Order of a symmetry group of a quadrilateral I am studying group theory. When reading text on the theory part, I do not feel difficult to understand the concepts, like what is a group, or what is Cauchy's Theorem. However, I often find it difficult to do problems, like the one below.

Let $Q$ be a plane quadrilateral. Show that its group $G (Q )$ of symmetries has order at most
  $8$. For each n in the set $\{ 1, 2,..., 8\}$, either give an example of a quadrilateral $Q$ with $G(Q)$
  of order $n$, or show that no such quadrilateral can exist.

What I find difficult is the following:


*

*I cannot decide which theorem is related to the question, and when I start doing the problem, I just think of examples, like a rhombus has a symmetry group of order 4.

*How to prove that for a particular value of $n$, say $n=7$, no such symmetry group exists?

*How to relate theorems I've learnt to problems?
 A: Not every problem can be solved by simply invoking a theorem. Some of them require reasoning from the basics.
Note that a symmetry will be completely determined by where it sends the vertices of the $Q$. Let the vertices be $A,B,C,D$  and edges $AB$, $BC$, $CD$, $DA$. There are 4 possiblities for where a symmetry can send $A$, namely $A,B,C,D$. Call the vertex where $A$ is sent $X$. Once we've determined where $A$ goes, there is only one possible place that $C$ can go, and that is to the only vertex that is not adjacent to $X$, because a symmetry must preserve edges. After that there are two possiblities for where we may send $B$, so there are at most $8$ possibilities.
Clearly if $Q$ is scalene the symmetry group has order $1$. For an isosceles trapezoid with two different base lengths the symmetry group has order $2$. For a rectangle that is not a square the symmetry group has order $4$, and for a square the symmetry group has order $8$ (you should verify all of these).
The symmetry group of an arbitrary $Q$ can be seen as a subgroup of the symmetry group of a square where some symmetries aren't allowed. Lagrange's theorem then implies that the order must divide $8$, hence no other orders are possible (here is an example where a theorem can be applied).
A: Let $f$ be a symmetry of the plane quadrilateral $A,B,C,D$ $f$ sends a diagonal to a diagonal, if $(AC)$ is the diagonal through $A$ and $C$, 
$f((AC))=f((AC))$ In this case we have:
$Id$
$f(A)=A, f(C)=C, f(B)=D$
$f(A)=C, f(B)=B$
$f(A)=C, f(B)=D$
If $f((AC))=f((BD))$ we have:
$f(A)=B, f(C)=D, f(B)=A, f(D)=C$
$f(A)=B, f(C)=D, f(B)=C, f(D)=A$
$f(A)=D, f(C)=B, f(B)=A, f(D)=C$
$f(A)=D, f(C)=B, f(B)=C, f(D)=A$
The group of symmetries of $A,B,C,D$ is a subgroup of the symmetric group $S_4$ (since it is contained in the group of permutations of $\{A,B,C,D\}$) which has 24 elements, thus Lagrange theorem implies that its order divides 24 and thus can't be 7.
A: *

*Show that $G(Q)^+$ (the isometries of determinant $1$) acts $\textit{freely}$ on the set of vertices of $Q$. Deduce from this that $|G(Q)^+|$ will divide $4$. 

*Show that either $G(Q)^+=G(Q)$ or $[G(Q):G(Q)^+]=2$ (hint use the group morphism $det:G(Q)\rightarrow \{\pm 1\}$). 

*Deduce from 1 and 2 that $|G(Q)|\leq 8$, and $|G(Q)|$ is a power of $2$ (in particular the rectangle does not have $6$ symmetries).


*For $1$, $2$, $4$, $8$ give a corresponding quadrilateral.


