Is $SL_n(\mathbb{Z}_p)$ profinite?

102 Views Asked by At

Serre's Galois Cohomology states that $SL_n(\mathbb{Z}_p)$ is a profinite group. How can one see this? Is there a way to identify the open normal subgroups of finite index of $SL_n(\mathbb{Z}_p)$ (when we see it with the product topology, for example)?

EDIT: I am considering the definition:

$$SL_n(\mathbb{Z}_p) = \left\{M \in \text{Mat}_{n\times n}(\mathbb{Z}_p) | \det(M) = 1\right\}$$

and the topology induced by the product topology in $\prod_{n^2} \mathbb{Z}_p$. We know it's a topological group but it's not obvious to us why we have the characterization $$SL_n(\mathbb{Z}_p) = \varprojlim SL_n(\mathbb{Z}/p^k\mathbb{Z})$$ which would surely make it the limit of finite discrete groups, and thus, profinite.