A separable field extension of degree a product of two primes The inspiration for asking this question is due to this question and its first answer; see also this same question:
Let $F$ be a field and let $E/F$ be a separable field extension with $[E:F]=n=p_1p_2$, where $p_1$ and $p_2$ are primes (not necessarily different primes).
Let $\alpha_1$ be a primitive element: $E=F(\alpha_1)$.
Assume that $\alpha_1 \neq \alpha_2$ is conjugate to $\alpha_1$
and that $\alpha_2 \in E$.
In the above quoted question $[E:F]=p$ is prime, and it was shown that $E/F$ is Galois. 
My question: What additional conditions are necessary in our case
$[E:F]=p_1p_2$, in order to get that $E/F$ is Galois? 
It seems that our assumption that a conjugate $\alpha_1 \neq \alpha_2 \in E$ is not enough. Is it true that further assuming (=in addition to the existence of a conjugate $\alpha_1 \neq \alpha_2 \in E$) that $\alpha_2$ is not equivalent to $\alpha_1$ will imply that $E/F$ is Galois?  (We say that $\alpha_1$ and $\alpha_2$ are equivalent if $F(\alpha_1)=F(\alpha_2)$.)
Thank you very much.
Edit: Truly, I do not see why assuming, in addition, that $\alpha_2$ is not equivalent to $\alpha_1$ will imply that $E/F$ is Galois; it only implies that $F(\alpha_2) \subsetneq F(\alpha_1)$.
 A: Well, assuming $\alpha_1$ and $\alpha_2$ are not equivalent does imply $E/F$ is Galois, but only vacuously, because it is impossible for them not to be equivalent.  If $f$ is the minimal polynomial of $\alpha_1$ (and thus also of $\alpha_2$), then $[F(\alpha_1):F]=[F(\alpha_2):F]=\deg f$.  But $F(\alpha_2)\subseteq E=F(\alpha_1)$, so the only way they can have the same (finite) dimension is if they are equal, so $F(\alpha_2)=E$ as well.
For an example where $E$ contains two conjugate primitive elements but is not Galois, consider $F=\mathbb{Q}$ and $E=\mathbb{Q}(\sqrt[4]{2})$.  Then $\alpha_1=\sqrt[4]{2}$ is a primitive element and its conjugate $\alpha_2=-\sqrt[4]{2}$ is also in $E$, but $E/F$ is not Galois. 
One correct way to generalize the case where the degree is prime is that $E/F$ is Galois if $E$ contains more than $\max(p_1,p_2)$ different conjugates of a primitive element.  This is because the cardinalities of the equivalence classes of the conjugates must divide $n=p_1p_2$ (since the Galois group acts transitively on them and so each equivalence class has the same size).
