Prove $\operatorname{Gal}(L/K)$ is Abelian

Question

I am given that $L/K$ is a finite extension of fields and that $[L : K] = 25$ and want to prove that $\operatorname{Gal}(L/K)$ is abelian.

I also had to show that $|\operatorname{Gal}(L/K)|$ divides $[L : K]$.

I did this by letting $L^G$ be the fixed field of $G$ with $[L: L^G] = |G| = |\operatorname{Gal}(E/F)|$ then by Tower Law $[L: K] = [L: L^G][L^G: K] = |G|[L^G : K]$.

Back to showing $\operatorname{Gal}(L/K)$ is abelian, I've noted that as $[L: K]$ is $25$ which means $|\operatorname{Gal}(L/K)|$ is either $1, 5$ or $25$.

If it has order $25$, then it is abelian as all groups of order $p^2$ (with p being prime) are abelian.

If it has order $1$ then it is the trivial group and therefore abelian.

I still need to show it for when $|\operatorname{Gal}(L/K)| = 5$.

Am I heading in the right direction?

Thanks

Answer 1

For Galois extensions $L/K$ the order of the Galois group is equal to the field degree. So we have $|{\rm Gal}(L/K)|=[L:K]=5^2$. Since all groups of order $p^2$ are abelian, as you have noted, we are done.

Reference: Proof of Order of Galois Group equals Degree of Extension