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?

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).
