Here is the full statement

$K/L/k/F$ the field extensions with $K/F$ Galois extension, and $L$ the normal closure of $k$ (i.e. the smallest field extension of $k$ within $K$ such that $L/F$ is normal). If $\sigma\in Gal(K/F)$, then $\sigma(k)\subseteq L$.

My Approach: By Galois correspondence, it is enough to show $Gal(K/L)\subseteq Gal(K/\sigma(k))=\sigma Gal(K/k)\sigma^{-1}$, i.e. to show $Gal(K/L)\sigma\subseteq \sigma Gal(K/k)$. Then I have no idea how to go on. Can anyone help? Thanks in advance

  • $\begingroup$ Since $L/F$ is Galois, $\sigma\in{\rm Gal}(K/F)$ fixes $L$, so $\sigma(k)$ remains in $L$. $\endgroup$ – anon Jun 4 '18 at 4:08
  • $\begingroup$ @anon how can I overlook the important fact that $L/F$ Galois, thx! $\endgroup$ – Unavailable Jun 4 '18 at 4:16

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