I have to decide if the following statement is true:
If $E/L$, $L/K$ are normal, then $E/K$ is normal
I have though of this proof but I don't know if it is correct, could someone help me?
Let $p(x) \in K[x]$ and let $\alpha \in E$ be a root of $p(x)$. If $\alpha \in L$ then all the other roots of $p(x)$ are in $L$ (because $L/K$ is normal) and therefore they are also in $E$. Thus $E/K$ is normal. If $\alpha \notin L$ then we can also see $p(x) \in K[x]$ as a polynomial in $L[x]$. If $p(x) \in L[x]$ has a root $\alpha \in E$ then all its other roots are also in $E$ because $E/L$ is normal. Therefore $E/K$ is normal
The statement is false: Consider $K = \mathbb{Q}$, $L = \mathbb{Q}[\sqrt{2}]$, and $E = \mathbb{Q}[\sqrt[4]{2}]$.
Indeed, in this case we have $E/L$ and $L/K$ are degree 2, and thus automatically Galois extensions. However, $E/K$ is not normal, as $x^4 - 2$ has a root in $E$, but does not split completely.