Suppose $L:K$ is a finite extension and $G(L:K)$ denotes the Galois group of $L:K$.
Since $L:K$ is finite it is algebraic and therefore any element of $a\in L$ is the root of some polynomial $f\in K[x]$.
The Galois group consists of permutations of $L$ that permute roots of $f$.
My question is do the permutations of the Galois group when applied to $a$ give all the roots of $f$?
I ask this because in my notes for one particular proof it mentions the set $\{\sigma(a): \sigma \in G(L:K)\}$ and says let the distinct elements of this set be $a_1,a_2,\ldots,a_n$ for some $n \in \mathbb{N}$. Later in the proof it says if $\phi \in G(L:K)$ then $\phi(\{a_1,a_2,\ldots,a_n\}=\{a_1,a_2,\ldots,a_n\}$ i.e. $\phi$ permutes the $a_i$’s.
I can’t see how $\phi$ permutes these $a_i$’s. My logic was that since $a$ is a root of $f$ then $\{a_1,a_2,\ldots,a_n\}$ are roots of $f$ as elements of the Galois group permute roots. But unless these are ALL the roots of $f$ I can’t see how $\phi$ necessarily permutes these roots as why can’t $\phi$ map some $a_i$ to another root of $f$ not in this set?
There is the important condition of $L/K$ being Galois, which is equivalent to being a normal and separable extensions, or that $|Aut(L/K)| = [L:K]$. Not all extensions are Galois. Further, the Galois group of a Galois extension $L/K$ is not simply the permutations of $L$ that permute the roots of some random polynomial $f$. The Galois group of $L/K$, $Gal(L/K) = Aut(L/K)$, which is the group of field automorphisms fixing the base field $K$.
Under these assumptions, it is a standard result of Galois theory that the Galois group acts transitively on the set of roots of any irreducible polynomial in $K[X]$.