Let $L/K$ be a normal number field extension with ring of integers $\mathcal O_L/\mathcal O_K$. Let $Q$ be a prime ideal of $\mathcal O_L$ and inertia group $E = \{g \in \operatorname{Gal}(L/K)|\forall\alpha\in\mathcal O_L, g(\alpha) \equiv \alpha \pmod Q \}$. Let $L_E$ be the fixed field of $E$ and $\mathcal O_{L_E}$ it's ring of integers.
Then show that $\mathcal O_L = \mathcal O_{L_E} + Q$(all possible pairs addition). My idea was to write $\alpha = \sum_{g\in E}g(\alpha) - R$ or $\prod_{g \in E}g(\alpha) - R$ and show that $R \in Q$ but I could not quite finish the proof.
The subgroup of Galois that you mention is usually called the inertia group. You have $L\supset L^E\supset K$, where the lower extension is unramified, i.e. that’s where the residue field extension happens; and the upper extension is totally ramified, with no residue field extension.
Now, you’re asking whether $\mathscr O_L/Q\cong\mathscr O_{L^E}/Q'$, where $Q'$ is the intersection of $Q$ with $L^E$. This is true: the residue fields of these two primes (one a prime of $L$, the other of $L^E$) are the same.