I am trying to wrap my head around the difference between a dihedral group and a symmetric group. If we think about the dihedral group as a set of vertices of some polygon labeled $1$ through $n$ and elements of the group being all the permutations of that set and if the same set is the source for all the permutations of a symmetric group then why is the order of dihedral group not also $n!$?
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1See also this question. – Dietrich Burde Aug 31 '19 at 09:17
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Where have you seen that the order of $D_n$ is $n!$ ? It is much less, $2n$ (https://en.wikipedia.org/wiki/Dihedral_group) – Jean Marie Aug 31 '19 at 11:17
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Consider the group $D_4.$ It represents the motions you can apply to a square. No matter how we rotate or reflect that square, there is no way diagonal vertices become adjacent. This is one way to think about why there are only 8 members to this group and not 24.
Doug M
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The dihedral group $D_n$ has only $2n$ elements, because it contains only the permutations arising from rigid motions of the plane preserving the regular $n$-gon. Also algebraically the groups $D_n$ and $S_n$ are quite different. Of course, the order is different for $n>2$, and $S_n$ is not solvable for $n>4$, whereas all dihedral groups $D_n$ are solvable, even $2$-step solvable. Also, $S_n$ for $n>3$ has trivial center $Z(S_n)=1$, whereas $D_n$ has non-trivial center for $n$ even. $D_n$ has a normal subgroup $C_n$ of index $2$, so that we can write $D_n=C_2\ltimes C_n$.
Dietrich Burde
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The dihedral group only contains permutations that result from rigid motions of the plane, preserving the distance between points. Equivalently, it only contains permutations that preserve the connectivity of the edges of the polygon.
Chris Culter
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