Show that for even integers $ n $ there exists an element $ g \in D_{n} $ : $ \operatorname{ord}(g)=2 $ and $ \operatorname{sgn}(\varphi(g))=1 $ Let $ n>3 $, let $ \Delta_{n} $ be a regular $n$-corner, and let $ D_{n} $ be the dihedral group which is the symmetry group of $ \Delta_{n} $. If we number the vertices of $ \Delta_{n} $, we obtain an injective homomorphism $ \varphi: D_{n} \rightarrow S_{n} $.
Show that for even integers $ n $ there exists an element $ g \in D_{n} $ such that $ \operatorname{ord}(g)=2 $ and $ \operatorname{sgn}(\varphi(g))=1 $.
Attempt:
We first considered what such an element with these properties might look like and concluded $g =
  \left[ {\begin{array}{cc}
    1 & 0 \\
    0 & -1 \\
  \end{array} } \right]$ (reflection on the x-axis) would be an element of $D_{n} $ with $ \operatorname{ord}(g)=2 $ and $ \operatorname{sgn}(\varphi(g))=1 $.
Now we know what such an element exists and looks like, would this already be sufficient proof for the existence of such an element? If not, how would one show this formally?
 A: It is not perfectly clear to me, why „reflection at the x-axis“ is a welldefined isometry, unless you explicitly state in which way you represent your $n$-gons in $\Bbb R^2$.
For example, if you choose to represent the square (ie. 4-gon) as $\{v\in \Bbb R^2\mid \Vert v \Vert_\infty\leq 1\}$ (the square with sides parallel to the $x$- and $y$–axis), then reflection at the $x$–axis does give a permutation of signum 1 (I assume $\operatorname{sign}: S_n \rightarrow \{\pm 1\}$) since we transpose the upper right with the lower right corner and the upper left with the lower left corner.
However, if you choose to represent the square as $\{v\in \Bbb R^2\mid \Vert v \Vert_1 \leq 1\}$ (the diamond), then reflection at the x-axis just swaps the upper tip with the lower tip, hence gives a permutation of signum $-1$.
So let‘s agree on a standard representation of the regular $n$-gon as the convex hull of the $n$-th primitive roots of unity in $\Bbb C$, ie. as $\operatorname{conv}(\{z\in\Bbb C\mid z^n =0\})$. Note that $1$ is always a primitive root of unity and that any root of unity $z$ also has $\overline{z}$ a root of unity. This means that in any case reflection at the x-axis (ie. Conjugation) gives a welldefined isometry of the $n$-gon.
If $n$ is odd this in fact induces a permutation of the vertices of signum 1, since conjugation has to fix $1$, while it transposes each other (automatically non-real) root of unity with its conjugate. This is an even number of transpositions.
However, if $n$ is even we run into the problem discussed with the diamond-shaped square. We have two real roots of unity, namely $1$ and $–1$ and thus are left with $\frac{n-2}{2}$ transpositions. This number decides whether reflection at the x-axis induces a permutation of signum $1$ or $–1$. In the case where $\frac{n-2}{2}$ is odd (as it is for the square) we have to use another isometry. I leave it as an exercise to check that reflection along the line spanned by $\frac{e^{\frac{2\pi i}{n}}+ 1}{2}$ does the job. (for the diamond shaped square this is the reflection along $(1,1)$).
