On subgroups of isometries and their respective fixed-points I am working on the following problemset:

Let $G < \DeclareMathOperator{\Iso}{Iso}\Iso(E)$ be a finite subgroup
  of the isometries in the euclidean plane. Denote: $$G = \{g_1, \ldots
 , g_n \} \qquad \text{where} \ n = \vert G \vert$$ define the mapping:
  $$s: E \rightarrow E$$ $$p \mapsto \frac{1}{n} \sum_{i=1}^{n} g_i(p)$$
  for each of the following subgroups define the mapping $s$
a) $G=\{\DeclareMathOperator{\Id}{Id}\Id\}$.
b) $G= \{ \Id ,s_w \}$ where as $s_w$ is the reflection on a straight
  line $w$.
c) $G= \left\langle d_{\frac{2 \pi}{n}} \right \rangle $ the subgroup
  which is created by the rotation around $0$ with the angle $\frac{2
 \pi}{n}$.
d) $G = D_n$ the dihedral group.

My Attempt: 
a) For $G=\{\Id\}$ it is clear to see:$$p \mapsto 1 \sum_{i=1}^1 \overbrace{g_i}^{= \Id}(p) = p$$
b) For $G= \{ \Id ,s_w \}$ we can observe: $$p \mapsto \frac{1}{2} \sum_{i=1}^{2} g_i(p) = \frac{1}{2}( \Id(p) + s_w(p))= \frac{1}{2}( p + s_w(p))$$
c) Here is where my problem starts. Since I know that all rotations have their respective center as fix point, the mapping $s$ has to map $p$ to $0$. Therefore the sum has to be $0$ - correct ? So that means every randomly chosen point of the euclidean plane maps to $0$ while using the mapping $s$ and the $g_i$ the elements of $G= \left\langle d_{\frac{2 \pi}{n}} \right \rangle $. I just don't know how to show that. I mean it seems just so obvious that rotations only have their center as fixed-point. HELP
d) Same problem here. Instead of just considering the rotations we will also take into account the reflections with respect to a straight line which intersects with $0$.
Im very thankful for help on c) and d)
 A: For (c), $G$ is a cyclic group generated by an element $g$ which rotates the least amount in the counterclockwise direction around the fixed point which I'll denote $q$. The sequence of points $g(p), g^2(p), g^3(p), …, g^{n-1}(p), g^n(p)=p$ trace out the vertices of a regular $n$-sided polygon $P$ centered on the point $q$. We have:
$$\frac{1}{n} \sum_{i=1}^n g^i(p) = q + \frac{1}{n} \sum_{i=1}^n (g^i (p) - q)
$$
That last sum is just the sum of the displacement vectors $v_1,v_2,\ldots,v_n$, where $v_i=g^i(p)-q$, pointing from the center $q$ to the vertices of the polygon $P$. Now if you translate $v_1,v_2,…,v_n$ so that the tail of each is incident to the tip of the previous, they close up to form the boundary of a regular polygon, and because of that it follows that their sum $v_1+…+v_n$ equals zero. Thus
$$\frac{1}{n} \sum_{i=1}^n g^i(p) = q
$$
For (d), here's a brief outline. Let $G$ be dihedral of order $2n$. I'll assume for simplicity that the point $q$ that is fixed by $G$ is the origin. There are $2n$ reflection rays for $G$, the group $G$ acts on those rays, and the rays come in two orbits. If the orbit of $p$ lies on one of the reflection ray orbits then the orbit of $p$ has size $n$ and is invariant under the index 2 rotation subgroup and the proof is the same as in (c). So assume the orbit of $p$ does not lie on one of the reflection ray orbits, so the orbit of $p$ has size $2n$. Choose one of the two reflection ray orbits. Using this choice, one can pair up the points in the orbit of $p$: for each of the $n$ chosen reflection rays $R$, pair up the two points in the orbit of $p$ that are closest to $R$. For each of these $n$ point pairs, take their sum. These $n$ sums form $n$ points which are invariant under the index 2 rotation subgroup, and now the proof finishes as in (c).
A: A rotation around $O$ by an angle in $(0, 2\pi)$ changes the angle of any point $\ne O$ and hence does not fix any point $\ne O$.
