Subgroup inavariant subextension of Galois extension

Question

Let $L/K$ be a Galois extension. Let $a$ be a generator of normal basis $ga$. Let $H$ be any subgroup of G. I need to prove that $M/K$ ($H$-invariant subextension) is generated by $x=\sum_{h\in H}ha$. The only thing i had managed to find out is that one of bases of $M/K$ is $\{\sum_{h\in Hg}ha\}$ and that subextension generated by $x$ is $H$-invariant, so one of ways is to prove that every element of this basis is lying inside generated subextension, but it doesn't feel like an obvious fact.

Answer 1

Ok. So it is given that $$\mathcal{B}=\{ga\mid g\in G\}$$ is a $K$-basis of $L$. And $$x=\sum_{h\in H} ha.$$

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Let's try and identify the elements of $G$ that keep $x$ fixed. More precisely, I claim that $gx=x$ if and only if $g\in H$. Anyway, automorphisms respect sums, so $$ gx=\sum_{h\in H}(gh)a. $$

Assume first that $g\in H$. Then $gh$ ranges over $H$ while $h$ does, so $gx=x$.

Assume then that $g\notin H$. By basic properties of cosets we then have $gH\cap H=\emptyset$. Given that $\mathcal{B}$ is linearly independent over $K$ this implies that $gx\neq x$.

From this it follows immediately that $g\in G$ fixed the elements of $K(x)$ pointwise if and only if $g\in H$. In other words, $H$ is the subgroup of $G$ the Galois correspondence associated with the intermediate field $K(x)$. But, the Galois correspondence is a bijection, so $K(x)=\operatorname{Inv}(H)$.

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If you want a $K$-basis of $M=K(x)$, then, indeed, one consists of the sums $x_g:=\sum_{h\in H}hga$ with $g$ ranging over a set of representatives of the cosets $Hg$. All such sums are fixed points of $H$, so they are in $M$. They are obviously linearly independent over $K$, and there are $[G:H]=[M:K]$ of them, so they form a basis.

Answer 2

You mean $L/K$ is a Galois extension, $\{ ga,g\in G\}$ is a normal basis, $M$ is the subfield of $L$ fixed by $H$ a normal subgroup, then $\{ g\sum_{h\in H}ha,g\in G/H\}$ is a normal basis of the Galois extension $M/K$.

This follows from that $$M = Tr_{L/M}(L) = Tr_{L/M}(\sum_{g\in G} ga K)= \sum_{g\in G} (\sum_{h\in H} hg a)K=\sum_{g\in G} (\sum_{h\in H} gh a)K$$ where the last step is because $gH=Hg$.

Answer 3

Lets check H-fixed elements in $K[G] \cong L/K$ (isomorphism as $K[G]$-modules is given by normal basis). It would be elements looking like $k\sum_{h\in Hg}h$ because only cosets of $H$ are stable under $H$ multiplications on the left and $H$ acts transitively on such cosets. Subspace $K$-linearly generated by such elements has $K$-dimension $[G:H]$.

From the fact that cardinality of orbit under actions of Galois group is equal to dimension of corresponding generated subextension it follows that $dim \:K(x) = [G:H]$ because all $gH$ cosets are distinct and $ga$ forms a basic. From $x$ structure it's easy to see that all powers $x^n$ are $H$-fixed hence $K(x) \leq M$. From dimension argument it follows that $K(x) = M$