Abelian Galois group of even order I'm stuck at the following problem.

Let $a$ and $b$ be algebraic real numbers over $\mathbb{Q}$. Let $K= \mathbb{Q}(a+bi)$ be a simple extension of $\mathbb{Q}$. Suppose that $K$ is a Galois extention over $\mathbb{Q}$ and the the Galois group $G(K/\mathbb{Q})$ is abelian. Moreover, assume that $a^2+b^2 \in \mathbb{Q}$ and $b \neq 0$.
  (1) Prove that the order of $G(K/\mathbb{Q})$ is even.
  (2) Let the order of $G(K/\mathbb{Q})$ is $2m$. Prove that for each positive divisor $d$ of $m$, there exists an irreducible polynomial over $\mathbb{Q}$ whose zeros are all real and degree $d$.

This is an attempt to solve this problem. 


*

*Let $\rho$ be the complex conjugate transformation and $f(x)$ be the minimal polynomial of $a+bi$. Since $f(x)$ is invariant under $\rho$, $\rho$ restricted on $K$ is a member of the Galois group.

*Let $E$ be the fixed field of $\rho$. We have an intermediate field $K-E-\mathbb{Q}$ where $K=E[i]$.

*Degree of $K$ over $E$ is $2$. Hence the order of the Galois group is even. Moreover the Galois group of $E$ over $\mathbb{Q}$ is abelian. Therefore from the converse of the Lagrange's theorem, for each divisor $d$ of $m$, there corresponds a subgroup of the group and an irreducible polynomial.


But I have not used the condition $a^2+b^2 \in \mathbb{Q}$. This should be wrong.
 A: Let $a$ and $b$ be algebraic real numbers over $\mathbb{Q}$ and let $K = \mathbb{Q}(a+bi)$. Assume that $K$ over $\mathbb{Q}$ is a Galois extension whose Galois group is abelian.
Since $K-\mathbb{Q}$ is Galois extension, we have $a-bi \in K$ and so $a \in K$. Consider an intermediate field $K-E-\mathbb{Q}$ where $E = \mathbb{Q}(a)$. Consider $x^2-2ax+a^2+b^2 \in E[x]$. This is irreducible since $b \neq 0$ and $E \subseteq \mathbb{R}$. This is the minimal polynomial of $a+bi$ over $E$. Therefore $K-E$ is degree $2$. This tells us that the order of the Galois group is even, say $2m$ where $m = [E:\mathbb{Q}]$.
Since $G=G(K/\mathbb{Q})$ is abelian, it follows that $G(E/\mathbb{Q})$ is also abelian. Hence by the inverse of the Lagrange's theorem, for each divisor $d$ of $m$, we have a subgroup $H$ of $G$ of order $m/d$. Let $F$ be the fixed field in $E$ by $H$. Then $[F:\mathbb{Q}] = d$. Since characteristic is zero, $F = \mathbb{Q}(\alpha)$, we have that  $\hbox{irr}(\alpha, \mathbb{Q})$ has exactly $d$ real solutions degree $d$ polynomial. One more thing we can require is that the polynomial has zeros on $K \cap \mathbb{R}$.
