I do know that, given an extension $F\big/K$, if there is a normal intermediate extension, the corresponding subgroup of $\mbox{Gal}\left(F\big/K\right)$ is normal. The problem is that I don't see why because the definition of a normal subgroup and the definition of a normal extensions mean very different to me. I don't know why they are related. I hope there is someone that can put some light on it. Thank you.

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    $\begingroup$ Are you looking for a complete proof? Are you looking into having some details on a proof you might have explained? That theorem should be stated and proved on any Galois Theory book. $\endgroup$
    – Git Gud
    Jan 25 '13 at 19:56
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    $\begingroup$ Have you ever looked at a proof of this theorem? I think it's a little difficult helping you without some very concrete questions of yours to start with. $\endgroup$ Jan 25 '13 at 20:06
  • $\begingroup$ I didn't have the proof but I have found it. I'll take a look, thanks for your comments. $\endgroup$
    – synack
    Jan 25 '13 at 20:08

Let $G$ be the Galois group of $F/K$, and let $L$ be an intermediate extension. Then for each $\sigma \in G$ we have a conjugate extension $\sigma L$ which is also an intermediate extension of $F/K$. Normality of $L/K$ means that $\sigma L \subseteq L$ for all $\sigma \in G$, or equivalently $\sigma L = L$ for all $\sigma \in G$. By the Galois correspondence this is equivalent to $\operatorname{Gal}(F/L) = \operatorname{Gal}(F/\sigma L)$ for all $\sigma \in G$. But $\operatorname{Gal}(F/\sigma L) = \sigma \operatorname{Gal}(F/L)\sigma^{-1}$, as one sees easily. So normality of $L/K$ is equivalent to $\operatorname{Gal}(F/L) = \sigma \operatorname{Gal}(F/L) \sigma^{-1}$ for all $\sigma \in G$, which is exactly saying that $\operatorname{Gal}(F/L)$ is a normal subgroup of $\operatorname{Gal}(F/K)$.

  • $\begingroup$ +1 For making a rather concise shorcut of a rather involved theorem...which I'm afraid no one who hasn't yet studied some Galois Theory won't understand anyway. $\endgroup$
    – DonAntonio
    Jan 25 '13 at 21:52

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