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












2














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 $Gamma_{i+1}$.






abstract-algebra group-theory profinite-groups pro-p-groups







share|cite|improve this question






















  • 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






  • 1




    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






  • 2




    $Gamma_i$ is a pro-$p$-group (for $ige 1$, 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
















2














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 $Gamma_{i+1}$.






abstract-algebra group-theory profinite-groups pro-p-groups







share|cite|improve this question






















  • 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






  • 1




    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






  • 2




    $Gamma_i$ is a pro-$p$-group (for $ige 1$, 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














2












2








2







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 $Gamma_{i+1}$.






abstract-algebra group-theory profinite-groups pro-p-groups







share|cite|improve this question













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 $Gamma_{i+1}$.






abstract-algebra group-theory profinite-groups pro-p-groups




abstract-algebra group-theory profinite-groups pro-p-groups






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Oct 13 '18 at 1:51









MathStudent

573420




573420












  • 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






  • 1




    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






  • 2




    $Gamma_i$ is a pro-$p$-group (for $ige 1$, 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


















  • 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






  • 1




    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






  • 2




    $Gamma_i$ is a pro-$p$-group (for $ige 1$, 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
















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




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




Can you explain a bit more why its true?
– MathStudent
Oct 13 '18 at 2:25




1




1




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






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




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




2




2




$Gamma_i$ is a pro-$p$-group (for $ige 1$, 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




$Gamma_i$ is a pro-$p$-group (for $ige 1$, 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















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