# $\text{Gal}(L|K)$ not abelian [duplicate]

$f \in K[X]$ is irreducible and separable. $L$ the splitting field of $f$. Show:

$[L:K] > \text{deg}(f)$ $\Rightarrow$ $\text{Gal}(L|K)$ is not abelian

• You ask a question and gives the answer 2 minutes after? Moreover, what you say does not seem to answer the question. May 11 '14 at 21:29
• A duplicate (contrapositive version). May 11 '14 at 21:42

Denote $G={\rm Gal}(L/K)$. We know that $G$ acts transitively on the set $X$ of roots of $f$.
Let $x\in X$ and $G_x$ the corresponding stabilizer. Since $|G|>|X|$ we must have $|G_x|>1$. Since the stabilizers of a transitive action are conjugate, if $G$ were abelian, the subgroup $H=G_x$ would be constant (not depending on $x$). Let $h\neq1$ in $H$. This element fixes all the roots of $f$ and thus must be the trivial automorphism of $L$ which is clearly a contradiction.
Thus $G$ is not abelian.