Subgroup inavariant subextension of Galois extension 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.

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 $K(x) = M$
