Let $k$ be a $\mathfrak{p}$-adic field, $K/k$ be a cyclic extension with Galois group $G = \langle \sigma \rangle$, and $U_K$ be the group of units in (the valuation ring of) $K$. One of the results of Chapter IX of Lang's Algebraic Number Theory is the computation of the Herbrand quotient $Q(G, U_K) := (U_k : N^K_kU_K)/(\ker N^K_k : (1-\sigma)U_K)$ where $\ker N^K_k$ is the kernel of the norm as a homomorphism from $U_K$ to $U_k$. Knowing the value of this quotient is useful, e.g. in computing the index $(k^* : N^K_kK^*) = [K:k]$.
The trick Lang uses is as follows (from p. 188):
Let $\{\omega_\tau\}$ be a normal basis for $K$ over $k$. After multiplying the elements of this basis by a high power of a prime element $\pi$ in $k$, we can assume that they have small absolute value. Let $M = \sum_{\tau \in G} \mathcal{O}_k\cdot \omega_\tau$.
As far as I can tell, the reason for this first step is to ensure that the exponential map is well-defined on $M$, since the power series defining it only converges on sufficiently small neighborhoods of $0$.
The $G$ acts on $M$ semilocally, with trivial decomposition group. Furthermore, $\exp M = V$ is $G$-isomorphic to $M$ (the inverse is given by the log), and $V$ is an open subgroup of the units, whence of finite index in $U_K$. Therefore $1 = Q(G, V) = Q(G, U_K)$.
It's clear to me why this implies that $V$ has finite index ($U_K$ is compact). However, I don't understand exactly why $V$ is open in $U_K$. Should this be obvious? Also, why does this depend on $G$ being cyclic?