# 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.

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 $$f(x)$$.

• 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/…
– user775699
Sep 26, 2020 at 16:27
• @Tim Sure, done. Does this answer or the other one help at all? Sep 26, 2020 at 20:05
• sorry I wasn't able to see due to some personal reasons, I will surely see it today.
– user775699
Oct 2, 2020 at 12:06