I've been trying to solve this problem for $L:K$ finite with char $K = 0$:
$\beta_1,...,\beta_n$ basis for $L$ over $K$, $H$ subgroup of $\Gamma(L:K)$ $\implies$ $\phi(H) = K(\gamma_1,...,\gamma_n)$, where $\gamma_i = \sum_{\sigma \in H} \sigma(\beta_i)$
Note that $L:K$ is finite but not necessarily normal.
I think I am close because I can show that
$\Gamma(L:\phi(H)) = \Gamma(L:K(\gamma_1,...,\gamma_n))$
But does this necessarily imply $\phi(H) = K(\gamma_1,...,\gamma_n)$, given the conditions in the problem?
If I could show $L:K(\gamma_1,...,\gamma_n)$ normal, the result would follow.
Any ideas?
Okay, I worked out the problem. I was making it too complicated by going through $\Gamma(L:\phi(H)).$ Instead I look at the fixed field $\phi(H)$ directly.
I'll post my answer here, and maybe people can check it.
For each $\tau \in H, \tau(\gamma_i) = \tau(\sum_{\sigma \in H}\sigma(\beta_i)) = \sum_{\sigma \in H}\tau\sigma(\beta_i) = \sum_{\sigma \in H}\sigma(\beta_i) = \gamma_i, 1 \le i \le n.$
So $\gamma_i \in \phi(H), 1 \le i \le n$. Thus $K(\gamma_1,...,\gamma_n) \subseteq \phi(H).$
On the other hand, consider $\alpha \in \phi(H)$. For each $\sigma \in H, \sigma(\alpha) = \alpha$. Write $\alpha = k_1\beta_1 + \dots + k_n\beta_n$. Then
$\begin{align} \textstyle{\sum_{\sigma \in H}\sigma(\alpha)} &= \textstyle{\sum_{\sigma \in H}\sigma(k_1\beta_1 + \dots + k_n\beta_n)} \\ &= \textstyle{k_1 \sum_{\sigma \in H}\sigma(\beta_1) + \cdots + k_n \sum_{\sigma \in H}\sigma(\beta_n)} \\ &= k_1\gamma_1 + \cdots + k_n\gamma_n \end{align}$
Also
$\sum_{\sigma \in H}\sigma(\alpha) = \sum_{\sigma \in H}\alpha = |H|\alpha$
So $\alpha = \cfrac{1}{|H|}(k_1\gamma_1 + \cdots + k_n\gamma_n) \in K(\gamma_1,...,\gamma_n).$ Thus $\phi(H) \subseteq K(\gamma_1,...,\gamma_n).$