I am currently reading "Ramanujan Complexes of type $\tilde{A}_d$" by Alexander Lubotzky. There he claims, (on page 273) that $\text{GL}_d(\mathbb{Z}_p)$ is a maximal compact subgroup of $\text{GL}_d(\mathbb{Q}_p)$. I see that this is a subgroup and I proved, that $\mathbb{Z}_p$ is a compact subset of $\mathbb{Q}_p$. But why is $\text{GL}_d(\mathbb{Z}_p)$ compact in $\text{GL}_d(\mathbb{Q}_p)$?
I tried to embed $\mathbb{Q}_p$ continuously into $GL_d(\mathbb{Q}_p)$ s.t. the image of $\mathbb{Z}_p$ is $GL_d(\mathbb{Z}_p)$. Then the image would also be compact.
I also tried to use the determinant, which should still be continuous here (it's a polynomial). $\mathbb{Q}_p$ is a metric space, so it's a Hausdorff space and all points are closed. So $\{ 0 \}$ is closed in $\mathbb{Q}_p$ and the pre-image of it should be closed in $\text{GL}_d(\mathbb{Q}_p)$. Then I thought I could use a similar argument to prove that $\text{GL}_d(\mathbb{Z}_p)$ is compact in $\mathbb{Q}_p$. But this did not seem to work.
Can someone give me an argument for this?