Consider the directed graphs $G_n^p$ with node set $\{0,1,\dots,p-1\}$, $0 \leq n < p$ with an arrow from $a$ to $b$ if $na=b\operatorname{mod}p$. These graphs are graphical multiplication tables for $\mathbb{Z}_p := \mathbb{Z}/p\mathbb{Z}$.

For example $G^{12}_n$:

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My question is:

Is it just by coincidence that $G^{12}_4$ looks like a 2-dimensional drawing of a 3-dimensional cube

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– or might there be something "deep" in it, assuming that there are some deep and hard to see relations between projective geometry and prime numbers?

Note that $p = 2 \cdot 2\cdot 3$, $n = 2 \cdot 2$.

But also note that $G^{12}_{10}$ does the same job, while in this case $n=2\cdot 5$.

Note especially that it is important to distribute the nodes of the graph equally on a circle – like the $p$th roots of unity – to get the "desired" graph.

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    $\begingroup$ Maybe there's an interpretation of your observation in terms of isometry groups. The graph that looks like the projection of a cube has the symmetries of an equilateral triangle. But you can get such a triangle by cutting a cube in two equal parts, with the center of gravity of this triangle being the isobarycenter of the vertices of the cube. This triangle is invariant under the rotation whose axis is the large diagonal orthogonal to the plane it defines. $\endgroup$ – Sylvain Julien Nov 12 '18 at 17:24
  • $\begingroup$ Rotation of angle $ 2k\pi/3 $ for $ k $ an integer, of course. $\endgroup$ – Sylvain Julien Nov 12 '18 at 17:28
  • $\begingroup$ Also $ n=10 $ and $ n=4 $ are "dual" in the sense that the difference of two diametrally opposed nodes is $ 6=10-4=12/2 $ . $\endgroup$ – Sylvain Julien Nov 12 '18 at 17:33
  • $\begingroup$ The graph $G^{12}_{4} $ is then the projection of the 12-edges 3-dimensional polytope on the disk whose border contains the vertices of the equilateral triangle I mentioned above. $\endgroup$ – Sylvain Julien Nov 12 '18 at 17:37

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