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I want to prove that the automorphism group of the cube is $\mathbb{Z}_2 \times S_4$, by using information about the congruences of a cube. By the cube, I mean the graph of the platonic solid, i.e. the graph whose vertices are $\{0,1\}^3$, with two vertices being adjacent iff their bit strings differ in exactly one coordinate. This is an exercise in [Lovasz, Combinatorial Problems and Exercises]. The solution given there says "each automorphism of the cube is, in fact, induced by an a congruence and that the whole group is the direct product of $\mathbb{Z}_2$ and the group of direct congruences of the cube... the group of direct congruences of the cube is $S_4$...". What is the definition of "congruence"? And why is it a direct product and $S_4$ for the cube?

Other proofs would be great too. I am able to show that the automorphism group of the cube is isomorphic to $\mathbb{Z}_2^3 \rtimes S_3$ (in fact, this is true for the $n$-dimensional cube also, with 3 replaced by $n$), but is this semidirect product as good a description of the automorphism group as $\mathbb{Z}_2 \times S_4$? It seems in general there could be more than one semidirect product and a direct product description would be more complete.

Martin Sleziak
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    Note that $S_4\cong \mathbb{Z}_2^2 \rtimes S_3$. The $\mathbb{Z}_2^2$-part consists of the identity and the double transpositions in $S_4$, which is the famous normal Klein four subgroup of $S_4$. – azimut Mar 10 '13 at 18:13

2 Answers2

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There is a mirror symmetry inverting the cube, changing it from left handed to right handed. This has order 2. We can factor out this symmetry by identifying the opposite points of the cube:

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Then we are really considering the symmetries of the 4 diagonals rather than the cube itself. For example there is an order 3 symmetry of spinning around the pink diagonal $$(\color{red}{red}\;\color{green}{green}\;\color{blue}{blue})=(b\;d\;g)(a\;h\;f)$$ you can find more symmetries too for example an order 4 one $$(\color{magenta}{pink}\;\color{\green}{green}\;\color{red}{red}\;\color{blue}{blue})=(c\;d\;h\;g)(b\;a\;e\;f)$$

It's "obvious" that the symmetries of the diagonals give $S_4$ acting on the diagons (any 3 cycle and 4 cycle gives that)! Furthermore inverting the cube has no effect on this object so it's really a direct product. So from the dictionary between symmetries of the diagonals and of the cube we have $C_2 \times S_4$ on the cube.

  • What does it mean to invert the cube? And why is it a direct product? We have that the four diagonals form a partition $\Sigma={\Delta_1,\ldots,\Delta_4}$ of the vertex set into blocks, for eg $\Delta_1={a,g}$. The automorphism group $G$ of the cube acts on $\Sigma$, and the image of this action is all of $Sym(\Sigma)$. The kernel of this action is the set of automorphisms that fixes each diagonal, i.e. is isomorphic to $\mathbb{Z}_2$. By the first isomorphism theorem, $G/\mathbb{Z}_2 \cong S_4$. How do I get an internal direct product? – 64647a Mar 16 '13 at 17:17
  • invert the cube means swap a with g etc. it is a directly product because the permutations that generate the $S_4$ are manifestly invariant with respect to inverting. –  Mar 16 '13 at 17:19
  • What does it mean that "each automorphism of the cube is, in fact, induced by an a congruence and that the whole group is the direct product of Z2 and the group of direct congruences of the cube"? I think your proof is as follows: Let $H \le S_8$ be the subgroup generated by the two permutations you provided. Then, $|H|=24$ (any short proof for this?), whence $H \trianglelefteq G$. Also, since the action of $H$ on $\Sigma$ is faithful, the kernel $K$ of the action of $G$ on $\Sigma$ satisfies $K \cap H=1$. Also, $K \cong \mathbb{Z}_2$, so we have the internal direct product. – 64647a Mar 16 '13 at 18:41
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Let $G \le S_8$ be the automorphism group of the cube graph drawn above on vertex set $\{a,\ldots,h\}$. Observe that $|G|=48$ (there are many ways to show this), and we would like to identify this group. We show $G \cong S_4 \times \mathbb{Z}_2$. To get the $S_4$ we need 4 elements that $G$ acts on. Let $\Sigma$ denote the set of 4 main diagonals of the cube; for example, one of the diagonals is the subset $\{a,g\}$. Then $G$ acts on $\Sigma$, and the corresponding permutation representation $f: G \rightarrow Sym(\Sigma)$ is onto and has kernel $K:=\{1,\rho\}$, where $\rho$ denotes reflection about the center of the cube. Hence $K \trianglelefteq G$.

Let $H \le G$ be the set of direct congruences of the cube. (The platonic solid has both direct congruences, which are defined to be those isometries that preserve the orientation, as well as indirect/opposite isometries, which are isometries that change the order from a clockwise to anticlockwise labeling. The 24 rotations of the cube are direct (i.e. orientation preserving) congruences, whereas reflection does not preserve orientation. The 24 direct congruences of the cube are also called rotational symmetries or rigid motions of the cube.) Then $|H|=24$, whence $H \trianglelefteq G$. Also, $f$ restricted to $H$ is injective. To prove injectivity, note that if $1 \ne h \in H$ is a rotational symmetry that fixes two diagonals setwise, say $\{a,g\}$ and $\{b,h\}$, then it must interchange $e$ and $f$ and hence does not fix the remaining two diagonals setwise; equivalently, any rotational symmetry of the cube that fixes each of the four diagonals (setwise) must be the identity. Hence, $H \cap K=1$ and $f$ restricted to $H$ is an isomorphism from $H$ to $Sym(\Sigma)$, whence $H \cong S_4$. Thus, the automorphism group of the cube graph is the internal direct product of $H$ and $K$.

Note that since $K$ is a normal subgroup of $G$ of order 2, it is in the center of $G$. Thus, the reflection $\rho$ commutes with every automorphism of the cube graph.

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