I'm trying to understand how to calculate the Hausdorff dimension of some profinite groups and came across the following paper (here) which defines the Hausdorff dimension for a closed subgroup $H \leq G$ as $$dim(H)=\liminf_{n \rightarrow \infty} \frac{\log{ \vert H:H \cap G_n \vert }}{\log{\vert G:G_n \vert}}$$ where $ \{G_n\}_{n=0}^\infty$ is a filtration of the profinite group $G$.
The given example says that:
Let $G=\mathbb{Z}_p \bigoplus \mathbb{Z}_p$ (where $\mathbb{Z}_p$ is the ring of p-adic integers) and $H=\{0\} \bigoplus \mathbb{Z}_p$. Then if the Hausdorff dimension is computed relative to the filtration $G_n=p^n \mathbb{Z}_p \bigoplus p^n \mathbb{Z}_p$, we obtain $\dim(H)=\frac{1}{2}$.
We have defined the p-adic integers as the set of infinite sums $\sum_{i=0}^\infty a_i p^i$ for $0 \leq a_i \leq p-1$.
However, I am confused about how to calculate $\vert H:H \cap G_n \vert$ (and $\vert G:G_n \vert$.
I think that:
$G$ is the set of ordered pairs $(\sum_{i=0}^\infty a_i p^i, \sum_{i=0}^\infty b_i p^i)$ for $0 \leq a_i,b_i \leq p-1$.
$H$ is the set of ordered pairs $(0, \sum_{i=0}^\infty b_i p^i)$ for $0 \leq b_i \leq p-1$.
$G_n$ is the set of ordered pairs $(\sum_{i=n}^\infty a_i p^i, \sum_{i=n}^\infty b_i p^i)$ for $0 \leq a_i,b_i \leq p-1$.
$H \cap G_n$ is the group of ordered pairs $(0, \sum_{i=n}^\infty c_i p^i)$ for $0 \leq b_i <p-1$
And $H: H \cap G_n = \{ h(H \cap G_n):h \in H \}$
But I can't see where I go from here?