Galois GroupsDate: 04/28/2005 at 20:15:47 From: Brian Subject: Galois groups Suppose K > Q (Q = rational numbers) is an extension of degree 4 which is not a Galois extension. Let L be the Galois closure of K over Q. Prove: 1. The Galois group Aut_Q(L) is isomorphic to S4,A4, or D8 (dihedral group of order 8). 2. Aut_Q(L) is isomorphic to D8 if and only if K contains a quadratic extension of Q. I'm completely confused! This Galois theory is so hard! I don't see how we would have any information about this Galois group. The only information I can gather is that this Galois group contains a subgroup of index 4, but I'm not sure how they narrow the groups down to only three. Please help! Date: 04/29/2005 at 06:49:36 From: Doctor Jacques Subject: Re: Galois groups Hi Brian, To simplify matters, let us write G(L/Q) for Aut_Q(L), the Galois group of L over Q; we will use similar notations for other groups and subfields. As K has degree 4, it can be generated by an irreducible polynomial of degree 4, and the Galois group of that polynomial is a subgroup of S4, this is the group G(L/Q). As this group has a subgroup of index 4, its order is divisible by 4, and not equal to 4, since this would imply K = L, contrary to the fact that K/Q is not Galois. This means that L has order 8, 12, or 24; in each case, there is only one subgroup (up to isomorphism for order 8) of the corresponding order in S4--these subgroups are isomorphic to D8, A4, and S4. It is easy to check that each of these subgroups contains a subgroup of index 4, corresponding to G(L/K). If M is a quadratic extension of Q, contained in K, we have the following inclusions of the corresponding groups: G(L/K) < G(L/M) < G(L/Q) where "<" means "is a subgroup of", and [G(L/Q) : G(L/M)] = 2. Assume first that G(L/Q) = D8; this means that G(L/K) is a subgroup of D8, of order 2. We can view D8 as the permutation group generated by (1,2,3,4) and (1,3) (we can also picture it as the symmetry group of a square whose vertices are 1,2,3,4). D8 has three subgroups of order 4: G1 = {(), (1,2,3,4), (1,2)(3,4), (1,4,3,2)} G2 = {(), (1,3), (2,4), (1,3)(2,4)} G3 = {(), (1,2)(3,4), (1,4)(2,3), (1,3)(2,4)} The subgroups of order 2 (G(L/K) is one of them) are those generated by (1,2)(3,4), (1,4)(2,3), (1,3), (2,4), and (1,3)(2,4), and you can check that each of these subgroups is contained in G2 or G3. (Note that none of these subgroups is normal in D8; this is compatible with the fact that K/Q is not Galois.) This shows that G2 or G3 is a subgroup of index 2 in G(L/Q) and contains G(L/K); therefore, the fixed field (M) of that group is an extension of degree 2 of Q contained in K, by the fundamental theorem. This disposes of the "only if" part: if G(L/Q) = D8, K contains a quadratic extension M of Q. Assume now that G(L/Q) = A4. In this case, G(L/M) should be a subgroup of order 6 in A4; however, A4 does not contain a subgroup of order 6 (there are various ways to show this, write back if you want to discuss it further). The last case is G(L/Q) = S4. In this case, G(L/K) has order 6, and G(L/M) has order 12. However, the only subgroup of order 12 in S4 is A4, and, once again, we are led to the contradiction that A4 (G(L/M)) should contain a subgroup (G(L/K)) of order 6. This shows that, if G(L/Q) is not isomorphic to D8, K cannot contain a quadratic extension of Q, and proves the "if" part of the statement. Does this help? Write back if you'd like to talk about this some more, or if you have any other questions. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/ |
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