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Let $F$ be a field of characteristic $0$ and let $f(x)\in F[x]$, and let $G$ be the Galois group of $f(x)$ over $F$.

Now, if $f(x)$ is irreducible, I know that $G$ acts transitively on the roots of $f(x)$. $(*)$

Suppose that $f(x)$ is not irreducible, and let $g(x)$ be an irreducible factor of $f(x)$. Does $G$ act transitively on the roots of $g(x)$?

I think we can show this by a restriction homomorphism but I'm not sure. My idea is this:

If $K$ is the splitting field of $f(x)$ and $L$ is the splitting field of $g(x)$ then $G = \text{Gal}(K/F)$ and $L$ is an intermediate field of $K/F$. Let $H = \text{Gal}(L/F)$. Define a map $$\phi : G\rightarrow H$$ $$\sigma \mapsto \sigma|_L$$

Now this map is surjective and so the image of $\phi$ is $H$ and so $\text{im}\phi$ acts transitively on the roots of $g(x)$ by $(*)$. But since $\phi$ is just a restriction, $G$ must also act transitively on the roots of $g(x)$.

It's quite a wordy proof but is the idea correct?

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That looks good. You can potentially cut down on the wordiness of the argument by "going in reverse" logically, by extending maps instead of restricting maps:

We know that the Galois group of $g$, which you call $\operatorname{Gal}(L/F)$, acts transitively on the roots of $g$. Since $L$ is a subfield of $K$ (which is an algebraic extension of $F$), every $F$-automorphism of $L$ can be extended to an $F$-automorphism of $K$ per the isomorphism extension theorem. Clearly, these extensions are elements of $\operatorname{Gal}(K/F)$ as this is the set of all $F$-automorphisms of $K$. Thus, we can conclude that $\operatorname{Gal}(K/F)$ acts transitively on the roots of $g$.

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