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 $\Gamma_{i+1}$.

  • $\begingroup$ 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 ? $\endgroup$ – AlexL Oct 13 '18 at 2:02
  • $\begingroup$ Can you explain a bit more why its true? $\endgroup$ – MathStudent Oct 13 '18 at 2:25
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    $\begingroup$ 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, $G/U \simeq H/V$. $\endgroup$ – AlexL Oct 13 '18 at 2:32
  • $\begingroup$ In my case, when $V=\Gamma_{i+1}/Z(\Gamma_{i+1})$, is $U=Z(\Gamma_i)\Gamma_{i+1}$? $\endgroup$ – MathStudent Oct 13 '18 at 2:46
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    $\begingroup$ $\Gamma_i$ 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.) $\endgroup$ – YCor Oct 13 '18 at 23:51

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