# How to construct a shape in the plane which has infinite group of symmetries?

A circle is one such example of a plane shape which has infinite symmetry.

How would I be able to construct other shapes on the plane having infinite group of symmetries.

• What are you including under the heading shapes? Must a shape be bounded, or can you consider, for instance, the set of all points whose Cartesian coordinates are both integers? – Brian M. Scott Sep 14 '13 at 4:02
• Preferably shapes which are bounded. – Astrolink Sep 14 '13 at 4:05
• If you’re interested only in bounded shapes and don’t take into account their absolute position in the plane, but only their orientation, then you’re looking at shapes that have the same symmetries as the circle: disk, annuli, etc. – Brian M. Scott Sep 14 '13 at 4:13
• This question has popped up twice recently. One is here. I cannot find the other one. It has a neat answer based around the unit disk with punctures at points on the boundary corresponding to rational numbers (thus the shape has a countable group of symmetries). – user1729 Sep 14 '13 at 11:26

Take a set of points $S$ and an infinite group $G$ which acts on the plane. Then the orbit of $S$ under $G$ (i.e. the collection of points $g\cdot s,g\in G,s\in S$) is such a set. For example, if $S=\{(0,1)\}$ and $G$ is the group of rotations in the plane, you get the circle. If $S$ is a vertical line and $G$ is the group of horizontal translations by an integer distance, you get the set of vertical lines with integer $x$-intercepts. Etc.
Take two points, one of them the origin. Might as well be $(0,0)$ and $(0,1)$ Now rotate around each point by some angle that doesn't divide $\pi$ rationally or the negative of that angle. Keep going countably many times. You have a set closed under a free group. It is uncountable.