Is this $SL_3(\mathbb{F_2})$ look like?

$SL_3(\mathbb{F_2})$ is the set of all invertible $3x3$ matrices over the field $F_2$ with determinant $1$. From what I understand $F_2$ is the set $\{0, 1\}$, ie.. $\mathbb{Z}$ mod $2$.

As far as I can see then, $SL_3(\mathbb{F_2})$ has to be $$I_3 = \begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix}$$

If the $1$'s were in any other location we would have zero determinant. So is this $SL_3(\mathbb{F_2})$?

-
No. Look at the (other) permutation or upper/lower triangular matrices for instance. (Also note that the special and general linear groups are the same over $\Bbb F_2$, and $\times$ can be typeset with \times.) Why exactly do you think $1$'s anywhere else will cause zero determinant? – anon Nov 14 '12 at 11:58

Here is a complete list of total 168 elements of $SL(3,2) = SL_{3}(\Bbb{F}_2)$. I hope you may get some ideas on what is going on in $SL(3,2)$ from this.
(Or you can just play 'Where's Waldo?' game with Waldo replaced by the identity matrix $I_3$!)
But isn't don't the elements of $SL(3,2) = SL_{3}(\Bbb{F}_2)$ have to have $1$ as their determinant? Many of those matrices have $-1$ as a determinant? – sonicboom Nov 14 '12 at 14:16
@sonicboom, In $\Bbb{F}_2$ we have $-1 = 1$. And I used Mathematica 8.0 to find them all. – Sangchul Lee Nov 14 '12 at 14:17