The question at hand:
The group $\mathrm{GL}_2(\mathbb F_p)$ of all $2\times 2$ invertible matrices with coefficients in the finite field $\mathbb F_p$ has order $(p^2 − 1)(p^2 − p)$. Let $\mathrm{SL}_2(\mathbb F_p)$ be the subgroup consisting of all matrices of determinant $1$, i.e. $\mathrm{SL}_2(\mathbb F_p)$ is the kernel of the group homomorphism $\det : \mathrm{GL}_2(\mathbb F_p) → \mathbb F_p$. Compute the order of $\mathrm{SL}_2(\mathbb F_p)$. Hint: $\mathrm{GL}_2(\mathbb F_p)$ is a disjoint union of fibers of $\det$.
I am feeling a little lost on this problem. I understand via my professor that $\mathrm{GL}_2(F_p)$ is the union of the fibres of det, that each fibre is a coset of $\mathrm{SL}_2(F_p)$, and that $\mathrm{GL}_2(F_p)$ is the order of $\mathrm{SL}_2(F_p)$ times the number of fibres. However, I am not sure how to show that the order is $(p^2-1)(p^2-p)$. Does anyone have any insight into this problem?
Best Answer
When you proved Lagrange's theorem about orders of subgroups you would have seen that all the cosets have the same number of elements (as that subgroup).
So the order of $\mathrm{SL}_2(F_p)$ should be got from $\mathrm{GL}_2(F_p)$ upon division by the number of cosets.
If $A,B$ are in the same coset it follows that the matrix $AB^{-1}$ has determinat 1. Conversely take two matrices $C,D$ having same determinant value $\Delta$. $CD^{-1}$ has determinant 1. So the number of cosets corresponds to number of possible values of determinants of elements in $\mathrm{GL}_2(F_p)$. As zero is excluded it is maximum $p-1$. SO it is $(p^2-1)(p^2-p)/(p-1)=p(p^2-1)$.
Still one minor point is left out. Can you show that for any $1\le k \le p-1$ there is matrix of determinant $k$?