Normalizer of $GL_2 (\mathbb{Z}_p)$ in $GL_2(\mathbb{Q}_p)$

Question

What is the normalizer of $GL_2 (\mathbb{Z}_p)$ in $GL_2(\mathbb{Q}_p)$? the definition of the normalizer $N$ is $\{g\in GL_2(\mathbb{Q}_p): g GL_2 (\mathbb{Z}_p) g^{-1} =GL_2 (\mathbb{Z}_p) \}$. Trivially $N$ contains $GL_2 (\mathbb{Z}_p) \left(\left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right] \mathbb{Q}_p^{\times}\right)$. Do they coincide? How about for $GL_n(\mathbb{Z}_p)$?

Answer 1

Let $G=GL_2(\Bbb Q_p)$ and $ H=GL_2(\Bbb Z_p)$. If $g\in G$ normalises $H$ then $ghg^{-1}\in H$ for all $h\in H$. But that implies $gh'g^{-1}\in H'$ for all $h'\in H'$ where $H'$ is the $\Bbb Z_p$-module generated by $H$. I'm pretty sure that $H'=M_2(\Bbb Z_p)$, so the question reduces to which $g\in G$ have $gM_2(\Bbb Z_p)g^{-1}\subseteq M_2(\Bbb Z_p)$.

Multiplying $g$ by a scalar (as in your factor $\Bbb Q_p^\times$) lets us consider non-singular $g\in M_2(\Bbb Z_p)$ and we can assume that $p$ does not divide all entries of $g=\pmatrix{a&b\\c&d}$. For $h=\pmatrix{1&0\\0&0}$ then $$ghg^{-1}=\frac1{\det g}\pmatrix{ad&-ab\\cd&-cb}$$ so that $ad/\det g\in\Bbb Z_p$ etc. Repeating this for the remaining one-entry matrices gives a bunch of similar relations. I'm sure these are enough to prove that $\det g $ is a unit, but I don't have time to look into the details right now...