Are the Dihedral Groups of order $n!$ isomorphic to $S_n$?

Question

It is well known that the symmetries of a triangle, which is the Dihedral Group of order 6, is isomorphic to $S_3$. It is clear that both of these have 6 elements. However, $D_4$, the symmetries of the square, are isomorphic to a subgroup of $S_4$, which has $24$ elements. If some Dihedral Group of order $n$ is equal to $m!$ for some integer $m$, then is that Dihedral Group isomorphic to $S_m$? If so, why?

Answer 1

$S_n$, unlike the dihedral group, has a unique subgroup of index two, $A_n$, that is never abelian ($n\gt3$).

The dihedral group, $D_m$, contains a subgroup of index two that is cyclic. A group of rotations.

Answer 2

All dihedral groups are solvable.

For $n \ge 5$, the symmetric group of degree $n$ is non solvable.

For $n=4$, note that every element of $S_4$ has order at most $4$, while $D_{12}$ has an element of order $12$, whence $S_4 \ncong D_{12}$.

Therefore, $D_m \cong S_n \iff n=m=3$.