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!
 A: 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.
A: Otherwise we can avoid Galois groups by taking the isomorphism $\sigma: Q(a_1) \to Q(a_n)$ and extend it to an isomorphism $\sigma :K \to \overline{K}$. 
$\sigma$ is defined by where is sends the roots of $f$ (since $K$ is generated over $Q$ by the roots of $f$), as $f \in Q[x]$ it must permute them which means $\sigma(K)=K$. 
The same argument holds with $\sigma|_E$. 
Thus $\sigma$ sends the minimal polynomial $\in E[x]$ of $a_1$ to the minimal polynomial $\in \sigma(E)[x]=E[x]$ of $a_n$.
