Given some finite $S\subseteq\mathbb R^2$, it is clearly possible for $S$ to have arbitrarily many lines of symmetry. However, it is not very clear if the same is necessarily true for subsets of $\mathbb{Z}^2$. Is it possible for finite subsets of $\mathbb{Z}^2$ with at least two elements to have an arbitrarily large number of lines of symmetry, or is there some upper bound on this number?
1 Answer
At first, all lines of symmetry have a common point $O$ (barycentre of $S$). Next, if $S$ has two lines of symmetry $\ell_1,\ell_2$, then $S$ is invariant under rotation on the angle $\phi=2\angle(\ell_1,\ell_2)$, which is the composition of two symmetries. If $1<|S|<\infty$, then the orbit of any point except $O$ is infinite unless $\phi$ is a rational multiple of $\pi$. In this latter case the orbit of a point $P\ne 0$ is a regular polygon (or 2 points set). But regular $n$-gon on the square lattice may exist only for $n=4$
(Proof. Consider for given $n$ a minimal regular $n$-gon $A_1\dots A_n$ on the grid. If $n\geq 7$, there exists a smaller polygon $B_1\dots B_n$, where $B_i=A_{i+1}-A_i$, indices modulo $n$. A contradiction. $n=6$ reduces to $n=3$, and if $ABC$ is an equilateral triangle with $A,B,C\in \mathbb{Z}^2$, then double area of $ABC$ equal to $\sqrt{3}AB^2/2$, this is irrational while double area of lattice polygon is always integer.
Another possible argument. If $n$ is odd, let $A_1=(0,0)$, $A_i=(x_i,y_i)$. If all $x_i$, $y_i$ are even, there exists smaller polygon with vertices $(x_i/2,y_i/2)$. If $A_1A_2^2=(x_{i+1}-x_i)^2+(y_{i+1}-y_i)^2$ is odd, then $x_{i+1}-x_i+y_{i+1}-y_i$ is odd for all $i$, summing up we get a contradiction. If $A_1A_2^2$ is divisible by 4, then $x_{i+1}-x_i$, $y_{i+1}-y_i$ are even, so $x_i$, $y_i$ are all even, this is considered above. If $A_1A_2^2=4k+2$, then $x_{i+1}-x_i$ is odd and summing up we get a contradiction. Case $n=8$ may be considered similarly to $n=3$ above.)
This proves that there may exist at most 4 lines of symmetry.