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 $Gamma_{i+1}$.
abstract-algebra group-theory profinite-groups pro-p-groups
asked Oct 13 '18 at 1:51
MathStudent
573420
|
show 1 more comment
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
asked Oct 13 '18 at 1:51
MathStudent
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
|
show 1 more comment
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
asked Oct 13 '18 at 1:51
MathStudent
573420
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
asked Oct 13 '18 at 1:51
MathStudent
573420
asked Oct 13 '18 at 1:51
MathStudent
573420
asked Oct 13 '18 at 1:51
MathStudent
573420
asked Oct 13 '18 at 1:51
MathStudent
573420
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
|
show 1 more comment
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
|
show 1 more comment
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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