abstract algebra - Find an example about splitting field

Let $f(x)$ be an irreducible polynomial in $\mathbb{Q}[x]$, and let $K$ be the splitting field of $f(x)$ over $\mathbb{Q}$.

Now, suppose that $E$ is a splitting field of some polynomial in $\mathbb{Q}[x]$ with $\mathbb{Q}\subseteq E\subseteq K$, and that $\alpha_{1},\alpha_{2},\ldots,\alpha_{n}$ are roots of $f(x)$ in $K$.

In this situation, prove or disprove : $$\deg(\alpha_{1},E)=\cdots=\deg(\alpha_{n},E)$$

I think that the splitting field $K=\mathbb{Q}(\sqrt[8]{2},i)$ of $f(x)=x^{8}-2\in\mathbb{Q}[x]$ over $\mathbb{Q}$ is a counterexample for this question when $E=\mathbb{Q}(\sqrt{2})$.

Am i correct? or is there an example rather than this?

If it is, under what conditions, the relation $\deg(\alpha_{1},E)=\cdots=\deg(\alpha_{n},E)$ could be true.

Give some comment or advice. Thank you!

Answer

Since $E$ is a splitting field, it is a Galois extension of $\mathbf{Q}$. Thus $N=\mathrm{Gal}(K/E)$ is a normal subgroup of $G=\mathrm{Gal}(K/\mathbf{Q})$. Since $f$ is irreducible, the group $G$ acts transitively on its roots. Let $\alpha$ and $\beta$ be roots of $f$, and choose $g \in G$ with $g(\alpha)=\beta$. Using normality of $N$ in $G$, the stabilizers $N_\alpha$ and $N_\beta$ of $\alpha$ and $\beta$ in $N$ are related by
$$g N_\alpha g^{-1}=N_\beta,$$ and in particular they have the same cardinality. Hence by the orbit stabilizer theorem the degrees over $E$ of $\alpha$ and $\beta$ are equal.