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While self studying Galois from Thomas Hunger Ford I have this particular question in theorem 4.2 on page 296.

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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:

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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.

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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.

  1. 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.)

  2. 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 $f(x)$.

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  • $\begingroup$ this is also a question by me asked many days ago which is not answered. Can you please answer it too if you have some time to spare? I shall be really thankful. math.stackexchange.com/questions/3834956/… $\endgroup$
    – user775699
    Sep 26, 2020 at 16:27
  • $\begingroup$ @Tim Sure, done. Does this answer or the other one help at all? $\endgroup$
    – Kenny Wong
    Sep 26, 2020 at 20:05
  • $\begingroup$ sorry I wasn't able to see due to some personal reasons, I will surely see it today. $\endgroup$
    – user775699
    Oct 2, 2020 at 12:06

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