I am trying to show that the elements $A:= \left( \begin{array}{ccc} 0 & 0 & 1 \\ 1 & 0 & 1\\ 0 & 1 & 0 \\ \end{array} \right)$, $B:= \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 1 \\ \end{array} \right)$ and $W:= \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ \end{array} \right)$ generate $SL_3(2)$. The order of the matrices are 7,3 and 2, respectively. By Lagrange's theorem, I am missing a subgroup of order 4, but I can't find it. Can you help me find Waldo?
Also, is there any general way when getting elements, and need to find all the subgroups they generate?
Thank you.
First, a quick note: You mentioned you are missing a subgroup of order $4$, yet to use Lagrange's theorem alone to obtain the order of the group you would need an element of order $8$ since of course the lowest common multiple of $2$, $3$, $4$ and $7$ is $84$ (though since $\operatorname{SL}_3(2)$ is simple and a subgroup of order $84$ has index $2$ and must be normal, this would also be sufficient). However, no elements of order $8$ exist in $\operatorname{SL}_3(2).$
One possible answer (and probably also the most elementary) may be found here, given that $\operatorname{SL}_3(2) \cong \operatorname{PSL}_3(2)$ as $Z(\operatorname{SL}_3(2))$ is trivial. Depending on why you want to know this, there are a lot of other ways to get around it as well.
For example, the maximal subgroups of $\operatorname{SL}_3(2)$ are two conjugacy classes of $S_4$ and one class of $C_7 \rtimes C_3$. One can then note that neither $S_4$ nor $C_7 \rtimes C_3$ have orders simultaneously divisible by $7$ and $2$ so $\left<A,W\right>$ cannot lie in any of these maximal subgroups. But a subgroup not contained in any maximal subgroup must then be the whole group.
Another possibility would be to note that $\operatorname{SL}_3(2)$ is a simple group and try to show that $\left<A,B,W\right>$ is normal.
You could also note that $\operatorname{SL}_3(2) \cong \operatorname{PSL}_2(7)$ and work in there instead, in case you found $2 \times 2$ matrices in characteristic $7$ easier to deal with.
Finally, if you just wanted to brute force check you could use some computational algebra software such as GAP, which is free, though of course this would require some effort to learn.