I want to solve the following exercise from Dummit & Foote. My attempt is down below. Is it correct? Thanks!

Show that the group of rigid motions of a cube is isomorphic to $S_4$.

My attempt: Let us denote the vertices of the cube so that $1,2,3,4,1$ trace a square and $5,6,7,8$ are the vertices opposite to $1,2,3,4$. Let us also denote the pairs of opposite vertices $d_1,d_2,d_3,d_4$, where vertex $i$ is in $d_i$. To each rigid motion of the cube we associate a permutation of the set $A=\{ d_i \}_{i=1}^4$. Denote this association by $\varphi:G \to S_4$, where $G$ is the group of those rigid motions, and we identified $S_A$ with $S_4$. By definition of function composition we can tell that $\varphi$ is a group homomorphism.

We prove that $\varphi$ is injective, using the trivial kernel characterisation:

Suppose $\varphi(g)=1$ fixes all of the the pairs of opposite edges (that is we have $g(i) \in \{i,i+4 \}$ for all $i$, where the numbers are reduced mod 8). Suppose $g$ sends vertex $1$ to its opposite $5$. Then the vertices $2,4,7$ adjacent to $1$ must be mapped to their opposite vertices as well. This is because out of the two seemingly possible options for their images, only one (the opposite vertex) is adjacent to $g(1)=5$. This completely determines $g$ to be the negation map which is not included in our group. The contradiction shows that we must have $g(1)=1$, and from that we can find similarly that $g$ is the identity mapping. Since $\ker \varphi$ is trivial $\varphi$ is injective.

In order to show that it is surjective, observe that $S_4$ is generated by $\{(1 \; 2),(1 \; 2 \; 3 \; 4) \}$ (this is true because products of these two elements allow us to sort the numbers $1,2,3,4$ in any way we like). We now find elements in $G$ with images under $\varphi$ being those generators. Observe that if $s$ is a $90^\circ$ rotation around the axis through the centres of the squares $1,2,3,4$ and $5,6,7,8$, such that $1$ is mapped to $2$, followed by a rotation by $120^\circ$ around the line through $2,6$ (so that $1$ is mapped to $3$), we have $\varphi(s)=(1 \; 2)$ .Observe also that if $t$ is $90^\circ$ rotation around the axis through the centres of the squares $1,2,3,4$ and $5,6,7,8$, such that $1$ is mapped to $2$, we have $\varphi(t)=(1 \; 2 \; 3 \; 4)$. Now if $\sigma \in S_4$ is any permutation, we express in as a product involving $(1 \; 2),(1 \; 2 \; 3 \;4)$, and the corresponding product involving $s,t$ is mapped to $\sigma$ by $\varphi$. This proves $\varphi$ is surjective. We conclude that $\varphi$ is an isomorphism, so $G \cong S_4$.

  • $\begingroup$ It is hard to follow your surjectivity argument without a diagram indicating labeling the vertices. If this is a homework assignment, I would suggest including one in your write up that you turn in. It sounds like your $s$ sends $4$ to $3$ (the first transformation sends $4$ to $1$ and the second transformation sends $1$ to $3$), so it does not appear to be what you want in order for $\phi(s) = (1, 2)$. There's an easier proof for surjectivity (see answer below). $\endgroup$ – Michael Joyce Apr 13 '14 at 7:58
  • $\begingroup$ @MichaelJoyce I'm sorry, it should have been $1$ mapped to $8$ in $s$. $\endgroup$ – user1337 Apr 13 '14 at 8:59

The standard way of proving that $\phi$ is surjective is the following. First, show that $G$ has $24$ elements. Then, since you've proven that $\phi : G \rightarrow S_4$ is an injective function between two sets of the same cardinality, it follows that $\phi$ must be surjective. (For if not, the image of $\phi$ would have $< 24$ elements in it, so by the pigeonhole principle, there would be some $\pi \in S_4$ with at least two distinct elements mapping to it via $\phi$, contradicting the injectivity of $\phi$.)

To show that $G$ has $24$ elements, use the orbit-stabilizer theorem. For example, $G$ acts on the set of $6$ faces of the cube, and the stabilizer of a face is a cyclic group of order $4$ generated by a $90^{\circ}$ degree rotation. Thus, $G$ has $6 \cdot 4 = 24$ elements. You could also use the action of $G$ on vertices ($8 \cdot 3$), edges ($12 \cdot 2)$ or diagonals ($4 \cdot 6$) instead of faces.

  • $\begingroup$ Thank you! This is great because I already have that $|G|=24$ from a previous exercise. $\endgroup$ – user1337 Apr 13 '14 at 9:00

Surjective: The group of rigid motions of a cube contains $24$ elements, same as $S_4$. Proof - A cube has $6$ sides. If a particular side is facing upward, then there are four possible rotations of the cube that will preserve the upward-facing side. Hence, the order of the group is $6\times 4 = 24$.

Injective: A cube has $4$ diagonals. For one of the diagonals's head and tale attach $1$, for another $2$ and so on. We choose tale and head of a diagonal same, since there is no way to have rigid motion for a cube to change head and tale and still the diagonal remains in same orientation. For first diagonal you can attach any one of $1$, $2$, $3$, or $4$, as you could choose for first place in $S_4$. For the second diagonal it remains $3$ numbers to choose to attach as same for the second place in $S_4$. And till the last one, and we are done.



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