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 ?
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  • Why is $X^2$ not part of the group? What do you think $X^2$ is?
    – 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 .
    – Hansie
    Aug 15 at 12:11










  • But the entries of the matrices are modulo $2$, so $2 = 0$, and there is no problem since $X^2 = left[beginsmallmatrix1&2\0&1endsmallmatrixright] = left[beginsmallmatrix1&0\0&1endsmallmatrixright]$
    – Arthur
    Aug 15 at 12:12















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.enter image description here







share|cite|improve this question






















  • Why is $X^2$ not part of the group? What do you think $X^2$ is?
    – 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 .
    – Hansie
    Aug 15 at 12:11










  • But the entries of the matrices are modulo $2$, so $2 = 0$, and there is no problem since $X^2 = left[beginsmallmatrix1&2\0&1endsmallmatrixright] = left[beginsmallmatrix1&0\0&1endsmallmatrixright]$
    – Arthur
    Aug 15 at 12:12













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.enter image description here







share|cite|improve this question














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.enter image description here









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 15 at 12:13

























asked Aug 15 at 12:06









Hansie

228




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  • Why is $X^2$ not part of the group? What do you think $X^2$ is?
    – 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 .
    – Hansie
    Aug 15 at 12:11










  • But the entries of the matrices are modulo $2$, so $2 = 0$, and there is no problem since $X^2 = left[beginsmallmatrix1&2\0&1endsmallmatrixright] = left[beginsmallmatrix1&0\0&1endsmallmatrixright]$
    – Arthur
    Aug 15 at 12:12

















  • Why is $X^2$ not part of the group? What do you think $X^2$ is?
    – 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 .
    – Hansie
    Aug 15 at 12:11










  • But the entries of the matrices are modulo $2$, so $2 = 0$, and there is no problem since $X^2 = left[beginsmallmatrix1&2\0&1endsmallmatrixright] = left[beginsmallmatrix1&0\0&1endsmallmatrixright]$
    – Arthur
    Aug 15 at 12:12
















Why is $X^2$ not part of the group? What do you think $X^2$ is?
– Arthur
Aug 15 at 12:09





Why is $X^2$ not part of the group? What do you think $X^2$ is?
– 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 .
– 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 .
– Hansie
Aug 15 at 12:11












But the entries of the matrices are modulo $2$, so $2 = 0$, and there is no problem since $X^2 = left[beginsmallmatrix1&2\0&1endsmallmatrixright] = left[beginsmallmatrix1&0\0&1endsmallmatrixright]$
– Arthur
Aug 15 at 12:12





But the entries of the matrices are modulo $2$, so $2 = 0$, and there is no problem since $X^2 = left[beginsmallmatrix1&2\0&1endsmallmatrixright] = left[beginsmallmatrix1&0\0&1endsmallmatrixright]$
– Arthur
Aug 15 at 12:12











1 Answer
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1
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$$detbeginpmatrix1&1\0&1endpmatrix^2 = detbeginpmatrix1&0\0&1endpmatrix = 1 ne 0$$






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  • a,b,c,d should be Integer modulo 2 . b cannot be 2 .
    – Hansie
    Aug 15 at 12:12










  • @neraj That's not what "integer modulo $2$" means.
    – 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 .
    – Hansie
    Aug 15 at 12:14










  • Sorry i got the problem . Problem is right .
    – 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.
    – Arthur
    Aug 15 at 12:16











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










$$detbeginpmatrix1&1\0&1endpmatrix^2 = detbeginpmatrix1&0\0&1endpmatrix = 1 ne 0$$






share|cite|improve this answer




















  • a,b,c,d should be Integer modulo 2 . b cannot be 2 .
    – Hansie
    Aug 15 at 12:12










  • @neraj That's not what "integer modulo $2$" means.
    – 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 .
    – Hansie
    Aug 15 at 12:14










  • Sorry i got the problem . Problem is right .
    – 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.
    – Arthur
    Aug 15 at 12:16















up vote
1
down vote



accepted










$$detbeginpmatrix1&1\0&1endpmatrix^2 = detbeginpmatrix1&0\0&1endpmatrix = 1 ne 0$$






share|cite|improve this answer




















  • a,b,c,d should be Integer modulo 2 . b cannot be 2 .
    – Hansie
    Aug 15 at 12:12










  • @neraj That's not what "integer modulo $2$" means.
    – 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 .
    – Hansie
    Aug 15 at 12:14










  • Sorry i got the problem . Problem is right .
    – 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.
    – Arthur
    Aug 15 at 12:16













up vote
1
down vote



accepted







up vote
1
down vote



accepted






$$detbeginpmatrix1&1\0&1endpmatrix^2 = detbeginpmatrix1&0\0&1endpmatrix = 1 ne 0$$






share|cite|improve this answer












$$detbeginpmatrix1&1\0&1endpmatrix^2 = detbeginpmatrix1&0\0&1endpmatrix = 1 ne 0$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 15 at 12:10









Kenny Lau

18.9k2157




18.9k2157











  • a,b,c,d should be Integer modulo 2 . b cannot be 2 .
    – Hansie
    Aug 15 at 12:12










  • @neraj That's not what "integer modulo $2$" means.
    – 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 .
    – Hansie
    Aug 15 at 12:14










  • Sorry i got the problem . Problem is right .
    – 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.
    – Arthur
    Aug 15 at 12:16

















  • a,b,c,d should be Integer modulo 2 . b cannot be 2 .
    – Hansie
    Aug 15 at 12:12










  • @neraj That's not what "integer modulo $2$" means.
    – 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 .
    – Hansie
    Aug 15 at 12:14










  • Sorry i got the problem . Problem is right .
    – 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.
    – Arthur
    Aug 15 at 12:16
















a,b,c,d should be Integer modulo 2 . b cannot be 2 .
– Hansie
Aug 15 at 12:12




a,b,c,d should be Integer modulo 2 . b cannot be 2 .
– Hansie
Aug 15 at 12:12












@neraj That's not what "integer modulo $2$" means.
– Arthur
Aug 15 at 12:13





@neraj That's not what "integer modulo $2$" means.
– 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 .
– 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 .
– Hansie
Aug 15 at 12:14












Sorry i got the problem . Problem is right .
– Hansie
Aug 15 at 12:16




Sorry i got the problem . Problem is right .
– 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.
– 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.
– Arthur
Aug 15 at 12:16













 

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