Happy #GL_2(Z/7Z)

Ahhhhh it’s almost 2016 (only three more days)! I’m incredibly excited, mainly because I’ll be going to college this fall. In this blog post, I wanted to point out something which Noam Elkies informed me about, because I hadn’t recognized it and found it pretty awesome!

As the title suggests, we’ll take a look at the general linear group over a finite field. First, let p be a prime, and let x\in\mathbf{Z}/p\mathbf{Z}. Then the equation xy\equiv 1\bmod p always has a unique solution. More generally, if x\in\mathbf{Z}/n\mathbf{Z}, xy\equiv 1\bmod n always has a unique solution if \gcd(x,n)=1. Thus \#\mathrm{GL}_1(\mathbf{F}_p)=p-1.

What about \#\mathrm{GL}_2(\mathbf{F}_p)? If M=\begin{smallmatrix} a&b\\ c&d \end{smallmatrix}, then a necessary and sufficient condition for M to be invertible is that ad\neq bc. We could proceed via a case-by-case analysis, but that is more difficult. Alternatively, we could simply note a matrix’s determinant is zero if and only if the columns are linearly independent that the vector \begin{smallmatrix} a\\ c \end{smallmatrix} must be nonzero (which gives us p^2-1 choices), and \begin{smallmatrix} b\\ d \end{smallmatrix} can’t be a multiple of \begin{smallmatrix} a\\ c \end{smallmatrix}, which gives us p^2-p choices. Thus \#\mathrm{GL}_2(\mathbf{F}_p) is (p^2-1)(p^2-p).

In general, \#\mathrm{GL}_n(\mathbf{F}_p) is \prod^{n-1}_{k=0}(p^n-p^k). We can use basic group theory to determine \#\mathrm{SL}_n(\mathbf{F}_p) as well. Note that \mathrm{SL}_n(\mathbf{F}_p) is the kernel of the determinant function \mathrm{det}:\mathrm{GL}_n(\mathbf{F}_p)\to\mathbf{F}_p^\times. In particular, using the isomorphism theorems, \mathrm{GL}_n(\mathbf{F}_p)/\mathrm{SL}_n(\mathbf{F}_p)\simeq\mathbf{F}_p^\times, so rearranging gives \#\mathrm{SL}_n(\mathbf{F}_p)=\#\mathrm{GL}_n(\mathbf{F}_p)/\#\mathbf{F}^\times_p.

This is all fun stuff, but what we’re really here for is the star of the day (or rather, year): \#\mathrm{GL}_2(\mathbf{Z}/7\mathbf{Z})(=\#\mathrm{GL}_2(\mathbf{F}_7)). Using the above formula, we get: \#\mathrm{GL}_2(\mathbf{F}_7)=(7^2-1)(7^2-7)=48\times 42=2016 . Excluding the trivial cases (i.e. \mathrm{GL}_1), we’re not going to have this for another 1512 years! So enjoy 2016, and, as always, have fun!

Happy \#\mathrm{GL}_2(\mathbf{F}_7),


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