# Index of the congruence subgroups of $PGL_2(\mathbb{Z}_p)$

Let $$\Gamma_n$$ be the $$n$$-th congruence subgroup of $$GL(2,\mathbb{Z}_p)$$. So $$\Gamma_n$$ consists of matrices in $$GL(2,\mathbb{Z}_p)$$ which are congruent to the identity matrix modulo $$p^n$$. Let $$Z(\Gamma_n)$$ the the center of $$\Gamma_n$$.

My question is to compute the index $$[\Gamma_n/Z(\Gamma_n):\Gamma_{n+1}/Z(\Gamma_{n+1})]$$

I know that the index of $$[\Gamma_n:\Gamma_{n+1}]=p^4$$. I also know that if $$G_n$$ is the $$n$$-th congruence subgroup of $$SL(2,\mathbb{Z}_p)$$, then the index of $$[G_n:G_{n+1}]$$ is probably $$p^3$$

(because the topological generators of $$G_n$$ are $$1+p^nE_{2,1}, 1+p^nE_{1,2}$$ and $$(1+p^n)E_{1,1}+(1+p^n)^{-1}E_{2,2}$$).

Here $$E_{i,j}$$ is the elementary matrix having $$1$$ at $$(i,j)$$-th place and $$0$$ elsewhere.

I can't figure out the the index $$[\Gamma_n/Z(\Gamma_n):\Gamma_{n+1}/Z(\Gamma_{n+1})]$$, that is the index of the congruence subgroups of $$PGL_2(\mathbb{Z}_p)$$.

Any help is welcome. Thanks in advance.

• (Better asked on Math Stack Exchange...) – paul garrett Jan 22 '19 at 20:59
• Garrett: I asked on Math Stack Exchange on Oct 13, without any reply. Here is the link math.stackexchange.com/questions/2953359/… – MathStudent Jan 22 '19 at 21:21
• The centre of a finite index subgroup $\Gamma$ of $GL(2,{\mathbb Z}_p)$ is the intersection of $\Gamma$ with the group of scalar matrices since $\Gamma$ is Zariski dense in $GL(2)$ ). If $p$ is an odd prime, then $\Gamma _n$ is the product of scalars in $\Gamma _n$ and the intersection $\Delta_n$ of $\Gamma _n$ with $SL(2,{\mathbb Z}_p)$. This will reduce the computation to the index $[\Delta _n:\Delta _{n+1}]$ and that is $p^3$. For odd primes, $[\Gamma _n:\Gamma _{n+1}]$ should be $p^4$. – Venkataramana Jan 23 '19 at 1:43
• It's true, but not obvious, that $\Gamma_n/\operatorname Z(\Gamma_n)$ is a subgroup of $\operatorname{PGL}_2(\mathbb Z_p)$. (A priori it need not embed in any quotient of $\operatorname{GL}_2(\mathbb Z_p)$.) Also true is that $\operatorname Z(\Gamma_n) \cap \Gamma_{n + 1}$ equals $\operatorname Z(\Gamma_{n + 1})$, so that you're just computing $[\Gamma_n : \Gamma_{n + 1}]\cdot[\operatorname Z(\Gamma_n) : \operatorname Z(\Gamma_{n + 1})]^{-1}$. – LSpice Jan 24 '19 at 20:25
• It may help to realise that most of these computations can be reduced to the Lie algebra; $X \mapsto 1 + X$ induces an isomorphism of $\operatorname{Lie}(\Gamma_n)/\operatorname{Lie}(\Gamma_{n + 1})$ with $\Gamma_n/\Gamma_{n + 1}$, with the hopefully obvious meaning of the notation. – LSpice Jan 24 '19 at 20:26