TITLE: Showing $\mathbb{Q}(\sqrt[4]{2},i)=\mathbb{Q}(\sqrt[4]{2}+i)$ using the Galois orbit of $\sqrt[4]{2} + i$
QUESTION [6 upvotes]: The following is a problem from I. Martin Isaac's Algebra. Let $E=\mathbb{Q}(\sqrt[4]{2}+i)$. I am trying to show $\mathbb{Q}(\sqrt[4]{2},i)=E$ with the following hint:

Find at least five different elements in the orbit of $i+\sqrt[4]{2}$ under $\text{Gal}(E/\mathbb{Q})$.

I have solved the problem in the usual manner i.e. by showing $\mathbb{Q}(\sqrt[4]{2}+i)\subseteq \mathbb{Q}(\sqrt[4]{2},i)$ and $\mathbb{Q}(\sqrt[4]{2},i)\subseteq \mathbb{Q}(\sqrt[4]{2}+i)$ with a few calculations. 
My question is the following: 

What is the theoretical framework behind Isaacs' hint?

REPLY [5 votes]: It's easy to see that $\mathbb{Q}(\sqrt[4]{2} + i) \subseteq \mathbb{Q}(\sqrt[4]{2}, i)$.  Additionally, $\mathbb{Q}(\sqrt[4]{2}, i)$ has degree $8$ over $\mathbb{Q}$.  So the degree of $\mathbb{Q}(\sqrt[4]{2} + i)$ over $\mathbb{Q}$ must be a divisor of $8$.  If you can show that it has degree at least $5$, you know that it has to be $8$ and so the fields are equal.

REPLY [5 votes]: Let $K/k$ be a Galois extension with group $G$. Let $\alpha$ be an element of $K$ with minimal polynomial $f$ over $k$. Then $G$ acts transitively on the set of roots of $f$ in $K$, which all have multiplicity $1$ because our extension is separable. Hence
$$
[k(\alpha) : k] = \deg f = \#(G\alpha) \quad \text{divides} \quad \#(G) = [K : k].
$$
Is clear to you how one might find five conjugates in this case, by the way?