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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.enter image description here

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  • $\begingroup$ Why is $X^2$ not part of the group? What do you think $X^2$ is? $\endgroup$
    – Arthur
    Commented Aug 15, 2018 at 12:09
  • $\begingroup$ @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 . $\endgroup$
    – Flintoff
    Commented Aug 15, 2018 at 12:11
  • $\begingroup$ But the entries of the matrices are modulo $2$, so $2 = 0$, and there is no problem since $X^2 = \left[\begin{smallmatrix}1&2\\0&1\end{smallmatrix}\right] = \left[\begin{smallmatrix}1&0\\0&1\end{smallmatrix}\right]$ $\endgroup$
    – Arthur
    Commented Aug 15, 2018 at 12:12

1 Answer 1

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$$\det\begin{pmatrix}1&1\\0&1\end{pmatrix}^2 = \det\begin{pmatrix}1&0\\0&1\end{pmatrix} = 1 \ne 0$$

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  • $\begingroup$ a,b,c,d should be Integer modulo 2 . b cannot be 2 . $\endgroup$
    – Flintoff
    Commented Aug 15, 2018 at 12:12
  • $\begingroup$ @neraj That's not what "integer modulo $2$" means. $\endgroup$
    – Arthur
    Commented Aug 15, 2018 at 12:13
  • $\begingroup$ 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 . $\endgroup$
    – Flintoff
    Commented Aug 15, 2018 at 12:14
  • $\begingroup$ Sorry i got the problem . Problem is right . $\endgroup$
    – Flintoff
    Commented Aug 15, 2018 at 12:16
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    $\begingroup$ @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. $\endgroup$
    – Arthur
    Commented Aug 15, 2018 at 12:16

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