Permutation and group isomorphism question. $S_3$ is isomorphic to $D_3$ using an equilateral triangle.
Similarly, the book uses a square to represent $D_4$. Now, I know the geometries (rotations, reflections) on this square which comes up to a total of  8 transformations. Now, I am confused because $|S_4| = 24$. Does this mean $S_4$ is not isomorphic to $D_4$? 
If not, which symmetric group is $D_4$ isomorphic to? Thank you!
 A: You're correct, $D_4$, which is of order $8$, is NOT isomorphic to $S_4$ which is of order $4! = 24$. 
$S_4$ is isomorphic to the group of rigid motions (rotations) of the cube, acting on the vertices, and/or the faces, and/or the diagonals of the cube).
$D_4$ is isomorphic to a subgroup of the Symmetric group $S_n$, by Cayley's Theorem. Indeed, is is generated by two elements of $S_4$: $(1234)$ and $(13)$, and is hence, isomorphic to a subgroup of $S_4$, a subgroup of order $8$ in $S_4$: So $D_4 = \langle (1234)(13)\rangle \lt S_4$.
You can learn more about $D_4$, the dihedral group (sometimes called $D_8$) and its relation to other groups, at the Groupprops ("group properties") Wiki site : a nice source to learn about groups, in general!! See, in particular: Dihedral Group $D_4$, 
A: An isomorphism between groups is in particular a bijection between the underlying sets. For two finite sets there is a bijection between them if and only if they have the same number of elements.
Therefore there is no bijection between a set of $8$ and a set of $24$ elements. In particular, if so, there is no isomorphism between these two groups. 
A: Just to supplement the other posts, $D_4 < S_4$ because, for instance, if you keep to adjacent vertices of the square fixed, and swap the other two, you aren't applying a rigid motion.
