ok, so G = Aut(F/K), and thus L' = Aut(F/L). we want to show that N_{G}(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 N_{G}(L'), and h is any (arbitrary) element of L'. this means g^{-1}hg(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^{-1}hg(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^{-1}hg(a) = g^{-1}(h(g(a))) = g^{-1}(h(b)) = g^{-1}(b) = a, so

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

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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).