Is $GL_2(\mathbb{Z}_p)$ a maximal compact subgroup of $GL_2(\mathbb{Q}_p)$?

Question

Let $\mathbb{Q}_p$ be the set of all p-adic numbers and $\mathbb{Z}_p$ the set of all p-adic integers.

Is $GL_2(\mathbb{Z}_p)$ a maximal compact subgroup of $GL_2(\mathbb{Q}_p)$? Thank you very much.

Answer 1

Yes, $GL_n(\mathbf{Z}_p)$ is a maximal compact subgroup of $GL_n(\mathbf{Q}_p)$ for any $n$. This follows immediately from the Smith normal form (elementary divisors) theorem: any $M \in GL_2(\mathbf{Q}_p)$ can be written as $M = U D V$ where $U, V$ are in $GL_n(\mathbf{Z}_p)$ and $D$ is diagonal, with its diagonal entries all powers of $p$. So any subgroup strictly containing $GL_n(\mathbf{Z}_p)$ must contain a diagonal matrix with some entry a nontrivial power of $p$, and the subgroup generated by such an element is not compact.

(It's a general fact that if $K$ is a finite extension of $\mathbf{Q}_p$ with ring of integers $\mathcal{O}$, and $\mathcal{G}$ is a reductive group scheme over $\mathcal{O}$, then $\mathcal{G}(\mathcal{O})$ is maximal compact in $\mathcal{G}(K)$.)