Let $K$ be an abelian extension of $\mathbb{Q}$ with $[K:\mathbb{Q}] = p^m.$ Suppose $q$ is a prime $\neq p$ which is ramified in $K$. Fix a prime $Q$ of $K$ lying over $q$, and set $e = e(Q|q)$ (e is the ramification index of $Q$ over $q$).
(a) Prove that $V_1(Q|q) = {1}$ (Hint: $e$ is a power of $p$)
So far I have:
Since $K/\mathbb{Q}$ is Galois we have that $p^m = ref$ where $r$ is the number of primes and $f$ is their inertial degree. Therefore we have that $e|p^m$ and since $q$ is ramified $e>1 \Rightarrow e$ is a power of $p$.
I have $V_1(Q|q) := \{\sigma \in Gal(K/\mathbb{Q}) : \sigma (x) = x \pmod {Q^2}\ \forall x \in \mathcal{O_K}\}$ and I'm thinking that an argument based on the fact that the order of $\sigma$ has to divide the order of the group meaning it is either trivial or a power of $p$ might work but I can't quite make it work.