Permutations Isomorphic to $S_4$ Prove that the group generated by permutations $(0 2 6 4)(1 3 7 5)$ and $(4 2 1)(6 3 5)$ are isomorphic to the symmetric group $S_4$. 
I approached this problem by labeling the vertices of a cube. Since each of these permutations fix all possible rotations of the cube does that imply it is isomorphic to $S_4?$
I also think that the fact $(1 2 3 4)(1 2)$ generates $S_4$ can be useful if we consider $1$, $2$, $3$ and $4$ as diagonals of the cube. 
 A: Your idea looks fine to me. You can label vertices of the cube in such a way that 0264 form the bottom floor and 1375 the top floor such that a 90 degree rotation permutes the vertices via $r=(0264)(1375)$. In this labeling the space diagonals correspond to  pairs of vertices $A=07$, $B=25$, $C=61$, $D=43$. The permutation $r$ acts on the set of diagonals via as the 4-cycle $(ABCD)$. The permutation $s=(421)(635)=(142)(635)$ is a 120-degree rotation about the diagonal $07$, and as $s(6)=3$, $s(1)=4$ maps the diagonal $C=61$ to diagonal $D=43$. Continuing this we see that $s$ acts as the 3-cycle $(CDB)$.
Anyway, the two permutations map diagonals to diagonals, hence so do all the permutations in the group generated by $r$ and $s$.
It is easy to show that $(CDB)$ and $(ABCD)$ generate all of $Sym(\{A,B,C,D\})\cong S_4$.
For example you get all the eight 3-cycles by conjugating powers of $(CDB)$ by powers of $(ABCD)$. Those 3-cycles generate all of $A_4$. But $(ABCD)$ is an odd permutation, so we necessarily get all of $S_4$.
This proves your claim.
