Let $L,K$ be number fields, with $L/K$ a Galois extension, and $Q$ a prime ideal in $L$. Let $E$ be the inertia group, and $H=\{\sigma \in Gal(L/K) : \sigma(x) \equiv x \pmod {Q^2}\,\forall x\in O_L\}$. I want to prove:
(a) For $p\in Q-Q^2$, for each $\sigma \in H$ there is $\alpha_\sigma \in O_L$ such that $\sigma(p)\equiv \alpha_\sigma p \pmod{Q^2}$
(b) For $\sigma, \tau \in H$, $\alpha_{\sigma\tau}\equiv \alpha_\sigma \alpha_\tau \pmod {Q^2}$
I did the first one as follows: write $(p)=QI$. Then from the Chinese remainder theorem there is $a$ s.t. $a\equiv\sigma(p) \pmod {Q^2}$, $a \equiv 0 \pmod I$.
It is also known that $Q/Q^2$ is isomorphic to $O_L/Q$, and in particular there is $z\in Q$ s.t. $z\equiv 1\pmod {Q^2}$. Then $za=up$ for some $u\in O_L$, and we can take $\alpha_\sigma=u$.
I don't know how to solve $(b)$. I understand you can use valuations, but I rather avoid it.