# Index of congruence subgroups in $GL_2(\mathbb{Z}_p)$ modulo their centers.

Let $$\Gamma_i$$ be the set of matrices in $$GL_2(\mathbb{Z}_p)$$ which are congruent to $$1$$ modulo $$p^i$$, that is they are the congruence subgroups. I know that $$\Gamma_i$$ is a pro-$$p$$ group and $$\Gamma_i/\Gamma_{i+1}$$ has cardinality $$p^4$$.

Is $$\Gamma_{i}/Z(\Gamma_{i})$$ again a pro-$$p$$ group?

Also, I would like to know the index of $$\Gamma_{i+1}/Z(\Gamma_{i+1})$$ inside $$\Gamma_{i}/Z(\Gamma_{i})$$ where $$Z(\Gamma_{i+1})$$ is the center of $$\Gamma_{i+1}$$.

• If you take the quotient of $\Gamma_i/Z(\Gamma_i)$ by a normal open subgroup, isn't it isomorphic to a quotient of $\Gamma_i$ by a normal open subgroup ? – AlexL Oct 13 '18 at 2:02
• Can you explain a bit more why its true? – MathStudent Oct 13 '18 at 2:25
• Let $G=\Gamma_i$, $H=\Gamma_i / Z(\Gamma_i)$ and $p:G \to H$ the projection. For every open normal subgroup $V$ of $H$, you can show that $U=p^{-1}(V)$ is an open normal subgroup of $G$ containing $Z(\Gamma_i)$. So one one hand, $G/U$ is a $p$-group, and on the other hand, $G/U \simeq H/V$. – AlexL Oct 13 '18 at 2:32
• In my case, when $V=\Gamma_{i+1}/Z(\Gamma_{i+1})$, is $U=Z(\Gamma_i)\Gamma_{i+1}$? – MathStudent Oct 13 '18 at 2:46
• $\Gamma_i$ is a pro-$p$-group (for $i\ge 1$, not for $i=0$). A quotient of a pro-$p$-group by a closed normal subgroup is still a pro-$p$-group. (The center of $\Gamma_i$ is not open.) – YCor Oct 13 '18 at 23:51