# Normalizer of a subgroup of a Galois group

I wanted to check whether my solution for this problem was correct. Let $k \subseteq L \subseteq K$ be a finite extension of fields, with $K/k$ Galois $H$ the normalizer of $Aut(K/L)$ in $Aut(K/k)$.

(i) Show that $H = \{ \phi \in Aut(K/k) : \phi(L) = L\}$.

(ii) Describe the group $H/Aut(K/L)$ as an automorphism group.

(i): Let $\phi \in Aut(K/k)$. It's required to show that $\phi(L) = L$ if and only if $\phi^{-1} Aut(K/L) \phi = Aut(K/L)$. If $\phi(L) = L$, let $\psi \in Aut(K/L)$ and $l \in L$. Since $\phi(l) = l$, we have $\psi \phi(l) = \phi(l)$, whence $\phi^{-1} \psi \phi(l) = l$, as required.

Conversely suppose $\phi(L) \neq L$. Then $\phi(l) \not\in L$ for some $l \in L$. Since $K/k$ is Galois, so is $K/L$, so there is some $\psi \in Aut(K/L)$ for which $\psi \phi(l) \neq \phi(l)$. But then $\phi^{-1} \psi \phi(l) \neq l$.

(ii) Let $E$ be the fixed field of $H$, so $H = Aut(K/E)$. Since $Aut(K/L)$ is normal in $Aut(K/E)$, we have that $E \subseteq L$ with $L/E$ Galois. As a normalizer, $H$ contains every subgroup $Aut(K/E_1)$ in $Aut(K/k)$ in which $Aut(K/L)$ is normal, i.e. $E$ is contained in every intermediate field of $K/k$ over which $L$ is Galois.

Also $H/Aut(K/L) = Aut(K/E)/Aut(K/L) \cong Aut(L/E)$, the isomorphism being given by restriction to $L$. So, there is a unique smallest intermediate field of $K/k$ over which $L$ is Galois, and $H/Aut(K/L)$ can be regarded as the Galois group of $L$ over this field.

A problem in your answer to $(i)$. You don't know that $\phi(l)=l$, because $\phi$ stabilizes $L$ only setwise - not elementwise. You do know that $l'=\phi(l)\in L$ and, consequently, that $\psi(l')=l'$. Using these bits you can surely show that $\phi^{-1}\psi\phi(l)=l$ as required. The argument for the converse part looks fine. You may want to add a conclusion that $\phi^{-1}\psi\phi\notin Aut(K/L)$. It is somewhat a matter of taste. If you explain this to a classmate who is not quite up to speed, you might still get an empty stare before adding that.
I did not spot any problems in your answer to $(ii)$.