the order of a group of matrices over $\mathbb{Z}/2\mathbb{Z}$ Let $G$ be the group of 3 by 3 matrices over $\mathbb{Z}/2\mathbb{Z}$ of determinant 1. Determine the order of $G$.
So does this mean all the elements of matrix $A$ in $G$ are 0 or 1(and the determinant is 1)? and we need to find the order of this group?
So the operation should be multiplication, right?
This is not a large set. But is there a way to find the order instead of check it one by one?
Thank you very much!
 A: It means that not only are the elements of the matrix 0 or 1, but that when multiplying matrices (yes multiplication is the group operation), you treat the entries of the matrix as lying in $\mathbb{Z}/2\mathbb{Z}$. So $1+1 = 0$...etc.
Here's a way you can find the order.
Split the matrix into its three columns. You know that a matrix has nonzero determinant if and only if its columns are linearly independent. Thus, the first column can be anything except the zero vector. Thus, you have $2^3 - 1 = 7$ choices for the first column.
The second column can be anything except a multiple of the first column, so you have $2^3 - 2$ choices (there are only two possible multiples of the first column!) The third column can be anything except a linear combination of the first two columns. I'll let you figure out how many choices there are for the third column.
Lastly note that if a matrix over $\mathbb{Z}/2\mathbb{Z}$ has nonzero determinant, it must have determinant 1. 
A: Did you know that $\mathbb Z/2\mathbb Z$ is field which is normally denoted $\mathbb F_2$?  And did you know that everything you learned about matrices in linear algebra is valid so long as the entries of those matrices come from a field?
In particular a matrix has nonzero determinant if and only if it's columns are linearly independent.  Well the only nonzero element in $\mathbb F_2$ is $1$ so you're just trying to count how many sets of $3$ linearly independent vectors there are.
Well you have $2^3 - 1$ choices for the first vector cause you can choose anything that's nonzero.  For the second vector you can choose anything that's not a scalar multiple of the first vector.  Well the only scalar multiples of the first vector are $0$ and the vector so that's $2^3 - 2$ choices.  I'll let you figure out how many choices you have for the third vector (don't get fooled by the obvious pattern, $2^3 - 3$ isn't the right answer!).
A: Define
$$H:=\left\{\;A\in M_3(\Bbb F_2)\;:\;\;\det A\neq 0\;\right\}$$
or in other words: $\;H\;$ is the group of all invertible $\;3\times3\;$ matrices over the field $\;\Bbb F_2\;$ with two elements.
(1) Show that $\;|H|=(2^3-1)(2^3-2)(2^3-2^2)=168\;$
(2) Define
$$\phi: H\to\Bbb F_2^*\;,\;\;\phi(A):=\det A$$
Show that $\;\phi\;$ is a surjective group homomorphism
(3) Prove that $\;G=\ker\phi\;$
(4) Use the first isomorphism theorem now to deduce $\;|G|\;$
If you need a little more help read the next spoiler after you try to work out the above.

spoiler Can you see that $\;G=H\;$?

