Numer of elements in SL_n(F_p)?

Question

Mathematics

Answer 1

Let's start with GL_n(F_p), the group of invertible n by n matrices over F_p.

An n by n matrix is invertible if and only if its rows are linearly independent. When forming an invertible n by n matrix over F_p, there are (p^n - 1) choices for the first row, because we exclude the zero vector. There are (p^n - p) choices for the second row, because we exclude the p vectors that are scalar multiples of the first row. There are (p^n - p^2) choices for the third row, because we exclude the p^2 vectors that are linear combinations of the first two rows. The pattern continues, and there are (p^n - p^(k-1)) choices for the k'th row. Therefore, the number of elements in GL_n(F_p) is (p^n - 1) * (p^n - p) * (p^n - p^2) * ... * (p^n - p^(n-1)).

Now, SL_n(F_p) is a subgroup of index (p-1) in GL_n(F_p), because it is the kernel of the determinant map from GL_n(F_p) to (F_p)*. So we can find the order of SL_n(F_p) by dividing the number of elements in GL_n(F_p) by (p-1).