Let \(F\) be a finite field of order \(p\). Compute the order of \[ SL_3(F), \] the group of \(3\times3\) matrices over \(F\) with determinant \(1\).
Solution
1. The order of \(GL_3(F)\) is\[ \abs{GL_3(F)}=(p^3-1)(p^3-p)(p^3-p^2). \]
An invertible \(3\times3\) matrix is equivalent to an ordered basis of \(F^3\), given by its columns.
There are \(p^3-1\) choices for the first nonzero column. Once it is chosen, its span has \(p\) elements, so there are \(p^3-p\) choices for the second column. The first two independent columns span a plane with \(p^2\) elements, leaving \(p^3-p^2\) choices for the third column.
2. The determinant map\[ \det:GL_3(F)\longrightarrow F^\times \]is surjective and has kernel \(SL_3(F)\).
The kernel statement is the definition of \(SL_3(F)\). For every \(a\in F^\times\), the diagonal matrix \[ \operatorname{diag}(a,1,1) \] has determinant \(a\), so the determinant map is surjective.
3. One has\[ \abs{SL_3(F)} = \frac{(p^3-1)(p^3-p)(p^3-p^2)}{p-1}. \]
Since \(F\) has \(p\) elements, its multiplicative group \(F^\times\) has order \(p-1\). By step 2 and the first isomorphism theorem, \[ GL_3(F)/SL_3(F)\cong F^\times. \] Thus \[ [GL_3(F):SL_3(F)]=p-1, \] and step 1 gives the displayed formula.
4. Therefore\[ \boxed{\abs{SL_3(F)}=p^3(p^3-1)(p^2-1)}. \]
Factor \[ p^3-p=p(p^2-1), \qquad p^3-p^2=p^2(p-1) \] in step 3 and cancel \(p-1\).
5. Q.E.D.
Step 4 gives the requested order.