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### Galois Groups

```Date: 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
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

```

```
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

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.

more, or if you have any other questions.

- Doctor Jacques, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Modern Algebra

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