Exercise 1: Galois Theory (J. Rotman) Definition: Let $F$ a figure in the plane, its symmetry group is defined by $\Sigma(F):=\{\sigma \in O(2,\Bbb R)\mid \sigma(F)=F\}$. Here $O(2,\Bbb R)$ denotes the real orthogonal group.
Exercise 1:
$\quad$ a) If $F$ is a square, prove that $\Sigma(F)\cong D_8$, the dihedral group of order $8$.
$\quad$  b) If $F$ is a rectangle that is not a square, prove that $\Sigma(F)\cong\Bbb Z_2\times \Bbb Z_2$.
$\quad $ c) Give an example of quadrilaterals $Q$ and $Q'$ with $\Sigma(Q)\cong \Bbb Z_2$ and $\Sigma(Q')=1$.
Context: As said in the title, this exercise comes at the end of a chapter in a book. I summarized this chapter in this sandbox.
My thoughts:
$\quad$ a) It is clear to me that there is $4$ symmetry axis in a square. The induced symmetry $\sigma$ are linear, and preserve distances. This makes $8$ elements, since $\sigma$ is of order $2$. Moreover, we know from Theorem 3 (in the summary) that $\Sigma(F)$ is isomorphic to a subgroup of the symmetric group $S_4$ and $D_8$ has eight elements. But I can't really conclude from here (and the beginning proof doesn't sound very rigorous).
$\quad$ b) Here, we have $2$ symmetry axis each symmetry is of order $2$ and we can compose them so it makes sens to me that $\Sigma(F)\cong\Bbb Z_2\times \Bbb Z_2$. However, I am struggling (again) to write that down in a more mathematical way.
$\quad$ c) 
$\quad$ 
PS: diversity of approaches is very welcome 
 A: a) May be we can describe them.  
Let denote $$\sigma :P\mapsto Pe^{i\frac{\pi}{4}}\quad\text{and}\quad\rho:P\mapsto -P$$ You clearly have the identity, $\sigma$, $\sigma ^2$ and $\sigma ^3$ and $\rho$ that are in $\Sigma(F)$. Now if you apply $\rho$, then you can also apply $\sigma, \sigma ^2$ and $\sigma ^3$ on $\rho(P)$, therefore $\sigma \rho,\sigma^2\rho$ and $\sigma ^3\rho$ are also in $\Sigma(F)$, we thus have $8$ elements. Now remark that $$\sigma ^4=1,\quad \rho^2=1,\quad\text{and}\quad\sigma \rho\sigma \rho=1,$$ therefore, $\Sigma(F)\cong \mathcal D_8$.
b) Here, you only have the identity, $\varphi:=\sigma ^2$, $\rho$ and $\varphi\rho$. It's easy to prove that all these element has order 2 and that they commute. You know that a commutative group with 4 elements is isomorphic to $\mathbb Z_4$ or $\mathbb Z_2\times \mathbb Z_2$. But as all these elements has order $2$, we conclude that $\Sigma(F)\cong \mathbb Z_2\times \mathbb Z_2$.
c) Your figure are correct.  
A: Part (a): let $r$  be the rotation of $\frac \pi2$ around the centre of the square and $s$ the symmetry with respect to one of the diagonals of the square. The other (axial) symmetries are $sr, sr^2, sr^3$. Symmetry with respect to the centre of the square is $r^2$ and the inverse rotation is $r^3$.
Now the dihedral group $D_4$ is generated by two elements $a$ and $b$ that satisfy the following relations: $$a^4=1,\quad b^2=1,\quad ab=ba^3.$$
These are exactly the relations satisfied by $r$ and $s$. So you just have to define a morphism from $\Sigma(\text{square})$ to $D8$ (or $D_4$ according to other standards) by sending $r$ to $a$, $s$ to $b$, extending to the other elements so that it is compatible with the group laws.
Part (b): same method. $\Sigma(\text{rectangle})$ contains four elements: $\operatorname{id}$, the symmetry  with respect to the centre of the rectangle $s_O$ and the symmetries with respect to the horizontal and vertical axes, $s_H$  and $s_V$. Define a map from $\Sigma(\text{rectangle})$ to $\mathbf Z_2\times \mathbf Z_2$ by sending $\operatorname{id}$ to $(0,0)$, $s_H$ to $(1,0)$, $s_V$ to $(0,1)$ and $s_0$ to $(1,1)$, then check it is compatible with the group laws
