Marcus, Number Fields, Problem 19(b), Chapter 4 It would be really helpful if somebody could check my argument! Marcus suggests something else but I (think) I found another proof. So here goes:
First of all let's set the stage: Let $L/K$ be a Galois extension of number fields with Galois group $G$. Let $\mathfrak{q}$ be a prime of the ring of integers $\mathcal{O}_L$ of $L$ lying above a prime $\mathfrak{p}$ of the ring of integers $\mathcal{O}_K$ of $K$. We define the ramification groups for $m \geqslant 0$ as follows: $$ V_m = \{\sigma \in G: \sigma(\alpha) \equiv \alpha \pmod{\mathfrak{q}^{m+1}}\quad \forall \alpha \in \mathcal{O}_L \}  .$$
Fix any element $\pi \in \mathfrak{q}-\mathfrak{q}^2$ (such an element exists by unique factorization of ideals in a Dedekind domain). Let $\sigma \in V_{m-1}$, $m>0$ (this will be referred to as Assumption 1). We assume that $\sigma(\pi) \equiv \pi\pmod{\mathfrak{q}^{m+1}}$ (this will be referred to as Assumption 2).
In the first part of the problem we proved that $ \sigma(\alpha) \equiv \alpha \pmod{\mathfrak{q}^{m+1}}\quad\forall \alpha \in \pi\mathcal{O}_L $.
Now we want to prove $\sigma(\alpha) \equiv \alpha\pmod{\mathfrak{q}^{m+1}}\quad \forall \alpha \in \mathfrak{q}$. So now I'll start writing my proof.
So let $\alpha \in \mathfrak{q}$. Of course, $\pi \alpha \in \pi \mathcal{O}_L$. Applying what we proved in the first part of the problem for the element $\pi \alpha$ we get $\sigma(\pi \alpha) \equiv \pi \alpha \pmod{\mathfrak{q}^{m+1}}$. That it we get: $$ \sigma(\pi \alpha) - \pi \alpha \in \mathfrak{q}^{m+1}. \tag{1} $$
On the other hand, Assumption 1 implies that $\sigma(\pi) \equiv \pi\pmod{\mathfrak{q}^m}$ (since $\pi \in \mathcal{O}_L$). That is, $\sigma(\pi) - \pi \in  \mathfrak{q}^m$. By assumption though, $\alpha \in \mathfrak{q}$. Hence, $\alpha \cdot (\sigma(\pi) - \pi) \in \mathfrak{q}^{m+1}$. Hence,
$$ \alpha \sigma(\pi) - \alpha \pi \in \mathfrak{q}^{m+1}. \tag{2}$$
Combining relations $(1)$ and $(2)$  yields $\sigma(\pi \alpha) - \alpha \sigma(\pi) \in \mathfrak{q}^{m+1}$. That is $$ \sigma(\pi) \cdot \sigma(\alpha) - \sigma(\pi) \alpha \in \mathfrak{q}^{m+1}.\tag{3}$$
Now, we know that $\sigma^{-1}(\mathcal{O}_L)=\mathcal{O}_L$. Therefore $\sigma^{-1}(\pi) \in \mathcal{O}_L$. Multiplying relation $(3)$ with $\sigma^{-1}(\pi)$ (and taking of course advantage of the fact that $\mathfrak{q}^{m+1}$ is an ideal of $\mathcal{O}_L$ ) gives us: $\sigma^{-1}(\pi) \cdot (\sigma(\pi) \sigma(\alpha) - \sigma(\pi) \alpha) \in \mathfrak{q}^{m+1} $. That is $\sigma(\alpha) - \alpha \in \mathfrak{q}^{m+1}$. Of course, this is equivalent to $\sigma(\alpha) \equiv \alpha \pmod{\mathfrak{q}^{m+1}}$, which is what we wanted to prove.
If something doesn't seem right please let me know! Thanks !
 A: You wrote :

Multiplying relation $(3)$ with $\sigma^{-1}(\pi)$ (and taking of course advantage of the fact that $\mathfrak{q}^{m+1}$ is an ideal of $\mathcal{O}_L$ ) gives us: $\sigma^{-1}(\pi) \cdot (\sigma(\pi) \sigma(\alpha) - \sigma(\pi) \alpha) \in \mathfrak{q}^{m+1} $. That is $\sigma(\alpha) - \alpha \in \mathfrak{q}^{m+1}$. Of course, this is equivalent to $\sigma(\alpha) \equiv \alpha \pmod{\mathfrak{q}^{m+1}}$, which is what we wanted to prove.

Note that it is not true in general that $\sigma^{-1}(\pi) \cdot \sigma(\pi) =1$.
For example, if you take $\sigma = \operatorname{id}$, then $\sigma^{-1}(\pi) \cdot \sigma(\pi) = \pi^2 \neq 1$, so that argument doesn't work.
What you can do, following the suggestion from Marcus, is to consider the decomposition $(\pi)=\mathfrak{q} \mathfrak{j}$, with $\mathfrak{j}$ another ideal of $\mathcal{O}_L$, and take $\beta \in \mathcal{O}_E \cap (\mathfrak{j} - \mathfrak{q})$, where $\mathcal{O}_E$ is the ring of integers corresponding to the subfield of $L$ fixed by the inertia group $E$. Then $\sigma(\beta) = \beta$ by definition of $\mathcal{O}_E$ since $\sigma \in E$, and moreover $\alpha \beta \in \mathfrak{q} \mathfrak{j} = (\pi)$, so $\alpha \beta=s \pi$ and by part a) of the exercise $\sigma(\alpha \beta) \equiv \alpha \beta$, from where we obtain: $$\sigma(\alpha) \sigma (\beta) = \beta \sigma(\alpha) \equiv \alpha \beta \implies \beta ( \sigma(\alpha) - \alpha) \in \mathfrak{q}^{m+1}$$
But $\beta \not\in \mathfrak{q}$, so using a unique factorization argument we can conclude $\sigma(\alpha) - \alpha \in \mathfrak{q}^{m+1}$. It would only remain to show that $\mathcal{O}_E \cap (\mathfrak{j} - \mathfrak{q}) \neq \emptyset$.
A: Take $z\in O_L/q^{m+1}$ such that $z$ has of order $N(q)-1$ in $(O_L/q)^\times$,  then $\zeta=z^{N(q)^{m+1}}$ has order $N(q)-1$ in $(O_L/q^{m+1})^\times$.
Let $S=\{ \zeta^n, n \ge 0\}\cup 0$.
As a set $$O_L/q^{m+1}=\{ \sum_{j=0}^m s_j \pi^j, s_j\in S\}$$
If $\sigma\in V_0$ then $\sigma(q)=q$ so that $\sigma(q^{m+1})=q^{m+1}$ and $\sigma\in Aut(O_L/q^{m+1})$. Moreover $\sigma(\zeta)=\zeta$.
Therefore $\sigma(a)=a$ for all $a\in O_L/q^{m+1}$ iff $\sigma(\pi)=\pi\in O_L/q^{m+1}$.
