# 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?

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.

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

$$\det\begin{pmatrix}1&1\\0&1\end{pmatrix}^2 = \det\begin{pmatrix}1&0\\0&1\end{pmatrix} = 1 \ne 0$$
• @neraj That's not what "integer modulo $2$" means. Commented Aug 15, 2018 at 12:13
• @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. Commented Aug 15, 2018 at 12:16