# Normal closure $L/k$ contains the Galois image of $k$

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

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