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The question at hand:

The group $GL_2(\mathbb F_p)$ 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 $SL_2(\mathbb F_p)$ be the subgroup consisting of all matrices of determinant $1$, i.e. $SL_2(\mathbb F_p)$ is the kernel of the group homomorphism $\det : GL_2(\mathbb F_p) → \mathbb F_p$. Compute the order of $SL_2(\mathbb F_p)$. Hint: $GL_2(\mathbb F_p)$ is a disjoint union of fibers of $\det$.

I am feeling a little lost on this problem. I understand via my professor that $GL_2(F_p)$ is the union of the fibres of det, that each fibre is a coset of $SL_2(F_p)$, and that $GL_2(F_p)$ is the order of $SL_2(F_p)$ 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?

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  • $\begingroup$ Hint: an element of $GL_{2}(F_{p})$ is invertible if and only if its columns are linearly independent over $F_{p}$. There are $p^{2}-1$ choices for the first column, because only the $0$ column must be excluded. How many choices are there for the second column? $\endgroup$ – Alex Wertheim Oct 18 '16 at 3:40
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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 $SL_2(F_p)$ should be got from $GL_2(F_p)$ 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 $GL_2(F_p)$. 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 $k$?

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