I'm facing this question:
Let $E/F$ be a finite field extension such that its Galois group $\text{Gal}(E/F)$ is abelian. Is it necessary that $E/F$ is a Galois extension?
Attempt:
My first guess is no, but apparently I can't come up with any counter example. Maybe choosing $F$ not to be a perfect field would help, but I'm not sure in what direction to move to get an abelian Galois group.
My attempt to prove the statement was also unfruitful; I tried showing that the field of fixed elements of $E$ through $\text{Gal}(E/F)$ is precisely $F$, namely $E^{\text{Gal}(E/F)}=F$, which I know is equivalent to $E/F$ being a Galois extension. The inclusion $F\subset E^{\text{Gal}(E/F)}$ is always true; for the other inclusion: $[E:F]=[E:E^{\text{Gal}(E/F)}]\cdot[E^{\text{Gal}(E/F)}:F]$, but since $E/F$ is finite, $[E:E^{\text{Gal}(E/F)}]=|\text{Gal}(E/F)|$. I suppose that, if this is the right way, the fact that the Galois group is abelian comes in here. But I can't prove that $[E:F]=|\text{Gal}(E/F)|$ with only this assumption.
Any counterexample or hint is greatly appreciated.