Let $L/\mathbb{Q}$ be a finite Galois extension with Galois group $G$. It is well known that the ring of integers $\mathcal{O}_L$ is free over its associated order $\mathfrak{A}_{L/\mathbb{Q}}=\{x\in \mathbb{Q}[G]\mid x\mathcal{O}_L\subseteq \mathcal{O}_L\}$ if

 1. $G$ is abelian (Leopoldt, 1959);
 2. $G$ is dihedral of order $2p$, where $p$ is a rational prime (Bergé, 1972);
 3. $G$ is the quaternion group of order $8$, and the extension is wild (Martinet, 1972).

In some other papers, I have found written that also the local counterparts are true ($L/\mathbb{Q}_p$ finite with the same Galois group as before), and it seems that the authors suggest that these results are naturally implied by the global ones. But while they seem "folklore", these implication are not immediate to me. A couple of observations (I wish to thank Fabio Ferri for the precious discussion).

 1. It seems to me that only the number fields case is considered in the papers.
 2. Probably the key is to repeat the proof as it is for the $p$-adic case; for example, this works in the Martinet's case. 
3. There are other ways to get these results, like for example using Lettl's work on absolutely abelian extensions for Leopoldt's local case, or realise local extensions as completion of global ones with the same Galois group (here $p\ne 2$, we refer to Henniart, 2001). But I am interested in knowing if there is a "direct way".
 
Summarising, my question is: is it immediate that the global cases imply the local cases?

(More generally, a question could be: if every Galois extension of number fields with a certain type of Galois group admits freeness of the ring of integers over the associated order, then the same also holds for every Galois extension of local fields with the same Galois group?)