A question in proof of a theorem related to Galois Group of polynomial While self studying Galois from Thomas Hunger Ford I have this particular question in theorem 4.2 on page 296.




Question: How does in Last line of (II) of proof author wrote that G is isomorphic to transitive subgroup of $S_n$?  ie how 3.8 implies it?

The problem is that I am unable to understand how 3.8 implies G is isomorphic to transitive subgroup of $S_n$.
Theorem 3.8:

and a subgrouo of $S_n$ is called transitive if given any i$\neq$ j there exists $\sigma \in G $ such that $\sigma(i) =j$.
Kindly help.
 A: The $S_n$ acts on the roots $u_1, \dots, u_n$ of $f(x)$ by permutation. Each $K$-automorphism in the Galois group $G$ of $f(x)$ is determined by the way in which it permutes the $u_1, \dots, u_n$ (these being the roots of $f(x)$ in a splitting field $F$ for $f(x)$ over $K$); thus $G$ can be viewed as a subgroup of $S_n$.
The statement that $G$ acts transitively on $u_1, \dots, u_n$ is the statement that, for any roots $u_i, u_j$ of $f(x)$ in the splitting field $F$, there exists an $K$-automorphism $\sigma \in G$ such that $\sigma(u_i) = u_j$.
The author proves the transitivity of the action of $G$ in two steps.

*

*Since $f(x)$ is irreducible over $K$, there exists a $K$-isomorphism $\widetilde {\sigma} : K(u_i) \to K(u_j)$ such that $\widetilde{\sigma}(u_i) = u_j$ for any roots $u_i$ and $u_j$ of $f(x)$. (When I say $K$-isomorphism, I mean that $\widetilde\sigma$ leaves elements in $K$ invariant.)


*By 3.8, any $K$-automorphism $\widetilde{\sigma} : K(u_i) \to K(u_j)$ extends to a $K$-isomorphism $\sigma : F \to F$ (where $F$ is the splitting field of $f(x)$ over $K$ that contains $u_i$ and $u_j$). [To spell this out, $F$ is the splitting field of $f(x)$ over $K$, but it is also the splitting field of $f(x)$ over $K(u_i)$, and over $K(u_j)$. To map my notation across to the notation in Theorem 3.8: my $K(u_i)$ plays the role of the $K$ in Theorem 3.8; my $K(u_j)$ plays the role of the $L$ in 3.8; my $F$ plays the role of both the $F$ and the $M$ in 3.8.]
Thus, for any roots $u_i, u_j$ of $f(x)$, there exists a $K$-automorphism $\sigma \in G$ such that $\sigma(u_i) = u_j$, i.e. the Galois group $G$ acts transitively on the roots of $f(x)$.
