As shown in this note, the symmetry group $S_4$ for a cube has $3$ subgroups that are isomorphic to $D_4$, the dihedral group of order $2 \times 4 = 8$. How to geometrically illustrate this fact? Specifically, where are the squares embedded in the cube?

Related post: How to geometrically show that there are $4$ $S_3$ subgroups in $S_4$?

  • $\begingroup$ $D_4$ is the symmetry group of a square. A cube has three axes. Does this help? $\endgroup$ – Kenny Wong Mar 14 '17 at 3:32
  • $\begingroup$ @KennyWong My first attempt is to pair up the opposite faces. I am not sure how to use the "three axes" hint. $\endgroup$ – hengxin Mar 14 '17 at 3:36
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    $\begingroup$ That's exactly what @KennyWong means: each axis of the cube corresponds to a pair of opposite faces, i.e. their common symmetric axis. Consider two faces as one, and we have a copy of $D_4$ in $S_4$. $\endgroup$ – Quang Hoang Mar 14 '17 at 3:45
  • $\begingroup$ The pictures on this page help, "the eightfold cube" one in particular. Here everything we care about is centrally-symmetric so we don't have to identify things, we can find nice copies right in the center. See if you can figure out how to label things to get the exact subgroups mentioned in the document. $\endgroup$ – pjs36 Mar 14 '17 at 3:52
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    $\begingroup$ @pjs36 Thanks for your help. I will try it. $\endgroup$ – hengxin Mar 14 '17 at 3:57

Thanks for another good question; I didn't expect the solution to be so pleasant!

Take your three squares to be those half way between opposite faces, as in this picture:

Cube with central squares; in the xy, xz, and yz planes

For the labeling, we use the four diagonals of the cube. For each diagonal, we only label one of the vertices, in such a way that no neighboring vertices are both labeled (the four labeled vertices are the vertices of a regular tetrahedron).

To label the vertices of the three squares, each vertex receives the label of the nearest labeled cube vertex. For example, the front left corner of the cube is labeled $1$. So all of the vertices of squares that are closest to this corner get labeled $1$ as well.

It's a fun check to verify that symmetries of the cube fixing a given square give rise to permutations of the three copies of $D_4$ listed in the document:

\begin{align*} H_{\rm green} &= \langle (1\ 2\ 3\ 4), (1\ 3) \rangle \\ H_{\rm blue} &= \langle (1\ 2\ 4\ 3), (1\ 4) \rangle \\ H_{\rm red} &= \langle (1\ 3\ 2\ 4), (1\ 2) \rangle \end{align*}

For example, consider the $180^\circ$ rotation fixing the top and bottom faces of the cube. It induces the permutation $(1\ 2)(3\ 4)$ on each of the red, blue, and green squares. But a $90^\circ$ rotation about the same axis doesn't fix the blue or green squares, and so it represents the permutation $(1\ 4\ 2\ 3)$, via the red square.

It's such a nice representation that we can even see $$H_{\rm green} \cap H_{\rm blue} \cap H_{\rm red} = \{1, (1\ 2)(3\ 4), (1\ 3)(2\ 4), (1\ 4)(2\ 3)\}.$$

I don't think the rotations that permute all of the squares (e.g., those whose axes are through diagonal vertices of the cube) are meaningful in this picture, but I'd like it very much if they were!

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