Let the polynomial $f(x) \in F[x]$ for a field $F$
Say $K = SF(f/F)$ [splitting field of the $f(x)$ over $F$] and Galois field of the $F$
Let the Galois group $G(K/F)$ and the root of the $f(x)$ are $A = \{\alpha_1, \alpha_2, ... \alpha_n\}$
Needless to say $\alpha_i$ can be elements of the $F$ not in $K$.
Then, Does this statement is true?
$\forall \sigma \in G(K/F)$, $\sigma$ are bijective fuction for $A \to A$
p.s.) Intuitively It looks like true. But I don't have any confidence that the statement is true or not.
Cause' Couldn't find any counterexample and proof. Any help would be appreciated. Thanks.
$\sigma\in\text{Gal}(K/F)\implies$ $\sigma:K\to K$ is a field isomorphism with $$\sigma(a)=a,\forall a\in F.$$
Now $\alpha_i$ be a root of $f(x)=a_0+a_1x+....+a_nx^n$,say. So that, $f(\alpha_i)=0$ i.e. $$a_0+a_1\alpha+....+a_n\alpha^n=0$$$$\implies\sigma(a_0)+\sigma(a_1)\sigma(\alpha_i)+...+\sigma(a_n)\big(\sigma(\alpha)\big)^n=\sigma(a_0+a_1\alpha+....+a_n\alpha^n)=0$$ $$\implies a_0+a_1\sigma(\alpha_i)+...+a_n\big(\sigma(\alpha)\big)^n=0\implies f\big(\sigma(\alpha_i)\big)=0$$ i.e. $\sigma(\alpha_i)$ is also a root of $f$.
Since $\sigma$ is already an isomorphism, it is bijective function on $K$. In particular, $\sigma$ is a permutes elemnts of $a$.