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 $G \cong S_4$.
Best Answer
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.