# Normalizer and automorphism group in a Galois group

I need help figuring out how to start on the following problem.

Let $$k \subset K$$ be a finite Galois extension with Galois group $$G(K/k)$$, let $$L$$ be a field with $$k \subset L \subset K$$, and set $$H = \{\sigma \in G(K/k)\mid \sigma(L) = L\}$$.

(a) Show that $$H$$ is the normalizer of $$G(K/L)$$ in $$G(K/k)$$.

(b) Describe the group $$H/G(K/L)$$ as an automorphism group.

Let $N$ be the normalizer of $G(K:L)$ in $G(K:k)$.

$H\subset N$

$h\in H, g\in G(K:L), x\in L, hgh^{-1}(x)=h(gh^{-1}(x))$, since $h(L)=L, h^{-1}(x)\in L$ and $g(h^{-1}(x))=h^{-1}(x)$ since $g\in Gal(K:L)$ we deduce that $hgh^{-1}(x)=h(h^{-1}(x))=x$.

$N\subset H$

Let $n\in N, g\in Gal(K:L), x\in L$,$g(n(x))=nn^{-1}gn(x)$ since $N$ is the normalizer of $Gal(K:L), n^{-1}gn\in Gal(K:L), ngn^{-1}(x)=x$ and $gn(x)=nn^{-1}gn(x)=n(x)$, hus $n(x)$ is in the subfield fixed by $Gal(K:L)$ which is $L$, by the Galois fundamental theorem.

Consider the map $\phi:H\rightarrow Gal(L:k)$ such that $\phi(h)$ is the restriction of $h$ to $L$, the kernel of $\phi$ is $Gal(K:L)$ and $\phi$ is surjective since the extension $K:L$ is separable.

• What do you mean by the restriction of $h$ to $L$? – user346096 Aug 2 '16 at 14:57
• The map $h':L\rightarrow L$ defined by $h'(x)=h(x)$. – Tsemo Aristide Aug 2 '16 at 14:59