If G is a group of 2 x 2 matrices under matrix multiplication and a,b,c,d are integers modulo 2 , ab - bc is not equal to 0 it's order is 6?
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EDIT - "Sorry i got the problem . Problem is right ."
The above question is from Herstein algebra .I think question is wrong .Suppose i take a=1 , b=1 , c=0 , d=1 in a matrix X , it satisfies the questions condition .But X.X does not belong to group.Because X.X will contain one element that is equal to 2 i.e X will contain a=1 , b=2 , c=0 , d=1.a,b,c,d should be Integer modulo 2 , b cannot be 2.Hence closure property fails . Is the question wrong or i am missing some thing ?
See the question 24 for further details.
abstract-algebra group-theory
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up vote
0
down vote
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EDIT - "Sorry i got the problem . Problem is right ."
The above question is from Herstein algebra .I think question is wrong .Suppose i take a=1 , b=1 , c=0 , d=1 in a matrix X , it satisfies the questions condition .But X.X does not belong to group.Because X.X will contain one element that is equal to 2 i.e X will contain a=1 , b=2 , c=0 , d=1.a,b,c,d should be Integer modulo 2 , b cannot be 2.Hence closure property fails . Is the question wrong or i am missing some thing ?
See the question 24 for further details.
abstract-algebra group-theory
Why is $X^2$ not part of the group? What do you think $X^2$ is?
â Arthur
Aug 15 at 12:09
@Arthur Because X.X will contain one element that is equal to 2 i.e X will contain a=1 , b=2 , c=0 , d=1 .
â Hansie
Aug 15 at 12:11
But the entries of the matrices are modulo $2$, so $2 = 0$, and there is no problem since $X^2 = left[beginsmallmatrix1&2\0&1endsmallmatrixright] = left[beginsmallmatrix1&0\0&1endsmallmatrixright]$
â Arthur
Aug 15 at 12:12
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
EDIT - "Sorry i got the problem . Problem is right ."
The above question is from Herstein algebra .I think question is wrong .Suppose i take a=1 , b=1 , c=0 , d=1 in a matrix X , it satisfies the questions condition .But X.X does not belong to group.Because X.X will contain one element that is equal to 2 i.e X will contain a=1 , b=2 , c=0 , d=1.a,b,c,d should be Integer modulo 2 , b cannot be 2.Hence closure property fails . Is the question wrong or i am missing some thing ?
See the question 24 for further details.
abstract-algebra group-theory
EDIT - "Sorry i got the problem . Problem is right ."
The above question is from Herstein algebra .I think question is wrong .Suppose i take a=1 , b=1 , c=0 , d=1 in a matrix X , it satisfies the questions condition .But X.X does not belong to group.Because X.X will contain one element that is equal to 2 i.e X will contain a=1 , b=2 , c=0 , d=1.a,b,c,d should be Integer modulo 2 , b cannot be 2.Hence closure property fails . Is the question wrong or i am missing some thing ?
See the question 24 for further details.
abstract-algebra group-theory
edited Aug 15 at 12:13
asked Aug 15 at 12:06
Hansie
228
228
Why is $X^2$ not part of the group? What do you think $X^2$ is?
â Arthur
Aug 15 at 12:09
@Arthur Because X.X will contain one element that is equal to 2 i.e X will contain a=1 , b=2 , c=0 , d=1 .
â Hansie
Aug 15 at 12:11
But the entries of the matrices are modulo $2$, so $2 = 0$, and there is no problem since $X^2 = left[beginsmallmatrix1&2\0&1endsmallmatrixright] = left[beginsmallmatrix1&0\0&1endsmallmatrixright]$
â Arthur
Aug 15 at 12:12
add a comment |Â
Why is $X^2$ not part of the group? What do you think $X^2$ is?
â Arthur
Aug 15 at 12:09
@Arthur Because X.X will contain one element that is equal to 2 i.e X will contain a=1 , b=2 , c=0 , d=1 .
â Hansie
Aug 15 at 12:11
But the entries of the matrices are modulo $2$, so $2 = 0$, and there is no problem since $X^2 = left[beginsmallmatrix1&2\0&1endsmallmatrixright] = left[beginsmallmatrix1&0\0&1endsmallmatrixright]$
â Arthur
Aug 15 at 12:12
Why is $X^2$ not part of the group? What do you think $X^2$ is?
â Arthur
Aug 15 at 12:09
Why is $X^2$ not part of the group? What do you think $X^2$ is?
â Arthur
Aug 15 at 12:09
@Arthur Because X.X will contain one element that is equal to 2 i.e X will contain a=1 , b=2 , c=0 , d=1 .
â Hansie
Aug 15 at 12:11
@Arthur Because X.X will contain one element that is equal to 2 i.e X will contain a=1 , b=2 , c=0 , d=1 .
â Hansie
Aug 15 at 12:11
But the entries of the matrices are modulo $2$, so $2 = 0$, and there is no problem since $X^2 = left[beginsmallmatrix1&2\0&1endsmallmatrixright] = left[beginsmallmatrix1&0\0&1endsmallmatrixright]$
â Arthur
Aug 15 at 12:12
But the entries of the matrices are modulo $2$, so $2 = 0$, and there is no problem since $X^2 = left[beginsmallmatrix1&2\0&1endsmallmatrixright] = left[beginsmallmatrix1&0\0&1endsmallmatrixright]$
â Arthur
Aug 15 at 12:12
add a comment |Â
1 Answer
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$$detbeginpmatrix1&1\0&1endpmatrix^2 = detbeginpmatrix1&0\0&1endpmatrix = 1 ne 0$$
a,b,c,d should be Integer modulo 2 . b cannot be 2 .
â Hansie
Aug 15 at 12:12
@neraj That's not what "integer modulo $2$" means.
â Arthur
Aug 15 at 12:13
But if you keep multiplying X with itself several times then we can have infinite elements .Order cannot be 6 . Can you explain me what question tries to tell .
â Hansie
Aug 15 at 12:14
Sorry i got the problem . Problem is right .
â Hansie
Aug 15 at 12:16
@neraj Yes, you can keep multiplying to get $X, X^2, X^3, X^4,ldots$, but it just so happens that $X^3 = X$ (since $3 = 1$ in the integers modulo $2$), so what you really get as you keep multiplying is $X, X^2, X, X^2, ldots$, which is not infinitely many different elements.
â Arthur
Aug 15 at 12:16
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$$detbeginpmatrix1&1\0&1endpmatrix^2 = detbeginpmatrix1&0\0&1endpmatrix = 1 ne 0$$
a,b,c,d should be Integer modulo 2 . b cannot be 2 .
â Hansie
Aug 15 at 12:12
@neraj That's not what "integer modulo $2$" means.
â Arthur
Aug 15 at 12:13
But if you keep multiplying X with itself several times then we can have infinite elements .Order cannot be 6 . Can you explain me what question tries to tell .
â Hansie
Aug 15 at 12:14
Sorry i got the problem . Problem is right .
â Hansie
Aug 15 at 12:16
@neraj Yes, you can keep multiplying to get $X, X^2, X^3, X^4,ldots$, but it just so happens that $X^3 = X$ (since $3 = 1$ in the integers modulo $2$), so what you really get as you keep multiplying is $X, X^2, X, X^2, ldots$, which is not infinitely many different elements.
â Arthur
Aug 15 at 12:16
add a comment |Â
up vote
1
down vote
accepted
$$detbeginpmatrix1&1\0&1endpmatrix^2 = detbeginpmatrix1&0\0&1endpmatrix = 1 ne 0$$
a,b,c,d should be Integer modulo 2 . b cannot be 2 .
â Hansie
Aug 15 at 12:12
@neraj That's not what "integer modulo $2$" means.
â Arthur
Aug 15 at 12:13
But if you keep multiplying X with itself several times then we can have infinite elements .Order cannot be 6 . Can you explain me what question tries to tell .
â Hansie
Aug 15 at 12:14
Sorry i got the problem . Problem is right .
â Hansie
Aug 15 at 12:16
@neraj Yes, you can keep multiplying to get $X, X^2, X^3, X^4,ldots$, but it just so happens that $X^3 = X$ (since $3 = 1$ in the integers modulo $2$), so what you really get as you keep multiplying is $X, X^2, X, X^2, ldots$, which is not infinitely many different elements.
â Arthur
Aug 15 at 12:16
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$$detbeginpmatrix1&1\0&1endpmatrix^2 = detbeginpmatrix1&0\0&1endpmatrix = 1 ne 0$$
$$detbeginpmatrix1&1\0&1endpmatrix^2 = detbeginpmatrix1&0\0&1endpmatrix = 1 ne 0$$
answered Aug 15 at 12:10
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Kenny Lau
18.9k2157
18.9k2157
a,b,c,d should be Integer modulo 2 . b cannot be 2 .
â Hansie
Aug 15 at 12:12
@neraj That's not what "integer modulo $2$" means.
â Arthur
Aug 15 at 12:13
But if you keep multiplying X with itself several times then we can have infinite elements .Order cannot be 6 . Can you explain me what question tries to tell .
â Hansie
Aug 15 at 12:14
Sorry i got the problem . Problem is right .
â Hansie
Aug 15 at 12:16
@neraj Yes, you can keep multiplying to get $X, X^2, X^3, X^4,ldots$, but it just so happens that $X^3 = X$ (since $3 = 1$ in the integers modulo $2$), so what you really get as you keep multiplying is $X, X^2, X, X^2, ldots$, which is not infinitely many different elements.
â Arthur
Aug 15 at 12:16
add a comment |Â
a,b,c,d should be Integer modulo 2 . b cannot be 2 .
â Hansie
Aug 15 at 12:12
@neraj That's not what "integer modulo $2$" means.
â Arthur
Aug 15 at 12:13
But if you keep multiplying X with itself several times then we can have infinite elements .Order cannot be 6 . Can you explain me what question tries to tell .
â Hansie
Aug 15 at 12:14
Sorry i got the problem . Problem is right .
â Hansie
Aug 15 at 12:16
@neraj Yes, you can keep multiplying to get $X, X^2, X^3, X^4,ldots$, but it just so happens that $X^3 = X$ (since $3 = 1$ in the integers modulo $2$), so what you really get as you keep multiplying is $X, X^2, X, X^2, ldots$, which is not infinitely many different elements.
â Arthur
Aug 15 at 12:16
a,b,c,d should be Integer modulo 2 . b cannot be 2 .
â Hansie
Aug 15 at 12:12
a,b,c,d should be Integer modulo 2 . b cannot be 2 .
â Hansie
Aug 15 at 12:12
@neraj That's not what "integer modulo $2$" means.
â Arthur
Aug 15 at 12:13
@neraj That's not what "integer modulo $2$" means.
â Arthur
Aug 15 at 12:13
But if you keep multiplying X with itself several times then we can have infinite elements .Order cannot be 6 . Can you explain me what question tries to tell .
â Hansie
Aug 15 at 12:14
But if you keep multiplying X with itself several times then we can have infinite elements .Order cannot be 6 . Can you explain me what question tries to tell .
â Hansie
Aug 15 at 12:14
Sorry i got the problem . Problem is right .
â Hansie
Aug 15 at 12:16
Sorry i got the problem . Problem is right .
â Hansie
Aug 15 at 12:16
@neraj Yes, you can keep multiplying to get $X, X^2, X^3, X^4,ldots$, but it just so happens that $X^3 = X$ (since $3 = 1$ in the integers modulo $2$), so what you really get as you keep multiplying is $X, X^2, X, X^2, ldots$, which is not infinitely many different elements.
â Arthur
Aug 15 at 12:16
@neraj Yes, you can keep multiplying to get $X, X^2, X^3, X^4,ldots$, but it just so happens that $X^3 = X$ (since $3 = 1$ in the integers modulo $2$), so what you really get as you keep multiplying is $X, X^2, X, X^2, ldots$, which is not infinitely many different elements.
â Arthur
Aug 15 at 12:16
add a comment |Â
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Why is $X^2$ not part of the group? What do you think $X^2$ is?
â Arthur
Aug 15 at 12:09
@Arthur Because X.X will contain one element that is equal to 2 i.e X will contain a=1 , b=2 , c=0 , d=1 .
â Hansie
Aug 15 at 12:11
But the entries of the matrices are modulo $2$, so $2 = 0$, and there is no problem since $X^2 = left[beginsmallmatrix1&2\0&1endsmallmatrixright] = left[beginsmallmatrix1&0\0&1endsmallmatrixright]$
â Arthur
Aug 15 at 12:12