The number of $n\times n$ matrix over integer modulo $p$ field with determinant equal $1$

Question

How to count the number of $n\times n$ matrix over integer modulo $p$ field with determinant equal $1$?

I know that the number of invertible matrices is GL$(n,p)$. Have any ideas?

Answer 1

A matrix will have determinant non-zero if and only if its columns are linearly independent. Choose a non-zero vector $v_1 \in \mathbb F_p^n$. Now $v_2$, the second column, must be linearly independent from $v_1$, i.e. must not lie in the subspace generated by $v_1$, which contains $p$ elements. Choosing $v_3$ outside of the subspace generated by $v_1,v_2$ means we cannot pick any of these $p^2$ elements. Keep going like this, and you will see that the answer is $$ \frac 1{p-1} (p^n-1)(p^n-p) \cdots (p^n-p^{n-1}). $$ (one can take any of these $n$-tuples of vectors $\{v_1,\cdots,v_n\}$ and construct a $(p-1)$-to-$1$ map to the matrices of determinant $1$ by normalizing the first vector).

Hope that helps,

Answer 2

In terms of group theory (perhaps easier to remember), one can define a surjective homomorphism $$\phi: GL(n,p) \rightarrow \mathbb F_p^{*}$$ by $$\phi(A)=det(A).$$ Obviously $ker(\phi)=SL(n,p)$ and since $|\mathbb F_p^{*}|=p-1$, index$[GL(n,p):SL(n,p)]=p-1$.