# Chain of fields that are Galois over all subfields

Is there an example of fields $F_1$, $F_2$, and $F_3$ such that $\mathbb{Q}\subset F_1\subset F_2\subset F_3$ such that $[F_3:\mathbb{Q}]=8$ and each field is Galois over all its subfields but $F_2$ is not Galois over $\mathbb{Q}$?

I know of $F_3$ is Galois over $\mathbb{Q}$, then it will automatically be Galois over all other subfields, but that's about it. What's a good field to investigate here? Thanks.

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Maybe we should be careful: $F_2$ being Galois over all its subfields and not being Galois over its subfield $\mathbf Q$ don't seem like compatible statements. –  Dylan Moreland Nov 25 '11 at 1:51
Do you mean, $F_i$ Galois over $F_j$ whenever $i\leq j$? –  Arturo Magidin Nov 25 '11 at 3:43

$${\bf Q}\subset{\bf Q}(\sqrt2)\subset{\bf Q}(\root4\of2)\subset{\bf Q}(\root4\of2,i)$$
There are infinitely many examples of Gerry's kind, since the question can be phrased in group theoretic terms as follows: we want the Galois group $G$ of $F_3/\mathbb{Q}$ to have a chain of subgroups $G=H_0>H_1>H_2>H_3=\{1\}$ such that each is normal in all the bigger ones except for $H_2$, which should not be normal in $G$.
Clearly, $G$ has to be non-abelian. There are only two non-abelian groups of order 8, the dihedral group $D_8$ and the quaternion group $Q_8$. $G=D_8$ with $H_2$ a non-central involution does it, as Gerry's example shows. Thus, any dihedral extension $F_3/\mathbb{Q}$ of order 8 will have such a chain of subfields. In $Q_8$ there is no such chain, since the only subgroup of order 2 is central and thus normal in $G$. So $D_8$ is the only possibility.