It should be obvious that the rotations form a normal subgroup of index 2 (the fact that it is of index 2 is sufficient to prove normality). If we call the rotation subgroup $R$, and the reflection coset $F$, we have:
$RR = R$
$RF = F$
$FR = F$
$FF = R$.
Subgroups containing only rotations are cyclic, due to the fact that $R$ is cyclic. We thus get exactly one subgroup of order $d$ contained in $R$, for each divisor $d$ of $n$.You should prove any (and all) of these subgroups are normal.
Subgroups containing only reflections and the identity must have order a power of 2. Since a reflection times a reflection is a rotation, with $(r^ks)(r^ms) = r^{k-m}$, it should be clear that any such subgroup is in fact of order 2 (we must have $k = m)$. These subgroups are typically NOT normal, but there is an exception for $n = 2$ ($D_2 = V$ is abelian).
Which brings us to "mixed subgroups", containing at least one rotation, and one reflection. These are going to be of the form $\langle r^k,s\rangle$ where $k|n$, that is, isomorphic to $D_m$, where $m = \dfrac{n}{k}$ . You can think of these as symmetries of an $m$-gon, which are also symmetries of an $n$-gon, since $n$ is a multiple of $m$ (the axes of symmetry of an $n$-gon include all the axes of symmetry of the $m$-gon, plus more).
These "mixed subgroups" aren't, in general, normal, but in some special cases they are: for example if $n$ is even and $k = \dfrac{n}{2}$, or $k = 2$.
Another case worth mentioning is when $n$ is an odd prime; in this case, any subgroup containing more than one reflection, or a reflection and a (non-trivial) rotation, is the entire group, which limits the possibilities for subgroups.
For a more complete analysis, see: https://in.answers.yahoo.com/question/index?qid=20091014113730AA7KJDt
John Smith Kyon, your idea is very good.
If we remove all unnecessary things from your reasoning,
we get a solution to our problem.
It looks something like this.
Let $n$ be odd and let $N$ be a normal subgroup in $D_n$ of order $2m$.
Step 0.
We know that $I=\{s,sr,\ldots,r^{n-1}\}$ is
a complete list of all elements of order $2$ of group $D_n$.
Step 1.
Since $N$ is of even order, we get $N\cap I\neq\varnothing$.
Let $sr^k\in N$.
Step 2.
Since $N$ is a normal subgroup of $D_n$, then
$$
r^t(sr^k)r^{-t}=sr^{k-2t}\in N \hbox{ for all $t\in\mathbb{Z}$}.
$$
Step 3.
Since $n$ is odd, we obtain that the set
$$
\{k-2t\mid t=0,1,\ldots,n-1\}
$$
is a complete residue system modulo $n$.
It follows that $I\subset N$. So $N=D_n$.
Best Answer
The index $2$ suggestion works, but you can also show this directly. One can check that the generators $R$ and $F$ of the dihedral group conform to the rule $RF = FR^{-1}$. From this, we see that any element in $D_n$ can be written as $R^jF^k$ where $0 \leq j \leq n-1$ and $0 \leq k \leq 1$.
A subgroup $N \leq G$ is normal whenever, given any $n \in N$ and $g \in G$, we have $gng^{-1} \in N$. In this case, any element of the rotation subgroup looks like $R^m$ for $1 \leq m \leq n-1$. Considering any arbitrary element $R^jF^k$ of $D_n$, we just need to show that $(R^jF^k)R^m(R^jF^k)^{-1} \in \langle R \rangle$. Clearly this is true if $k=0$, so assume $k=1$. Now look to the helpful rule in the first paragraph to conclude that this is indeed an element of $\langle R \rangle$.