjugacy). One can take the p-cycle in D_p to be s=(1,2,...,p). If t is an element of order 2 in D_p then tst^{-1} is in <s>, ie tst^{-1} = s^r for some r. But then t^2st^{-2} = s^{r^2} = s => r^2 = 1 mod p => r = +/-1. Thus r = -1 (since D_p is not abelian), and it follows that one has the natural embedding. I think this argument also applies if n = 2p, with p an odd prime. But in all other cases there are solutions of r^2 = 1 mod n other than +/-1, and so I think there are non-standard embeddings of D_n in S_n. -- Timothy Murphy e-mail: gayleard /at/ eircom.net tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland |