Let $P$ be any convex polygon with $n$-sides. Let $k$ be the number of axes of symmetry of $P$. Prove that if $k>0$, then $k$ divides $n$.
The first thing that came to my mind was Lagrange's theorem, which didn't seem to work. Then I found this but I'm not sure if I really understand what's going on. The proof goes as follows.
Let $G$ be a group of all symmetries of a polygon $P$ and $G_+$ be the subgroup of the group $G$ consisting of symmetries which do not change the orientation of the polygon $P$. Then $|G:G_+|=2$ and an element $g\in G$ is an axial symmetry iff $g\in G\setminus G_+$. So $|G|=2k$. By Brower theorem, each element $g\in G$ has a fixed point. Thus the elements of the group $G_+$ are rotations. Therefore the action of the group $G_+$ on the set $V$ of vertices of the polygon $P$ has no fixed points. Hence each $G_+$-orbit in the set $V$ consists of $k$ elements and so $|V|$ is divisible by $k$.
So $G$ is the symmetry group of $P$. Clearly $G$ consists of all possible rotations and reflections. We have to make sure that all rotational symmetries of $P$ have a common center so that their composition is an element of $G$. But this is guaranteed by the fact (I've seen this in numerous occasions but can someone prove it?) that all axes of symmetry of $P$ pass through a common point.
$G_+$ consists of elements of $G$ which are not reflections. So all rotations are elements of $G_+$. It is easy to see that $G_+$ is a subgroup of $G$. For if two symmetries of $P$ are both orientation-preserving, then so is their composition.
Then we determine the order of $G_+$. Let $g_1,\dots,g_k$ denote all possible reflections. Recall that a rotation followed by a reflection is a reflection. Consider any rotation $r$. Since $g_1^{-1}$ is a reflection, so is $g_1^{-1}\circ r$. Therefore $g_1^{-1}\circ r=g_i$ for some $i\in \{1,\dots,k\}$. Hence $r=g_1\circ g_i$. It follows that every rotation is the composition of $g_1$ and another reflection. Since $g_1\circ g_{i_1}\neq g_1\circ g_{i_2}$ whenever $i_1\neq i_2$, we must have $G_+=\{g_1\circ g_i:i\in \{1,\dots,k\}\}$, meaning that $|G_+|=k$.
Why is Brouwer fixed-point theorem needed here? Given $v\in V$, we know that ${G_+}_v$, the stabilizer subgroup of $G_+$ with respect to $v$, is trivial because a non-trivial rotation fixes no points. By the orbit-stabilizer theorem, the size of the orbit of $v$ is $|G_+|/|{G_+}_v|=k/1=k$. Since the collection of all orbits form a partition of $V$, we know that $k$ divides $|V|$.
Can anyone answer my questions or come up with a new proof? Thanks.