Number of elements of order P in GL(2,Zp)

Question

I want to find the number of elements of order p in GL(2,Zp). I know that GL(2,Zp) contains p order elements,but I want to know how many p order elements in the group.

Answer 1

Assuming you mean $GL_2(\mathbb{F}_p)$, there are a couple of ways to do this computation. One way is to argue that the eigenvalues of an element $X$ of order $p$ must be $1$ and $1$, and $X$ can't be diagonalizable, so (by the general theory of rational canonical form / Frobenius normal form) $X$ must be conjugate to a Jordan block $\left[ \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right]$. In particular $GL_2(\mathbb{F}_p)$ acts transitively on its elements of order $p$ by conjugation. Now by orbit-stabilizer it suffices to compute the size of the centralizer of a Jordan block.

In this case everything is small enough that we can just compute explicitly. We have

$$\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] \left[ \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right] = \left[ \begin{array}{cc} a & a+b \\ c & c+d \end{array} \right]$$

and

$$\left[ \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right] \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] = \left[ \begin{array}{cc} a+c & b+d \\ c & d \end{array} \right]$$

so the centralizer consists of matrices satisfying $c = 0$ and $a = d$, or in other words upper triangular matrices $\left[ \begin{array}{cc} a & b \\ 0 & a \end{array} \right]$ with constant diagonal. There are $p(p-1)$ of these so by orbit-stabilizer we conclude that $GL_2(\mathbb{F}_p)$ has

$$\frac{(p^2 - 1)(p^2 - p)}{p(p - 1)} = \boxed{ p^2 - 1 }$$

elements of order $p$.

Alternatively, since we know that the cyclic subgroups of order $p$ are in fact the Sylow $p$-subgroups of $GL_2(\mathbb{F}_p)$, we can instead count the Sylow $p$-subgroups and multiply by $p - 1$. It turns out that you can show that

• the upper triangular matrices $U_n(\mathbb{F}_p)$ are a Sylow $p$-subgroup, so the other ones are conjugates of this one, and
• the Sylow $p$-subgroups are in natural bijection with the complete flags in $\mathbb{F}_p^n$; the bijection sends a Sylow $p$-subgroup to the unique complete flag it preserves.

So to count Sylow $p$-subgroups we can count complete flags. In $\mathbb{F}_p^2$ a complete flag is just a line so the number of complete flags is the size of the projective line $\mathbb{P}^1(\mathbb{F}_p)$, so there are $p + 1$ lines and therefore $\boxed{ (p - 1)(p + 1) }$ elements of order $p$.