A claim says that two permutations are conjugate when they have the same type. I don't know how to show (using this claim) that a dihedral group $D_{2n}$ isn't a normal subgroup of a permutation group $S_n$.
Thank you.
Why a dihedral group isn’t a normal subgroup of a permutation group
dihedral-groups
Best Answer
First off, we need $n>3$, as $D_{2\cdot 3} = S_3$, and therefore trivially normal, and $D_{2\cdot 2}$ is larger than $S_2$.
With this assumption on $n$, consider the fact that $D_{2n}$ contains an $n$-cycle, but it doesn't contain all $n$-cycles.
Alternately, we may instead assume that the question wanted us to prove that no subgroup of $S_n$ isomorphic to $D_{2n}$ is normal, instead of just the canonical subgroup that arises from the symmetries of a regular $n$-gon. Let $G\subseteq S_n$ be a subgroup isomorphic to $D_{2n}$. Note that $D_{2n}$ and therefore $G$ has either exactly $n$ elements or exactly $n+1$ elements of order $2$ (depending on the parity of $n$).
If $n>5$, any possible cycle type of order $2$ in $S_n$ has more than $n$ elements. Thus there is a conjugation of $S_n$ that takes an order $2$ element of $G$ and sends it to a different permutation of the same cycle type not in $G$. So $G$ is not normal.
For $n = 4$ and $n = 5$ there are order-2 cycle groups with the correct number of elements. However, in those cases (as with any prime power), the argument at the top of the answer works: Any subgroup of $S_n$ which is isomorphic to $D_{2n}$ must contain an $n$-cycle. And there are $(n-1)!$ of those in $S_n$, which is too many for the subgroup to be normal.