Determine the elements in $GL(3)$ and $SL(3)$ on $\mathbb{F}_p$ $G=GL(3)$(3 by 3 matrices that is non-singular) and $H=SL(3)$(3 by 3 matrices that has determinant 1) on $\mathbb{F}_p$, where $p$ is a prime. What is the order of G? and H?
For $|G|$, the first row has $p^3-1$ choices. For the second row, it has $p^3-(p-1)-1$ choices(minus $p-1$ multiple of the first row). What about the third row? Is it $p^3-p*p+1$($p$ multiple for the two rows and add back zero vector)?
What about $|H|$? What is the trick to deal with determinant is 1? 
Thank you very much!
 A: Note that the third row needs to be linearly independent of the first two.  So your third row has to be anything in $\mathbb{F}_p^n$ outside of some two-dimensional subspace.  This will leave $p^n-p^2$ choices (note that your expression for the number of choices for the second row can be written as $p^n-p$).
Determining the order of $H$ once you know the order of $G$ can be done by considering the determinant as a homomorphism $G \to \mathbb{F}^\times$ and considering the First Isomorphism Theorem.
A: Consider the determinant map $\det:GL_3(F_p)\to F_p^\ast$, where $F_p^\ast$ dnotes the non-zero elements of $F_p$. This a surjective group homomorphism and the kernel is $SL_3(F_p)$. Hence, by the first isomorphism theorem we have: $$GL_3(F_p)/SL_3(F_p)\cong F_p^\ast\,.$$ But then $$\frac{|GL_3(F_p)|}{|SL_3(F_p)|}=\Bigl| GL_3(F_p)/SL_3(F_p)\Bigr|=|F_p^\ast|\,.$$ From this you can find the order of $SL_3(F_p)$.
Of course, if any of these statements are unknown to you, you should prove them :)
EDIT: Sorry, forgot to address $GL_3(F_p)$. I'm going to look at columns instead of rows. For the first column, we have $p^3-1$ choices. The second column cannot be a multiple of the first, giving us $p^3-p$ choices (since we can have $p$ possible multiples). The third column cannot be a multiple of the first or second column, giving us $p^3-p^2$ choices (Do you see why?). So all told, there are $(p^3-1)(p^3-p)(p^3-p^2)$ elements.
