1. Prove that if $f$ is separable and irreducible polynomial then the Galois group of $f$ is transitive.
  2. Prove also that even though the Galois group of $f$ is transitive not every permutation of the roots occur.

It is not difficult to give an example for the second statement e.g. Galois group of $x^4-2$ over $ \mathbb{Q}$, but I don't know how to prove the first statement, any help would be great.

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    $\begingroup$ Transitivity of a group is meaningful only if you specify the set the group is acting on (and the action). Ok, in this context you are clearly asking about its action on the zeros of $f$. Hint: If $S$ is any subset of the roots forming an orbit under the action of the Galois group, can you show that the coefficients of $p(x)=\prod_{s\in S}(x-s)$ is stable under the Galois group and hence, by the Galois correspondence its coefficients are ____. Furthermore, $p(x)\mid f(x)$, so _________ $\endgroup$ – Jyrki Lahtonen Aug 14 '15 at 16:33
  • $\begingroup$ So the orbit is the set of the roots of $f$ so the action is transitive. Thanks. $\endgroup$ – delueze Aug 14 '15 at 16:40
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    $\begingroup$ You are welcome. You are also welcome to flesh the argument to an answer yourself. That way you get more feedback, and we can check that no lingering holes remain in your reasoning. $\endgroup$ – Jyrki Lahtonen Aug 14 '15 at 16:42
  • $\begingroup$ @JyrkiLahtonen do we really need to invoke the Galois correspondence to conclude ? Can't it be more explicit ? $\endgroup$ – Weier Mar 20 at 10:02

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