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 $n!$?
[Math] Difference between dihedral and symmetric group
abstract-algebra
Related Solutions
Mathematically, a symmetry is a bijection of a set onto itself. In some cases, we restrict to bijections that preserve some structure (e.g. isometries, which preserve distances), but in the case of the symmetric group $S_n$ we just have a set with $n$ elements, and no additional structure to preserve, so all bijections (permutations) are allowed.
First we are going to deal with $\rm{D}_{2n}$ where $n$ is an odd number. Then $ \rm{D}_{2n} = <a,b> $ where $a^2=1$ is the reflection and $b^n$ is the rotation. Claim if $n=2m+1$ then the conjugacy classes, are the $\{ b^{i}, b^{-i} \}$ for $ 1 \leq i \leq m$ and $\{a b^{k} | 0\leq k < n \}$. $\\ \\$ We note that $\rm {D} _{2n}= \{a^{i}b^{j}| 0\leq i \leq 1 , 0 \leq j \leq n\}$. Take $1 \leq l <m$ and $0\leq k <n$ an element which is only rotation $ b^l $, then we see that $b^{k} b^l b^{- k}= b^l $ and $ a b^{k}b^l ( a b^{k} )^{-1}= a b^{k}b^l b^{-k}a = b^{-l} $ which proves that the orbit of $ b^l$ is just $\{ b^l , b^{-l} \}$. $\\ \\$ For the other type of conjugacy classes again we do the same computation for an element $ a$ then its orbit is $ \{a b^{k} | 0\leq k < n \}$ using the relation $ ab^{k}a= b^{-k} $ we get that $b^k a b^{-k} =a b^{-2k} $ and because $\gcd(2,n) = 1$, $<b^{-2k}>= <b>$ and by taking conjugation with $a b^k$ we get $a b^k a b^{-k} a =a b^{2k}$. Therefore we have prove the orbit always stays in the elements of the form $a b^{p}$ and that it covers all the elements which completes the analysis for $n$ odd. We get the conjugacy equation (and of course the identity element has orbit only itself) $2 n=1 + \underbrace {2 +2+ \cdots +2} _{m \rm{\,times}} +n.$
For the even case the analysis is the same but we get some different type of orbits, which is expected since the even dihedral has center whereas the odd does not. Take $n = 2 m $ for $b^{k}$ for $ 1 \leq k < m$ the same computation gives that the orbit of $b^{k}$ is the same namely $ \{ b^k , b^{-k}\}$. We have an inconsistency with $r^m$ because tihs element with the same computation we get it goes to $r^{-m}$ but $r^{-m} = r^{m}$ which proves it is only a one element orbit. Now again when we try to find the orbit of the reflection element $a$, again with the same computation as before we take elements of the form $ a b^{2k}$ but since $\gcd(2, n) =2$, $b^2$ has order $m$ and so its orbit take only the elements of the form $a b^l$ where $l$ is even. We are left with one more orbit namely $a b^{p} $ where $p$ is odd and by doing again the same computation this is indeed the orbit of $a b$. The class equation is now $$ 2n = 1+1+ \underbrace {2 +2 +\cdots +2 }_{m-1 \rm{times}} +m +m.$$
Best Answer
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.