Let P be a regular n-gon centered at the origin O, with vertices [v0,
v1, ..., v{n-1}].
Let S be the set of all the symmetries of P. That is, S consists of
all the mappings of the plane to itself that take P to itself.
S contains 2n elements. It contains n rotations and n reflections.
For each integer i in [0, 1, ..., n-2], you can rotate P clockwise by
360 * i / n degrees, and that is a symmetry of P. And for each vertex
v{i}, you can reflect P across the line through v{i} and O.
For example, consider n=3, the triangle. There are 3 rotational
symmetries, by 0, 120, and 240 degrees repectively, and there are
three reflections, one across each of the three altitudes of the
triangle.
We will call the rotations by the names r0, r1, ... r{n-1}, and the
reflections by the names R0, R1, ... R{n-1}. Then S is the set
{r_i, R_i} for i in [0, ..., n-1].
What is the group operation? It's usually written "o", and
a o b
is defined to be the symmetry that you get if you do b, and then do
a. Let's go back to the triangle for concreteness.
r2 o r1
means to rotate the triangle clockwise by 120 degrees, and then to
rotate it clockwise by 240 degrees. This makes a total of 360
degrees, which is equivalent to not rotating it at all, so we have
r2 o r1 = r0
and similarly
r1 o r2 = r0
Now consider
R0 o R0
we reflect the triangle acros a vertical line, and then reflect it
again; this puts everything back the way it was, so we have
R0 o R0 = r0