Result:
The number of elements in $D_n$ is $2n$.
I know that for a regular $n$-gon there are $n$ rotational symmetries and $n$ reflectional symmetries but why is it always true. Can anyone please give me any geometrical idea about it?
I feel this idea will help me in proving my original result. I hope I'm correct.
I think you have no doubt on the fact that there are $n$ rotations, that are the rotations of $\frac {2\pi k} n$ $\forall k=0,..,n-1$. About the symmetries we have to distinguish two cases: if $n$ is odd you have an axial symmetry for each segment that goes from the middle point of a side to the opposite vertex, instead if $n$ is even you have an axial symmetry for each couple of opposite sides and a punctual symmetry for each couple of opposite angle. So you have obtained that the dihedral group has at least $2n$ elements, why are we sure that are all the elements? Easily an isometry has to send one side to another (this can be done in $n$ possible choices) and then can have two orientation (for example the side $AB$ could go to $CD$ or with $A\to C$ and $B\to D$ or with $A\to D$ and $B\to C$) . So we conclude that dihedral group $D_n$ has $2n$ elements: $n$ rotations and $n$ symmetries.