# Is this a valid proof for the normalness of subgroups in galois theory?

Let $$F/\mathbb{Q}$$ be a finite galois extension, and $$Aut(F/\mathbb{Q})$$ be the corresponding galois group. What I want to show is that for all intermediate fields between $$F$$ and $$\mathbb{Q}$$, where $$\mathbb{Q} \subset F_{1} \subset F_{2} \cdots \subset F_{m} = F$$, if the field extensions are normal, then the corresponding subgroups of $$Aut(F/\mathbb{Q})$$ are also normal.

What I thought about was, because we can represent $$F$$ as a vector space over $$\mathbb{Q}$$, we can map every element in $$Aut(F/\mathbb{Q})$$ to matrices of $$|F|$$ dimension.

My question is, if all the field extensions are normal, can we describe elements of $$Aut(F/\mathbb{Q})$$ as diagonal matrices?

Because since all diagonal matrices are commutable, for any intermediate field, we would have $$g\cdot Aut(F/F_{k}) = Aut(F/F_{k}) \cdot g$$ for $$g \in Aut(F/F_{k})$$, hence $$Aut(F/F_{k})$$ is a normal subgroup of $$Aut(F/F_{k})$$.

What I'm currently not sure about is whether or not diagonal matrices can represent all possible automorphisms of $$F$$ that fixes $$\mathbb{Q}$$.

If this is a valid argument, is it because the normalcy of the field extensions implies that all automorphisms on the field can be written as a diagonal matrix?

If not, is it possible to prove the normalcy of subgroups using this line of thought?

• Note that there are groups (such as the quaternion group) all of whose subgroups are normal, but the group is not commutative. May 19 at 7:59

No, in general the automorphisms of $$F$$ are not diagonalizable. In fact, most of them are not: If $$n=[F:\mathbb{Q}]$$ and $$\sigma\in{\rm Gal}(F/\mathbb{Q})$$ has eigenvalue $$\lambda\neq 1$$ with eigenvector $$x$$, then $$x=\sigma^n(x)=\lambda^nx$$, so $$\lambda^n=1$$, but the only roots of unity in $$\mathbb{Q}$$ are $$\pm1$$. That means that $$x=\sqrt{a}$$ for some nonsquare $$a\in\mathbb{Q}$$, so for example as soon as $$[F:\mathbb{Q}]$$ is odd, no nontrivial automorphism of $$F$$ is diagonalizable.
Also, note that $${\rm Gal}(F_{k+1}/F_k)$$ is not at all a subgroup of $${\rm Gal}(F/\mathbb{Q})$$.