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
The phrase "degrees of freedom" refers to the number of independent real number valued parameters (as does the more formal term "dimension").
So, for example, each orientation preserving isometry of $n$-dimensional Euclidean space $E^n$ is a composition $T \circ O$ of a unique orthogonal transformation $O$ followed by a unique translation $T$, and these two factors are independent of each other. Translations are parameterized by vectors which have $n$ degrees of freedom, and rotations of $E^n$ are parameterized by orthogonal $n \times n$ matrices which have $\frac{(n-1)n}{2}$ degrees of freedom. Since the degrees of freedom of translations and of rotations are independent of each other, together they have $n + \frac{(n-1)n}{2}=\frac{n(n+1)}{2}$ degrees of freedom.
Now let's bring in reflections. Let me use $R_n$ to denote the reflection in the coordinate plane $x_n=0$. A general orientation reversing isometry can be expressed uniquely in the form $T \circ O \circ R_n$ where $T,O$ are independently chosen translation and orthogonal transformation. It follows that a general isometry can be written uniquely in the form $T \circ O \circ (R_n)^e$ where we independently choose three things: the translation $T$ with $n$ degrees of freedom; the orthogonal rotation $O$ with $\frac{(n-1)n}{2}$ degrees of freedom; and the exponent $e \in \{0,1\}$.
The choice of the exponent $e$ is discrete --- either $0$ or $1$ --- and this does not represent a "degree of freedom" because it is not a real number valued parameter.
By the way, what makes the calculations in my answer work correctly is the fact that the decomposition $T \circ O \circ (R_n)^e$ is uniquely determined by the isometry. In your question, where you use the fact that every isometry can be written as a composition of reflections, that composition is not unique, and so using it to count degrees of freedom can lead to errors.
Just as an example, picking any even number $n \ge 2$, every translation of the line can be written as a product of $n$ reflections. Since a reflection of the line has $1$ degree of freedom (namely, the reflection point), this would seem to be a proof that the translations of the line have dimension $n$. Since $n$ is arbitrary, there is an error in this argument... which I will leave you to ponder.
Best Answer
You're almost there. :) It may help to show that reflections across orthogonal lines commute: If $R_{1}$ and $R_{2}$ are reflections in orthogonal lines through a point $p$, then $R_{1} R_{2} = H_{p}$, the half-turn about $p$, and the same argument applies to $R_{2} R_{1}$.
Now if $R$ is an arbitrary reflection across a line through the origin, let $R^{\perp}$ denote reflection about the orthogonal line through the origin, decompose the half-turn about the origin as $H_{0} = RR^{\perp}$, and observe that each factor commutes with $R$.
(The algebraic version, of course, is that $H_{0}$ is induced by the scalar matrix $-I$, which commutes with every linear transformation, particularly with every linear reflection.)