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Want to do the $2$-faces action. We use the Orbit stabilizer theorem. Let $X$ be the set of faces (any face can go to any face), $X=\{1,2,3,4,5,6,7,8 \}$. Where $1,2,3,4$ are the front faces of the picture and the rest are the back faces of the picture.

$|X|=|G:G_1|=|G|/|G_1|$ so $|G|=8|G_1|$ so we need to find $G_1$, which is the stabilizer at $1$.

We can do three symmetries with planes, red to red, purple to purple and blue to blue. And then there is the identity rotation. So far we have $4$ symmetries. What are the last two? Because we need to find $6$ symmetries in total.


There are three rotational symmetries that fix the face $1$; one is the identity, one is $\sigma := (238)(457)$, and the other is $\sigma^2 = \sigma^{-1}$.

  • $\begingroup$ The rotation $\sigma$ permutes the three vertices of face $1$, so its axis must be the line through the center of that face (and the center of the octahedron). Since $\sigma$ has order $3$, it must be a rotation by $\frac{1}{3} \cdot 360^{\circ} = 120^{\circ}$. $\endgroup$ – Travis Feb 20 '16 at 13:08
  • $\begingroup$ So what degree rotation is $\sigma ^2$? $\endgroup$ – snowman Feb 20 '16 at 13:12
  • $\begingroup$ What is the order of $\sigma^2$? $\endgroup$ – Travis Feb 20 '16 at 13:15
  • $\begingroup$ Just to confirm $\sigma ^2 = (238)(457)(238)(457)$ right? $\endgroup$ – snowman Feb 20 '16 at 13:34
  • 1
    $\begingroup$ At a glance they look good to me. $\endgroup$ – Travis Feb 20 '16 at 13:57

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