Ok I swear this will be more or less the last topic on the group $GL_2(\mathbb F_3)$!

I'm searching for all its involutions.

I know that $ \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$ is the only involution in $N:=SL_2(\mathbb F_3)$.

Moreover there are four other involutions: $\pm\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$ and $\pm\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ which don't stay in $N$.

I think these five are the all and only involutions in $GL_2(\mathbb F_3)$. But how can I prove it without computes, systems etc? Is there a rapid way to see that?

Thank you all

  • 2
    $\begingroup$ Take any two linearly independent vectors $u,v\in{\Bbb F_3}^2$. Then the linear maps $u\mapsto -u,\ v\mapsto v$ and $u\mapsto v,\ v\mapsto u$ are also involutions. $\endgroup$ – Berci Jun 9 '14 at 18:24
  • $\begingroup$ Thanks a lot Berci. The first map corresponds to $\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$ and I already wrote it. Nevertheless the second corresponds to $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ and this is a new involution. But the question is: are there other involutions? Or these four are the all and only in $GL_2(\mathbb F_3)$? Thanks again! $\endgroup$ – Joe Jun 9 '14 at 18:46
  • $\begingroup$ I edited my post, adding your suggest and another one involution. $\endgroup$ – Joe Jun 9 '14 at 18:49
  • 1
    $\begingroup$ Well, the swap $u\mapsto v,\ v\mapsto u$ has eigenvectors $u\pm v$. Note also that any matrix similar to $\pmatrix{-1&0\\0&1}$ is also an involution. See my answer. $\endgroup$ – Berci Jun 9 '14 at 19:01
  • 3
    $\begingroup$ As Berci says, the involutions outside of ${\rm SL}_2(3)$ are all conjugate to $\left(\begin{array}{cc}-1&0\\0&1 \end{array}\right)$. It is easy to see that the centralizer of that matrix consists precisely of the diagonal matrices, of which there are $4$. So it has $48/4 = 12$ conjugates, and there are $13$ elements of order $2$ altogether. $\endgroup$ – Derek Holt Jun 9 '14 at 19:03

Let $M\in GL_2(\Bbb F_3)$ be an involution, i.e. $M^2=I$, i.e. $M$ is a root of the polynomial $x^2-1$. Then its minimal polynomial $f$ must be either one of $x+1,\ x-1,\ x^2-1$. The first two cases yield strictly $\pm I$.

So, assume that the minimal polynomial of $M$ is $x^2-1$, but this means that both $+1$ and $-1$ are eigenvalues, i.e. there eigenvectors $u$ and $v$ such that $Mu=-u$ and $Mv=v$.

So, after a change of base, the new matrix of $M$ is indeed of the form $\pmatrix{-1&0\\0&1}$, so all the involutions are of the form $$B\pmatrix{-1&0\\0&1}B^{-1}$$ for some $B\in GL_2(\Bbb F_3)$.

  • $\begingroup$ Thank you Berci, I should have though to work with minimal polynomials. Thanks again $\endgroup$ – Joe Jun 9 '14 at 19:06

It turns out your inventory is missing a couple. For example, $\begin{pmatrix} 1 & 0 \\ 1 & -1 \end{pmatrix}$ is an involution.

One way to attack this problem is to represent an arbitrary matrix as $M= \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and compute $M^2= \begin{pmatrix} a^2+bc & ab+bd \\ ac+cd & bc + d^2 \end{pmatrix}$. Set this equal to the identity matrix. Now start attacking cases:

  1. If $a=0$ then $bc=1$, whence $d^2=0$ as well. This leads to the solutions $M= \begin{pmatrix} 0 & \pm 1 \\ \pm 1 & 0 \end{pmatrix}$.
  2. If $a=1$ then $bc=0$, whence $d^2=1$ as well. This leads to different sub-cases, according to whether $d=1$ (in which case $b=c=0$) or $d= -1$, in which case $b=0$, $c=\rm{arbitrary}$ is another solution, as is $c=0$, $b=\rm{arbitrary}$.
  3. Finally the case $a=-1$ is very similar to case 2 above.

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