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.
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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.
Arthur
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Also, in my opinion, this exercise shows how silly it is to insist on the $D_{2n}$ naming scheme. Every other permutation group ($S_n, A_n, C_n$) indicates the size of the set it acts on, not the order of the group itself (although one could arague that $C_n$ does both). The full symmetry group acting on six elements is never $S_{720}$ and the alternating group is never $A_{360}$, but somehow the dihedral group is some times $D_{12}$ and some times $D_6$. – Arthur Sep 09 '19 at 08:58
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So if I consider an n-cycle r = (1 2 3 ... n) and some other n-cycle q which doesn't belong to D2n, I can find permutation a from Sn such that ara-1 = q, and the condition for D2n being a normal subgroup doesn't hold? – Cygne Sep 09 '19 at 10:14
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@Cygne That's exactly right. For instance, for $n = 4$, the dihedral group's only $4$-cycles are $(1234)$ and its inverse $(1432)$. However, $(12)(1234)(12)^{-1} = (1342)$. – Arthur Sep 09 '19 at 10:17
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@Arthur: I interpret the exercise as asking to show (for $n > 3$), that $D_{2n}$ is not isomorphic to a normal subgroup of $S_n$. In its current form, I don't think your answer shows that. – quasi Sep 09 '19 at 17:52
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@Arthur: Can you respond to my comment? – quasi Sep 10 '19 at 09:41
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@quasi Sure. Sorry. I wanted to give a correct response, so I waited, and then I forgot. I'll add that to my answer. – Arthur Sep 10 '19 at 10:05
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@Arthur: Thanks. – quasi Sep 10 '19 at 10:06
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@quasi I think something like that should do it. – Arthur Sep 10 '19 at 10:11
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@Arthur: If $n=4$, an element of $S_4$ which is a product of two disjoint $2$-cycles has $3$ conjugates. – quasi Sep 10 '19 at 10:16
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@Arthur: Also, doesn't $D_{2n}$ have exactly $n$ or $n+1$ elements of order $2$? – quasi Sep 10 '19 at 10:40
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@quasi $x^2 = e$ has exactly $n$ or $n+1$ solutions, but there is the identity element. – Arthur Sep 10 '19 at 10:41
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@quasi My mind is apparently everywhere today, except on the things I'm trying to actually do. – Arthur Sep 10 '19 at 10:45
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@Arthur: No rush. In the meantime, (+1). – quasi Sep 10 '19 at 10:46
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@quasi Thanks. And I think I have covered it all this time. – Arthur Sep 10 '19 at 10:49