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This is a follow up question of this other of mine.

Let $p$ be a prime. For every $i,j\in\{1,\dots,p-1\}$, there is unique $k_{ij}\in\{1,\dots,p-1\}$ such that $\varphi_{k_{ij}}(i)=j$$^\dagger$. Therefore, the square matrix $K=(k_{ij})_{\begin{smallmatrix}i=1,\dots,p-1\\j=1,\dots,p-1\end{smallmatrix}}$ is well defined. By naturally labelling the $\varphi_k$'s according to $\varphi_k(1)=k$, the routine ends up with, for example:

  • $p=5$: \begin{equation} K= \begin{pmatrix} 1 & \color\red2 & \color\green3 & 4 \\ \color\green3 & 1 & 4 & \color\red2 \\ \color\red2 & 4 & 1 & \color\green3 \\ 4 & \color\green3 & \color\red2 & 1 \\ \end{pmatrix} \end{equation}
  • $p=7$: \begin{equation} K= \begin{pmatrix} 1 & 2 & \color\red3 & 4 & \color\green5 & 6\\ 4 & 1 & \color\green5 & 2 & 6 & \color\red3\\ \color\green5 & \color\red3 & 1 & 6 & 4 & 2 \\ 2 & 4 & 6 & 1 & \color\red3 & \color\green5\\ \color\red3 & 6 & 2 & \color\green5 & 1 & 4\\ 6 & \color\green5 & 4 & \color\red3 & 2 & 1\\ \end{pmatrix} \end{equation}

(the colors identify the $k$'s such that $\varphi_k$ is a generator of $\operatorname{Aut}(\mathbb Z/p\mathbb Z)$). In both cases, $\det K=0 \pmod p$: I guess that this fact may be true for every $p$ and, if not for trivial reasons, related to the fact that $\mathbb F_p$ is a field. If true, and nontrivial, how goes the proof of this fact get along?

$^\dagger$$\varphi$ is the action by automorphisms of $(\mathbb Z/p\mathbb Z)^\times$ on $\mathbb Z/p\mathbb Z$.

RobPratt
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Kan't
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1 Answers1

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Well since we are working over a field then the theorems of linear algebra still hold. So as to why the determinant is zero, it is because the first row will just be $[1,2, \ldots,p-1]$ while the last row is simply $[-1 \cdot 1 , \ldots, -1 \cdot (p-1)]$ which are linearly dependent vectors and so the determinant must be zero.

Ito Yukis
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