# The group of rotation of a cube is isomorphic to $S_4$

Proof of my book is here

Since the group of rotations of a cube has the same order as $S_4$, (1)we need only prove that the group of rotations is isomorphic to a subgroup of $S_4$.To this end, observe that a cube has four diagonals and that the rotation group induces a group of permutations on the four diagonals. But we must be careful not to assume that different rotataions correspond to different permutations. To see that this is so, (2)all we need to do is show that all 24 permutiations of the diagonals arise froem rotations. Labeling the consecutive diagonals 1,2,3, ans 4, it is obvious that there is a $90^\circ$ rotation that yields the permutation $\alpha = (1234)$; another $90^\circ$ rotation about an axis perpendicular to our first axis yields the permutation $\beta=(1423)$ So, the group of permutations induced by the rotations contains the eight-element subgroup $\{\epsilon, \alpha,\alpha^2,\alpha^3,\beta^2,\beta^2\alpha,\beta^2\alpha^2,\beta^2\alpha^3\}$ and $\alpha\beta$, which has order. Clearly, then, the rotations yield all 24 permutaions, (3)since the order of the rotation group must be divisible by both 8 and 3

Q.1 I don't understand why we only need to check (1). and where can I find that (1) is satisfied ?

Q.2 Suddenly why do we need to show (2) instead of showing (1). what (2) is related to (1) ?

Q.3 How can we know (3) is satisfied, I think (3) only to show the order of the rotations group is multiple of 24, not exactly 24

(1) You know that $|G|=24$, hence any injective homomorphism $G\hookrightarrow S_4$ is automatically an isomorphism. The action of $G$ on the set of the four spacial diagonals gives us a homomoprhism $\phi\colon G\to S_4$, but is it injective?
(2) Now, contrary what (1) hints to, the author tries to show that $\phi$ is onto, not that it is 1-1. Then again, it does not matter: An onto homomorphism between two finite groups of same order is of course also an isomorphism. I agree that this is confusing. They should have formulated (1) differently.
(3) $\phi(G)$ contains a subgroup of order $8$ and a subgroup of order $3$, hence $\phi(G)$ has order a multiple of $24$. However, we already know that $|\phi(G)|\le |G|\le 24$.
Alternatively, 120° rotations around a diagonal fix that diagonal and cyclically permutes the other three. And a 180° rotation around the line through the midpoints of two opposing edges fixes two diagonals and swaps the other two. We conclude that the stabilizer of one diagonal is the full $S_3$ of the other three diagonals. As the rotations of the cube clearly act transitively on the diagonals, it follows that $\phi(G)$ has order $4\cdot |S_3|=24$, i.e., is the full $S_4$ of the four diagonals. Again, from $|G|=24$ we conclude that $phi$ is an isomorphism.
(It is not part of the quoted text, but I assume the author established $|G|=24$ in a similar argument: $G$ acts transitively on the six faces, and the stabilizer of one face is a cyclic group of four rotations, therefore $|G|=6\cdot 4$.)
• Thank you for your help. I understand what author want to show. but one thing I want more is that how the function $\phi : G \rightarrow S_4$ works in detail. I'm hard to imagine how to works the function. you mean for $R_{90^\circ} \in G$ s.t $\phi(R_{90^\circ})=\alpha =(1234)$? and then how to know that $\phi (G)$ contain all eight element by knowing only two element ($\alpha, \beta$) contained in $\phi (G)$? – fivestar May 4 '18 at 5:30