A bit of trouble with Galois groups

[Euclid]

Is the Galois group of F=Q(sqrt2,3i) the maps {id, tau , sigma, gamma}, where
(1) id is the identity
(2) tau maps sqrt2 to -sqrt2 and leave 3i alone
(3) sigma leaves sqrt2 alone and maps 3i to -3i
(4) gamma maps sqrt2 to -sqrt2 and 3i to -3i ?
If so, the what are the fixed fields of the subgroups?
If I'm not mistaken the (proper nontrivial) subgroups are H={id, tau}, J={id, sigma}, K={id, gamma}. It appears that the fixed field of H is Q(3i) and the fixed field of J is Q(sqrt2). But it also appears that the fixed field of K is just Q, which is also the fixed field of Gal(F/Q). But F/Q is Galois since F is the splitting field of a seperable polynomial, so we can't have two distinct groups associated to the same intermediate field.
What am I doing wrong?

 

[Hurkyl]

Try writing down an arbitrary element of F: it would be a Q-linear combination of 4 elements:

1, 3i, sqrt2, 3isqrt2

So, apply K to this arbitrary element and see if that gets you anywhere...

 

[M98Ranger]

It seems like you are on the right track, but there are a few things to clarify. First, the Galois group of F=Q(sqrt2,3i) is not just the maps {id, tau, sigma, gamma}, but rather the group generated by these maps. This means that the group also includes compositions of these maps and their inverses. So the Galois group is actually a larger group than just those four maps.

Next, the fixed field of a subgroup is the field that is fixed by all the elements in that subgroup. So for example, the fixed field of H={id, tau} would be the field that is fixed by both the identity and tau. In this case, that would be the field Q(3i) since both id and tau leave 3i unchanged. Similarly, the fixed field of J={id, sigma} would be Q(sqrt2) since both id and sigma leave sqrt2 unchanged.

However, the fixed field of K={id, gamma} is not just Q, but rather the field Q(sqrt2,3i) itself. This is because both id and gamma leave both sqrt2 and 3i unchanged, so the fixed field of K must contain both sqrt2 and 3i. It is not just Q because Q does not contain 3i.

In general, the fixed field of a subgroup will be the smallest field that contains all the elements that are fixed by that subgroup. This can sometimes be a larger field than expected, as in the case of K={id, gamma}.

So to summarize, the Galois group of F=Q(sqrt2,3i) is a larger group than just the four maps mentioned, and the fixed fields of the subgroups are Q(3i), Q(sqrt2), and Q(sqrt2,3i) respectively. This does not contradict the fact that F/Q is Galois, as the Galois group is generated by all possible automorphisms, not just the four mentioned.