4
$\begingroup$

Consider finite field extensions $L>K>F$ such that $L/F$ is Galois, and $K/F$ is separable. I am particularly interested in the case $F=\mathbb{Q}$.

By Galois theory, $K/F$ is normal iff $\mathrm{Gal}(L/K)$ is normal in $\mathrm{Gal}(L/F)$, in which case the automorphism group of $K$ over $F$ is $\mathrm{Gal(K/F)}=\frac{\mathrm{Gal}(L/F)}{\mathrm{Gal}(L/K)}$.

What if $K/F$ is not normal? All I know is that $\mathrm{Aut}(K/F)$ is determined by $\mathrm{Gal}(L/F)$ and its subgroup $\mathrm{Gal}(L/K)$ because $L$ and $K$ are determined by them. Suppoose I am given a finite group $G$ and its subgroup $H$, which are meant to be $\mathrm{Gal}(L/F)$ and $\mathrm{Gal}(L/K)$, is there a way to describe $\mathrm{Aut}(K/F)$, just group-theoretically, in terms of $G$ and $H$?

Thanks.

$\endgroup$
1
  • 2
    $\begingroup$ Yes, it's $N_G(H)/H$. $\endgroup$
    – Derek Holt
    May 11, 2014 at 20:20

1 Answer 1

1
$\begingroup$

Hint: consider the subgroup

$$D=\{\sigma \in G : \sigma(K) = K\},$$

namely the subgroup of $G$ which fixes $K$ non-pointwise. Remark that $D$ contains $H$. By the "lifting of automorphisms" lemma, the restriction map

$$D \to \mathrm{Aut}(K/F)$$

is surjective, and its kernel is precisely $H$. This map therefore identifies $\mathrm{Aut}(K/F)$ with $D/H$. All that you have left to do is to figure out what $D$ is, in terms of $G$ and $H$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .