Is Dihedral group just for small order? I'm sorry that the title is not really specifying a question, but i cannot think of a single sentence describing this question.
Dihedral group is defined to be rotations and reflections of $n$-agon in elementary abstract algebra texts. These groups are good examples for nonabelian groups.
For very small cardinals such as $6,8,10$, we can actually construct a dihedral group formally by checking a diagram.
However, for an arbitrary $N$, we cannot explicitly do this and the definition above is not formal at all.
Am i thinking correctly? If not, what would be a way to define Diheral group formally?
 A: There is a formal construction of the dihedral group for arbitrary $n$. Here is my favorite way to do it: let $r$ denote the rotation which moves the vertices of an $n$-gon by $2\pi/n$ radians and let $s$ be reflection about a fixed line. The elements $r$ and $s$ satisfy the properties
\begin{align}
r^n &= 1 \\
s^2 &= 1 \\
srs &= r^{-1}.
\end{align}
It turns out that these three properties completely characterize the dihedral group of order $2n$. In other words, I have given you a presentation for the dihedral group.
A: The dihedral group $D_n$ with $2n$ elements can be formally defined as follows: Start with $F_2$, the free group on two generators $a$ and $b$.  Let $S$ be the subgroup generated by $a^2, b^n, $ and $abab$. Since $S$ is normal in $F_2$ one can find the quotient $F_2/S$; this quotient is exactly $D_n$.   The element $a$ of the resulting group is one of the reflection symmetries (since one has $a^2=1$) and the element $b$ is one of the  primitive rotations (since one has $b^n=1$).
The other relation, that $abab=1$, is precisely what is required to get the behavior of the symmetries of an $n$-gon: if you reflect the $n$-gon across an axis, then rotate it by $\frac{2\pi}n$, then reflect it again, then rotate it again, you have the $n$-gon back in its original position.
Since we have $abab=1 = a^2b^n$, we also have $ba =  a^{-1}a^2b^nb^{-1} = ab^{n-1}$.  This gives us a method to put every element of the quotient group into the standard form  $a^ib^k$, where  $i\in\{0,1\}$ and $j\in\{0,\ldots n-1\}$. We do this as follows.  Suppose we start with any product of $a$s and $b$s. We can percolate all the $a$s to the left, using the relation $ba=ab^{n-1}$.  We then arrive at something of the form $a^pb^q$.  Since $a^2=1$ and $b^n=1$ we can reduce this to $a^ib^j$ where $0\le i<2, 0\le j<n$.
For example, the word $ab^3a^3b$ can be put into the $a^ib^j$ normal form as follows: $$\begin{align}ab^3a^3b &= ab^2(ba)a^2b \\&= ab^2(ba)b \\&= ab^2(ab^{n-1})b \\& = ab^2ab^n \\&= ab^2a \\&= ab(ba) \\& = ab(ab^{n-1}) \\& = a(ba)b^{n-1} \\& = a(ab^{n-1})b^{n-1}\\& = a^2b^{2n-2} \\&= b^{2n-2}\\& = b^nb^{n-2} \\&= b^{n-2}.\end{align}$$  
This shows that the quotient group has exactly $2n$ elements.
