Let $n \ge 1$ be an integer. Let $F \in D_{(2n+1)}$ be a reflection. Consider $F$ as an element of the permutation group of the vertices of a regular $(2n + 1)$-gon. Is this permutation even or odd?
My thinking is that the permutation would be even since if you have a permutation of odd order, then that permutation can be written as the product of an even number of two-cycles.
For example, if $\sigma = (1234)(57)$ in $S_7$, then $\sigma$ can be written as the product of 2-cycles, $(14)(13)(12)(57)$ Since $\sigma$ can be written as the product of an even number of 2-cycles, we conclude that it's an even permutation.
So, for this question, we're dealing with a polygon with an odd number of vertices and sides. $D_{(2n+1)}$ has $2n+1$ rotations and $2n+1$ reflections. Since the number of reflections is odd, doesn't that imply that they can be written as the product of an even number of 2-cycles? Or am I misinterpreting the question?
Any assistance would be most appreciated.
Kindest Regards
Hint
If there's one answer, the easiest would be to just look at a triangle. In that case two of the points get interchanged, one staying fixed. That implies a single transposition, and so odd order.
But to verify try a couple slightly more complicated examples. Yep. When I try a regular pentagon, I get reflection as a product of two transpositions. So even.
The pattern is easy to see. It's even when $n$ is even; odd when $n$ is odd.
That's if you draw a line through one point and split the opposite side, you then have $n$ pairs to interchange.. So $n$ transpositions.