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We know that every matrix of order three in $PGL(2,\mathbb Z)$ is conjugate to the following matrix $$ \left( \begin{array}[cc] &1 & -1 \\ 1&0 \end{array} \right) $$ I want to know if the same result holds in $PGL(2,k)$, where $k$ is any field, under some/no constraints.

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    $\begingroup$ Your matrix doesn't have order 3 because its trace is 1 so the eigenvalues can't be cube roots of unity. In general think about what the eigenvalues can be and use Cayley-Hamilton. $\endgroup$
    – eric
    Nov 8, 2015 at 13:12
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    $\begingroup$ To be fair, the question is asked about ${\rm PGL}(2,k)$, and the image of that element of ${\rm GL}(2,k)$ in ${\rm PGL}(2,k)$ is indeed an element of order $3$ (even when $k$ has characteristic $2$). $\endgroup$ Nov 8, 2015 at 13:36
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    $\begingroup$ Fields $k$ of characteristic 3 are likely to pose some added challenges. $\endgroup$ Nov 8, 2015 at 15:01

2 Answers 2

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The following argument does not need any assumption on the field $\Bbbk$. Let $[A]\in PGL(2,\Bbbk)$ be of order three. Then $A$ is invertible and not a multiple of $E=\binom{1\;0}{0\;1}$, so there exists a vector $v$ such that $v$ and $Av$ form a basis. With respect to this basis, $A$ is of cyclic normal form $$\begin{pmatrix}0&b\\1&a\end{pmatrix}\quad\text{with}\quad \begin{pmatrix}0&b\\1&a\end{pmatrix}^3=\begin{pmatrix}ab&b(a^2+b)\\a^2+b&ab+a(a^2+b)\end{pmatrix}\;.$$ So $[A]$ is of order three if and only if $b=-a^2$. Conjugating (in $PGL(2,\Bbbk)$) gives $$\begin{pmatrix}0&-a\\1&0\end{pmatrix}\begin{pmatrix}0&-a^2\\1&a\end{pmatrix}\begin{pmatrix}0&-a\\1&0\end{pmatrix}=\begin{pmatrix}a^2&-a^2\\a^2&0\end{pmatrix}\;,$$ which represents the matrix in the question.

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For $GL_2$, replace the first entry of your matrix with -1 and then the answer is yes. You can assume that the field is algebraically closed. A matrix $A$ with $A^3=\lambda I$ can be divided by a third root of lambda to get to $A^3=I$. Again by multiplication you can make sure the eigenvalues are the two different third roots of the unit (unless you are in characteristc 3 in which case there is only one third root of the unit, but that case is even easier). But then the characteristic polynomial is determined and this determines the conjugacy class in $GL_2$.

Now for $PGL_2$. Let $K$ be our field. Assume $A^3=\lambda I$ with some $\lambda\in K$. Then a third root of $\lambda$ must also lie in $K$, as the minimal polynomial of $A$ divides $x^3-\lambda$. So you can divide $A$ by that third root to obtain $A^3=I$. By the $GL_2$-case this matrix is conjugate to your standard matrix.

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    $\begingroup$ If two matrices over a filed $K$ are conjugate iver an extension field $L$, they are already conjugate over $K$. $\endgroup$
    – user1688
    Nov 8, 2015 at 14:15
  • $\begingroup$ Can you find any explicit examples of such a matrix $A$? What could its minimal polynomial be? $\endgroup$
    – Derek Holt
    Nov 8, 2015 at 15:11
  • $\begingroup$ Oh, no, this cannot happen. I edit my answer accordingly. $\endgroup$
    – user1688
    Nov 8, 2015 at 15:31
  • $\begingroup$ Note also that the first paragraph doesn't work in characteristic $3$, where there are no primitive third roots of $1$. $\endgroup$
    – Derek Holt
    Nov 8, 2015 at 15:34
  • $\begingroup$ Thanks Derek, that's right, but then the situation is even easier. $\endgroup$
    – user1688
    Nov 8, 2015 at 15:40

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