Principal congruence subgroup index in $SL(2,mathbb{Z})$
$begingroup$
Why has the principal congruence subgroup,
begin{equation}
Gamma(N)~=~Bigg{begin{pmatrix} a & b \ c & d end{pmatrix}in SL(2,mathbb{Z})~|~aequiv dequiv 1 ~text{and}~ bequiv cequiv 0~ text{mod}~NBigg}
end{equation}
in $SL(2,mathbb{Z})=Gamma(1)$ finite index,
begin{equation}
big[Gamma(1):Gamma(N)big]=varphi(N),
end{equation}
when $Gamma(N)$ and $SL(2,mathbb{Z})$ has infinite order, and why equal the index to the Euler-function?
group-theory
asked May 21 '15 at 12:22
AlízAlíz
428
$endgroup$
add a comment |
$begingroup$
Why has the principal congruence subgroup,
begin{equation}
Gamma(N)~=~Bigg{begin{pmatrix} a & b \ c & d end{pmatrix}in SL(2,mathbb{Z})~|~aequiv dequiv 1 ~text{and}~ bequiv cequiv 0~ text{mod}~NBigg}
end{equation}
in $SL(2,mathbb{Z})=Gamma(1)$ finite index,
begin{equation}
big[Gamma(1):Gamma(N)big]=varphi(N),
end{equation}
when $Gamma(N)$ and $SL(2,mathbb{Z})$ has infinite order, and why equal the index to the Euler-function?
group-theory
asked May 21 '15 at 12:22
AlízAlíz
428
$endgroup$
2
$begingroup$
It's the kernel of a homomorphism onto the finite group ${rm SL}(2,{mathbb Z}/N{mathbb Z})$, so it has finite index. I don't believe your claim about the index.
$endgroup$
– Derek Holt
May 21 '15 at 12:31
$begingroup$
So it isn't true: $big[Gamma(1):Gamma(N)big]=varphi(N)$ isn't it?
$endgroup$
– Alíz
May 21 '15 at 12:36
2
$begingroup$
The index is $|{rm SL}(2,{mathbb Z}/N{mathbb Z})| = N^3prod_{p|N}(1-frac{1}{p^2})$ with $p$ prime. See people.reed.edu/~jerry/332/mfms.pdf
$endgroup$
– Derek Holt
May 21 '15 at 15:50
$begingroup$
Thank you for the answer! :)
$endgroup$
– Alíz
May 21 '15 at 16:17
add a comment |
$begingroup$
Why has the principal congruence subgroup,
begin{equation}
Gamma(N)~=~Bigg{begin{pmatrix} a & b \ c & d end{pmatrix}in SL(2,mathbb{Z})~|~aequiv dequiv 1 ~text{and}~ bequiv cequiv 0~ text{mod}~NBigg}
end{equation}
in $SL(2,mathbb{Z})=Gamma(1)$ finite index,
begin{equation}
big[Gamma(1):Gamma(N)big]=varphi(N),
end{equation}
when $Gamma(N)$ and $SL(2,mathbb{Z})$ has infinite order, and why equal the index to the Euler-function?
group-theory
asked May 21 '15 at 12:22
AlízAlíz
428
$endgroup$
Why has the principal congruence subgroup,
begin{equation}
Gamma(N)~=~Bigg{begin{pmatrix} a & b \ c & d end{pmatrix}in SL(2,mathbb{Z})~|~aequiv dequiv 1 ~text{and}~ bequiv cequiv 0~ text{mod}~NBigg}
end{equation}
in $SL(2,mathbb{Z})=Gamma(1)$ finite index,
begin{equation}
big[Gamma(1):Gamma(N)big]=varphi(N),
end{equation}
when $Gamma(N)$ and $SL(2,mathbb{Z})$ has infinite order, and why equal the index to the Euler-function?
group-theory
group-theory
asked May 21 '15 at 12:22
AlízAlíz
428
asked May 21 '15 at 12:22
AlízAlíz
428
edited Jan 8 at 14:24
Alíz
asked May 21 '15 at 12:22
AlízAlíz
428
asked May 21 '15 at 12:22
AlízAlíz
428
asked May 21 '15 at 12:22
AlízAlíz
428
428
2
$begingroup$
It's the kernel of a homomorphism onto the finite group ${rm SL}(2,{mathbb Z}/N{mathbb Z})$, so it has finite index. I don't believe your claim about the index.
$endgroup$
– Derek Holt
May 21 '15 at 12:31
$begingroup$
So it isn't true: $big[Gamma(1):Gamma(N)big]=varphi(N)$ isn't it?
$endgroup$
– Alíz
May 21 '15 at 12:36
2
$begingroup$
The index is $|{rm SL}(2,{mathbb Z}/N{mathbb Z})| = N^3prod_{p|N}(1-frac{1}{p^2})$ with $p$ prime. See people.reed.edu/~jerry/332/mfms.pdf
$endgroup$
– Derek Holt
May 21 '15 at 15:50
$begingroup$
Thank you for the answer! :)
$endgroup$
– Alíz
May 21 '15 at 16:17
add a comment |
2
$begingroup$
It's the kernel of a homomorphism onto the finite group ${rm SL}(2,{mathbb Z}/N{mathbb Z})$, so it has finite index. I don't believe your claim about the index.
$endgroup$
– Derek Holt
May 21 '15 at 12:31
$begingroup$
So it isn't true: $big[Gamma(1):Gamma(N)big]=varphi(N)$ isn't it?
$endgroup$
– Alíz
May 21 '15 at 12:36
2
$begingroup$
The index is $|{rm SL}(2,{mathbb Z}/N{mathbb Z})| = N^3prod_{p|N}(1-frac{1}{p^2})$ with $p$ prime. See people.reed.edu/~jerry/332/mfms.pdf
$endgroup$
– Derek Holt
May 21 '15 at 15:50
$begingroup$
Thank you for the answer! :)
$endgroup$
– Alíz
May 21 '15 at 16:17
2
2
$begingroup$
It's the kernel of a homomorphism onto the finite group ${rm SL}(2,{mathbb Z}/N{mathbb Z})$, so it has finite index. I don't believe your claim about the index.
$endgroup$
– Derek Holt
May 21 '15 at 12:31
$begingroup$
It's the kernel of a homomorphism onto the finite group ${rm SL}(2,{mathbb Z}/N{mathbb Z})$, so it has finite index. I don't believe your claim about the index.
$endgroup$
– Derek Holt
May 21 '15 at 12:31
$begingroup$
So it isn't true: $big[Gamma(1):Gamma(N)big]=varphi(N)$ isn't it?
$endgroup$
– Alíz
May 21 '15 at 12:36
$begingroup$
So it isn't true: $big[Gamma(1):Gamma(N)big]=varphi(N)$ isn't it?
$endgroup$
– Alíz
May 21 '15 at 12:36
2
2
$begingroup$
The index is $|{rm SL}(2,{mathbb Z}/N{mathbb Z})| = N^3prod_{p|N}(1-frac{1}{p^2})$ with $p$ prime. See people.reed.edu/~jerry/332/mfms.pdf
$endgroup$
– Derek Holt
May 21 '15 at 15:50
$begingroup$
The index is $|{rm SL}(2,{mathbb Z}/N{mathbb Z})| = N^3prod_{p|N}(1-frac{1}{p^2})$ with $p$ prime. See people.reed.edu/~jerry/332/mfms.pdf
$endgroup$
– Derek Holt
May 21 '15 at 15:50
$begingroup$
Thank you for the answer! :)
$endgroup$
– Alíz
May 21 '15 at 16:17
$begingroup$
Thank you for the answer! :)
$endgroup$
– Alíz
May 21 '15 at 16:17
add a comment |
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$begingroup$
It's the kernel of a homomorphism onto the finite group ${rm SL}(2,{mathbb Z}/N{mathbb Z})$, so it has finite index. I don't believe your claim about the index.
$endgroup$
– Derek Holt
May 21 '15 at 12:31
$begingroup$
So it isn't true: $big[Gamma(1):Gamma(N)big]=varphi(N)$ isn't it?
$endgroup$
– Alíz
May 21 '15 at 12:36
2
$begingroup$
The index is $|{rm SL}(2,{mathbb Z}/N{mathbb Z})| = N^3prod_{p|N}(1-frac{1}{p^2})$ with $p$ prime. See people.reed.edu/~jerry/332/mfms.pdf
$endgroup$
– Derek Holt
May 21 '15 at 15:50
$begingroup$
Thank you for the answer! :)
$endgroup$
– Alíz
May 21 '15 at 16:17