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Is there a way to explicitly construct a matrix of (multiplicative) order $p + 1$ in $\mathrm{GL}_2(\mathbb{F}_p)$?

I am aware that by considering the cyclic multiplicative group $\mathbb{F}_{p^2}^\times \subseteq \mathrm{GL}_2(\mathbb{F}_p)$, there is in fact a matrix of order $p^2 - 1$, which gives us a matrix of order $p + 1$ (see e.g. this). However, this method is not exactly constructive, and I wonder if there's an easier way to get one of order $p + 1$.

Clement Yung
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  • You may want to compare this problem with that of finding an element of order $p-1$ in $GL_1(\Bbb{F}_p)\simeq\Bbb{F}_p^*$. That isn't "constructive" either! – Jyrki Lahtonen Sep 20 '23 at 02:59
  • On the other hand, it is easy to describe corresponding cyclic subgroups of orders $p\pm1$ respectively. $\Bbb{F}p^*$ itself is cyclic of order $p-1$. And if $\Delta$ is a quadratic non-residue modulo $p$, then, denoting $\epsilon=\sqrt{\Delta}\in\Bbb{F}{p^2}$, we see that $$G={1}\cup{\frac{x+\epsilon}{x-\epsilon}\mid x\in\Bbb{F}p}\le\Bbb{F}{p^2}^*$$ is a subgroup of order $p+1$. Observe that $\epsilon^p=-\epsilon$ as Frobenius is the only non-trivial automorphism. Therefore raising any element of $G$ to $p$th power gives its inverse. – Jyrki Lahtonen Sep 20 '23 at 03:07
  • (cont'd) A particularly simple case is when $p\equiv-1\pmod4$ and we can choose $\Delta=-1$, $\epsilon=i$, when the parametrization of $G$ in my comment above may be familiar from a parametrization of the unit circle on the complex plane, $x$ ranging over the reals (and $x=\pm\infty$ yielding $1$ as the limit). – Jyrki Lahtonen Sep 20 '23 at 03:10
  • (cont'd) In the case $p\equiv-1\pmod4$ we can describe a subgroup of order $p+1$ as consisting of matrices of the form $$\frac1{x^2+1}\pmatrix{x^2-1&2x\cr-2x&x^2-1\cr}$$ together with the identity matrix (corresponding to $x=\infty$). This is again something familiar from calculus and trig substitutions. In other words, use $$\epsilon=\pmatrix{0&1\cr -1&0\cr}$$ in the above recipe. I'm afraid this doesn't really shed any light to the problem of identifying a matrix of maximal order $p+1$ within this subgroup. – Jyrki Lahtonen Sep 21 '23 at 04:08

1 Answers1

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Let $X$ be any such matrix. It must be diagonalizable over $\overline{\mathbb{F}_p}$ (because non-diagonalizable matrices have order divisible by $p$ due to the Jordan block), so it has two eigenvalues $\alpha, \alpha^p \in \mathbb{F}_{p^2}$ of multiplicative order $p + 1$. This means they multiply to $1$, so the characteristic polynomial of $X$ has the form

$$\det(tI - X) = t^2 - (\alpha + \alpha^p) t + 1$$

and must be irreducible (or else $\alpha \in \mathbb{F}_p^{\times}$ would have order dividing $p - 1$). Conversely, if $t^2 - at + 1$ is an irreducible quadratic with constant term $1$, then its companion matrix

$$X = \left[ \begin{array}{cc} 0 & -1 \\ 1 & a \end{array} \right]$$

has two eigenvalues of order dividing $p + 1$ in $\mathbb{F}_{p^2}$ and has the same order as the order of its eigenvalues.

The conclusion I reach from this is that this is essentially exactly as difficult as finding an element of order $p + 1$ in $\mathbb{F}_{p^2}$. Already there is no "explicit" way of finding an element of order $p - 1$ in $\mathbb{F}_p$ so I don't think there's an "explicit" way to do this either. What you can say, at least, is that they must be roots of the cyclotomic polynomial $\Phi_{p+1}(x)$.

If $p$ is odd then $t^2 - at + 1$ is irreducible iff the discriminant $\Delta = a^2 - 4$ is a quadratic non-residue which means finding $X$ as above is also at least as difficult as finding a quadratic non-residue, and again as far as I know there's no "explicit" way to do this either.

Qiaochu Yuan
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