Mixing
Abstract algebra: Proving the matrix is a subgroup
$begingroup$
I'm confusing about this specific problem:
Here, $Gamma$ is a group of 2x2 - matrices with integer entries, with respect to the usual matrix multiplication, and $det(Gamma) = 1$.
But will the identity $I = begin{bmatrix}1&0\0&1end{bmatrix}$ fail to prove when $n = 1$ and therefore, $I notin Gamma_n$ ? (Because $1 equiv 0 pmod{1}$)
Please, correct me if I'm mistaken.
abstract-algebra matrices
asked Jan 25 at 6:17
Thai DoanThai Doan
487
$endgroup$
|
show 1 more comment
$begingroup$
I'm confusing about this specific problem:
Here, $Gamma$ is a group of 2x2 - matrices with integer entries, with respect to the usual matrix multiplication, and $det(Gamma) = 1$.
But will the identity $I = begin{bmatrix}1&0\0&1end{bmatrix}$ fail to prove when $n = 1$ and therefore, $I notin Gamma_n$ ? (Because $1 equiv 0 pmod{1}$)
Please, correct me if I'm mistaken.
abstract-algebra matrices
asked Jan 25 at 6:17
Thai DoanThai Doan
487
$endgroup$
$begingroup$
The set of all $2 times 2$-matrices is not a group with respect to matrix multiplication. You probably want $Gamma$ to be the group of invertible $2 times 2$-matrices over the integers.
$endgroup$
– Matthias Klupsch
Jan 25 at 6:23
2
$begingroup$
@hardmath : It seems to me the set $Gamma_1$ is not the trivial subgroup but the whole $Gamma$, since the conditions are trivially satisfied.
$endgroup$
– Matthias Klupsch
Jan 25 at 6:25
1
$begingroup$
Regarding your problem of proving $I in Gamma_1$, ask yourself: is $1 equiv 1 $ mod $1$ and is $0 equiv 0 $ mod $1$? This is all you need in order to have $I in Gamma_1$.
$endgroup$
– Matthias Klupsch
Jan 25 at 6:28
$begingroup$
@MatthiasKlupsch sorry for missing an important info, but it is invertible since $det(Gamma) = 1$
$endgroup$
– Thai Doan
Jan 25 at 6:29
$begingroup$
@Matthias: Yes, you are right. It's not clear to me what group $Gamma$ is, perhaps an additive group or a multiplicative group of matrices. I guess it doesn't matter as far as case $n=1$ goes.
$endgroup$
– hardmath
Jan 25 at 6:34
|
show 1 more comment
$begingroup$
I'm confusing about this specific problem:
Here, $Gamma$ is a group of 2x2 - matrices with integer entries, with respect to the usual matrix multiplication, and $det(Gamma) = 1$.
But will the identity $I = begin{bmatrix}1&0\0&1end{bmatrix}$ fail to prove when $n = 1$ and therefore, $I notin Gamma_n$ ? (Because $1 equiv 0 pmod{1}$)
Please, correct me if I'm mistaken.
abstract-algebra matrices
asked Jan 25 at 6:17
Thai DoanThai Doan
487
$endgroup$
I'm confusing about this specific problem:
Here, $Gamma$ is a group of 2x2 - matrices with integer entries, with respect to the usual matrix multiplication, and $det(Gamma) = 1$.
But will the identity $I = begin{bmatrix}1&0\0&1end{bmatrix}$ fail to prove when $n = 1$ and therefore, $I notin Gamma_n$ ? (Because $1 equiv 0 pmod{1}$)
Please, correct me if I'm mistaken.
abstract-algebra matrices
abstract-algebra matrices
asked Jan 25 at 6:17
Thai DoanThai Doan
487
asked Jan 25 at 6:17
Thai DoanThai Doan
487
edited Jan 25 at 6:39
Thai Doan
asked Jan 25 at 6:17
Thai DoanThai Doan
487
asked Jan 25 at 6:17
Thai DoanThai Doan
487
asked Jan 25 at 6:17
Thai DoanThai Doan
487
487
$begingroup$
The set of all $2 times 2$-matrices is not a group with respect to matrix multiplication. You probably want $Gamma$ to be the group of invertible $2 times 2$-matrices over the integers.
$endgroup$
– Matthias Klupsch
Jan 25 at 6:23
2
$begingroup$
@hardmath : It seems to me the set $Gamma_1$ is not the trivial subgroup but the whole $Gamma$, since the conditions are trivially satisfied.
$endgroup$
– Matthias Klupsch
Jan 25 at 6:25
1
$begingroup$
Regarding your problem of proving $I in Gamma_1$, ask yourself: is $1 equiv 1 $ mod $1$ and is $0 equiv 0 $ mod $1$? This is all you need in order to have $I in Gamma_1$.
$endgroup$
– Matthias Klupsch
Jan 25 at 6:28
$begingroup$
@MatthiasKlupsch sorry for missing an important info, but it is invertible since $det(Gamma) = 1$
$endgroup$
– Thai Doan
Jan 25 at 6:29
$begingroup$
@Matthias: Yes, you are right. It's not clear to me what group $Gamma$ is, perhaps an additive group or a multiplicative group of matrices. I guess it doesn't matter as far as case $n=1$ goes.
$endgroup$
– hardmath
Jan 25 at 6:34
|
show 1 more comment
$begingroup$
The set of all $2 times 2$-matrices is not a group with respect to matrix multiplication. You probably want $Gamma$ to be the group of invertible $2 times 2$-matrices over the integers.
$endgroup$
– Matthias Klupsch
Jan 25 at 6:23
2
$begingroup$
@hardmath : It seems to me the set $Gamma_1$ is not the trivial subgroup but the whole $Gamma$, since the conditions are trivially satisfied.
$endgroup$
– Matthias Klupsch
Jan 25 at 6:25
1
$begingroup$
Regarding your problem of proving $I in Gamma_1$, ask yourself: is $1 equiv 1 $ mod $1$ and is $0 equiv 0 $ mod $1$? This is all you need in order to have $I in Gamma_1$.
$endgroup$
– Matthias Klupsch
Jan 25 at 6:28
$begingroup$
@MatthiasKlupsch sorry for missing an important info, but it is invertible since $det(Gamma) = 1$
$endgroup$
– Thai Doan
Jan 25 at 6:29
$begingroup$
@Matthias: Yes, you are right. It's not clear to me what group $Gamma$ is, perhaps an additive group or a multiplicative group of matrices. I guess it doesn't matter as far as case $n=1$ goes.
$endgroup$
– hardmath
Jan 25 at 6:34
$begingroup$
The set of all $2 times 2$-matrices is not a group with respect to matrix multiplication. You probably want $Gamma$ to be the group of invertible $2 times 2$-matrices over the integers.
$endgroup$
– Matthias Klupsch
Jan 25 at 6:23
$begingroup$
The set of all $2 times 2$-matrices is not a group with respect to matrix multiplication. You probably want $Gamma$ to be the group of invertible $2 times 2$-matrices over the integers.
$endgroup$
– Matthias Klupsch
Jan 25 at 6:23
2
2
$begingroup$
@hardmath : It seems to me the set $Gamma_1$ is not the trivial subgroup but the whole $Gamma$, since the conditions are trivially satisfied.
$endgroup$
– Matthias Klupsch
Jan 25 at 6:25
$begingroup$
@hardmath : It seems to me the set $Gamma_1$ is not the trivial subgroup but the whole $Gamma$, since the conditions are trivially satisfied.
$endgroup$
– Matthias Klupsch
Jan 25 at 6:25
1
1
$begingroup$
Regarding your problem of proving $I in Gamma_1$, ask yourself: is $1 equiv 1 $ mod $1$ and is $0 equiv 0 $ mod $1$? This is all you need in order to have $I in Gamma_1$.
$endgroup$
– Matthias Klupsch
Jan 25 at 6:28
$begingroup$
Regarding your problem of proving $I in Gamma_1$, ask yourself: is $1 equiv 1 $ mod $1$ and is $0 equiv 0 $ mod $1$? This is all you need in order to have $I in Gamma_1$.
$endgroup$
– Matthias Klupsch
Jan 25 at 6:28
$begingroup$
@MatthiasKlupsch sorry for missing an important info, but it is invertible since $det(Gamma) = 1$
$endgroup$
– Thai Doan
Jan 25 at 6:29
$begingroup$
@MatthiasKlupsch sorry for missing an important info, but it is invertible since $det(Gamma) = 1$
$endgroup$
– Thai Doan
Jan 25 at 6:29
$begingroup$
@Matthias: Yes, you are right. It's not clear to me what group $Gamma$ is, perhaps an additive group or a multiplicative group of matrices. I guess it doesn't matter as far as case $n=1$ goes.
$endgroup$
– hardmath
Jan 25 at 6:34
$begingroup$
@Matthias: Yes, you are right. It's not clear to me what group $Gamma$ is, perhaps an additive group or a multiplicative group of matrices. I guess it doesn't matter as far as case $n=1$ goes.
$endgroup$
– hardmath
Jan 25 at 6:34
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
The excerpt explains for you that $Gamma_2 $ will be the matrices with main diagonal entries odd and minor diagonal entries even. ( This is another way of saying $aequiv dequiv1pmod2$ and $bequiv cequiv 0pmod2$).
Thus the identity, $I=begin{pmatrix} 1&0\0&1end{pmatrix}$ is an element of $Gamma_2 $.
answered Jan 25 at 7:05
Chris CusterChris Custer
14.2k3827
$endgroup$
$begingroup$
Thanks for your help, but it doesn't follow that $I in Gamma_n$ for any positive integers $n$
$endgroup$
– Thai Doan
Jan 25 at 7:11
1
$begingroup$
It actually does. For any $n$ , we have $1equiv 1pmod n$ and $0equiv0pmod n$.
$endgroup$
– Chris Custer
Jan 25 at 7:17
$begingroup$
Nice! I have upvoted your post and thank you for pointing out $1 equiv 1 pmod{n}$.
$endgroup$
– Thai Doan
Jan 25 at 7:29
add a comment |
$begingroup$
My problem is I was mistaken about the "congruence": $1 equiv 1 pmod{1}$. They are both in the same equivalence class for 0.
answered Jan 25 at 7:26
Thai DoanThai Doan
487
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
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oldest
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active
oldest
votes
$begingroup$
The excerpt explains for you that $Gamma_2 $ will be the matrices with main diagonal entries odd and minor diagonal entries even. ( This is another way of saying $aequiv dequiv1pmod2$ and $bequiv cequiv 0pmod2$).
Thus the identity, $I=begin{pmatrix} 1&0\0&1end{pmatrix}$ is an element of $Gamma_2 $.
answered Jan 25 at 7:05
Chris CusterChris Custer
14.2k3827
$endgroup$
$begingroup$
Thanks for your help, but it doesn't follow that $I in Gamma_n$ for any positive integers $n$
$endgroup$
– Thai Doan
Jan 25 at 7:11
1
$begingroup$
It actually does. For any $n$ , we have $1equiv 1pmod n$ and $0equiv0pmod n$.
$endgroup$
– Chris Custer
Jan 25 at 7:17
$begingroup$
Nice! I have upvoted your post and thank you for pointing out $1 equiv 1 pmod{n}$.
$endgroup$
– Thai Doan
Jan 25 at 7:29
add a comment |
$begingroup$
The excerpt explains for you that $Gamma_2 $ will be the matrices with main diagonal entries odd and minor diagonal entries even. ( This is another way of saying $aequiv dequiv1pmod2$ and $bequiv cequiv 0pmod2$).
Thus the identity, $I=begin{pmatrix} 1&0\0&1end{pmatrix}$ is an element of $Gamma_2 $.
answered Jan 25 at 7:05
Chris CusterChris Custer
14.2k3827
$endgroup$
$begingroup$
Thanks for your help, but it doesn't follow that $I in Gamma_n$ for any positive integers $n$
$endgroup$
– Thai Doan
Jan 25 at 7:11
1
$begingroup$
It actually does. For any $n$ , we have $1equiv 1pmod n$ and $0equiv0pmod n$.
$endgroup$
– Chris Custer
Jan 25 at 7:17
$begingroup$
Nice! I have upvoted your post and thank you for pointing out $1 equiv 1 pmod{n}$.
$endgroup$
– Thai Doan
Jan 25 at 7:29
add a comment |
$begingroup$
The excerpt explains for you that $Gamma_2 $ will be the matrices with main diagonal entries odd and minor diagonal entries even. ( This is another way of saying $aequiv dequiv1pmod2$ and $bequiv cequiv 0pmod2$).
Thus the identity, $I=begin{pmatrix} 1&0\0&1end{pmatrix}$ is an element of $Gamma_2 $.
answered Jan 25 at 7:05
Chris CusterChris Custer
14.2k3827
$endgroup$
The excerpt explains for you that $Gamma_2 $ will be the matrices with main diagonal entries odd and minor diagonal entries even. ( This is another way of saying $aequiv dequiv1pmod2$ and $bequiv cequiv 0pmod2$).
Thus the identity, $I=begin{pmatrix} 1&0\0&1end{pmatrix}$ is an element of $Gamma_2 $.
answered Jan 25 at 7:05
Chris CusterChris Custer
14.2k3827
answered Jan 25 at 7:05
Chris CusterChris Custer
14.2k3827
answered Jan 25 at 7:05
Chris CusterChris Custer
14.2k3827
answered Jan 25 at 7:05
Chris CusterChris Custer
14.2k3827
14.2k3827
$begingroup$
Thanks for your help, but it doesn't follow that $I in Gamma_n$ for any positive integers $n$
$endgroup$
– Thai Doan
Jan 25 at 7:11
1
$begingroup$
It actually does. For any $n$ , we have $1equiv 1pmod n$ and $0equiv0pmod n$.
$endgroup$
– Chris Custer
Jan 25 at 7:17
$begingroup$
Nice! I have upvoted your post and thank you for pointing out $1 equiv 1 pmod{n}$.
$endgroup$
– Thai Doan
Jan 25 at 7:29
add a comment |
$begingroup$
Thanks for your help, but it doesn't follow that $I in Gamma_n$ for any positive integers $n$
$endgroup$
– Thai Doan
Jan 25 at 7:11
1
$begingroup$
It actually does. For any $n$ , we have $1equiv 1pmod n$ and $0equiv0pmod n$.
$endgroup$
– Chris Custer
Jan 25 at 7:17
$begingroup$
Nice! I have upvoted your post and thank you for pointing out $1 equiv 1 pmod{n}$.
$endgroup$
– Thai Doan
Jan 25 at 7:29
$begingroup$
Thanks for your help, but it doesn't follow that $I in Gamma_n$ for any positive integers $n$
$endgroup$
– Thai Doan
Jan 25 at 7:11
$begingroup$
Thanks for your help, but it doesn't follow that $I in Gamma_n$ for any positive integers $n$
$endgroup$
– Thai Doan
Jan 25 at 7:11
1
1
$begingroup$
It actually does. For any $n$ , we have $1equiv 1pmod n$ and $0equiv0pmod n$.
$endgroup$
– Chris Custer
Jan 25 at 7:17
$begingroup$
It actually does. For any $n$ , we have $1equiv 1pmod n$ and $0equiv0pmod n$.
$endgroup$
– Chris Custer
Jan 25 at 7:17
$begingroup$
Nice! I have upvoted your post and thank you for pointing out $1 equiv 1 pmod{n}$.
$endgroup$
– Thai Doan
Jan 25 at 7:29
$begingroup$
Nice! I have upvoted your post and thank you for pointing out $1 equiv 1 pmod{n}$.
$endgroup$
– Thai Doan
Jan 25 at 7:29
add a comment |
$begingroup$
My problem is I was mistaken about the "congruence": $1 equiv 1 pmod{1}$. They are both in the same equivalence class for 0.
answered Jan 25 at 7:26
Thai DoanThai Doan
487
$endgroup$
add a comment |
$begingroup$
My problem is I was mistaken about the "congruence": $1 equiv 1 pmod{1}$. They are both in the same equivalence class for 0.
answered Jan 25 at 7:26
Thai DoanThai Doan
487
$endgroup$
add a comment |
$begingroup$
My problem is I was mistaken about the "congruence": $1 equiv 1 pmod{1}$. They are both in the same equivalence class for 0.
answered Jan 25 at 7:26
Thai DoanThai Doan
487
$endgroup$
My problem is I was mistaken about the "congruence": $1 equiv 1 pmod{1}$. They are both in the same equivalence class for 0.
answered Jan 25 at 7:26
Thai DoanThai Doan
487
answered Jan 25 at 7:26
Thai DoanThai Doan
487
answered Jan 25 at 7:26
Thai DoanThai Doan
487
answered Jan 25 at 7:26
Thai DoanThai Doan
487
487
add a comment |
add a comment |
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$begingroup$
The set of all $2 times 2$-matrices is not a group with respect to matrix multiplication. You probably want $Gamma$ to be the group of invertible $2 times 2$-matrices over the integers.
$endgroup$
– Matthias Klupsch
Jan 25 at 6:23
2
$begingroup$
@hardmath : It seems to me the set $Gamma_1$ is not the trivial subgroup but the whole $Gamma$, since the conditions are trivially satisfied.
$endgroup$
– Matthias Klupsch
Jan 25 at 6:25
1
$begingroup$
Regarding your problem of proving $I in Gamma_1$, ask yourself: is $1 equiv 1 $ mod $1$ and is $0 equiv 0 $ mod $1$? This is all you need in order to have $I in Gamma_1$.
$endgroup$
– Matthias Klupsch
Jan 25 at 6:28
$begingroup$
@MatthiasKlupsch sorry for missing an important info, but it is invertible since $det(Gamma) = 1$
$endgroup$
– Thai Doan
Jan 25 at 6:29
$begingroup$
@Matthias: Yes, you are right. It's not clear to me what group $Gamma$ is, perhaps an additive group or a multiplicative group of matrices. I guess it doesn't matter as far as case $n=1$ goes.
$endgroup$
– hardmath
Jan 25 at 6:34