# Normalizer subgroup corresponding to a closed intermediate field

• Sep 3rd 2012, 04:32 AM
Bingk
Normalizer subgroup corresponding to a closed intermediate field
Hello, this was another problem that I had difficulty with:

Let $F$ be a field extension of $K$, and let $L$ be a closed intermediate field. Let $G=Aut_kF$. Show that the normalizer of $L'$ in $G$ is the set of elements of $G$ which map $L$ onto itself.

$L'$ is the corresponding subgroup (in $G$) of $L$.

I'm not sure. I wanted to use FTGT (Fundamental Theoreom of Galois Theory), but this problem doesn't state that the extension is finite.
• Sep 4th 2012, 01:48 AM
Deveno
Re: Normalizer subgroup corresponding to a closed intermediate field
ok, so G = Aut(F/K), and thus L' = Aut(F/L). we want to show that NG(L') = {g in G: g(L) = L} (this is a looser requirement than g(a) = a, for all a in L).

so suppose g is (any fixed element) in NG(L'), and h is any (arbitrary) element of L'. this means g-1hg(a) = a, for all a in L.

now suppose g(a) = b (at the moment, all we know is b is some element of F).

then g(g-1hg(a)) = g(a) = b, so:

hg(a) = b, and thus h(b) = b, for all h in L'. this means b is in the fixed field of h, so b is in L. hence g(L) is contained in L, and since g is bijective, g(L) = L.

on the other hand, suppose g in G maps L to L. we want to show that g normalizes any h in L'.

so let g(a) = b, where a,b are in L, and let h be any element of Aut(F/L). then:

g-1hg(a) = g-1(h(g(a))) = g-1(h(b)) = g-1(b) = a, so

g-1hg is in L', thus g normalizes any h in L'.

**************

it may seem mysterious that the closure of L was not mentioned explicitly, here. so where did we use it? when we said that h(b) = b implies b is in L. since L is closed, the fixed field of L' is precisely L, no more, no less (that is: given any u in F, but not in L, some element h of L' has: h(u) ≠ u. so if b = g(a) is not in L, there is SOME h in L' with h(b) ≠ b. but the a in L we picked was arbitrary, as was h).

note that F itself may not have this property, since F might not be galois over K. in other words, the fixed field of G could be larger than K (it has to be at least K).
• Sep 8th 2012, 10:41 PM
Bingk
Re: Normalizer subgroup corresponding to a closed intermediate field
Thank you for that. I was actually thinking along similar lines. I knew that the normalizer contained L' and g(L)=L for g in L'. But I didn't know what to do with the other automorphisms in the normalizer, so I gave up on that track. Thanks again!