# 2D symmetry group with 1 or more rotations (nontrivial) but exactly 1 reflection (nontrivial)

For there to be only one reflection symmetry, the 2D polygon has to have one line of symmetry but no others, so maybe something like a Christmas tree or arrowhead shape. But the problem is that then there are no nontrivial rotational symmetries. If I consider a line segment, then there is only one reflection, but it is identical to a rotation in that case.

It seems impossible to have exactly one reflection and 1 or more rotations in 2D, but how could I prove that? Or maybe I am mistaken and it IS possible?

Suppose $$s$$ is a reflection and $$r$$ is a non-trivial rotation. Then $$rs$$ is a reflection and $$rs\ne s$$.

• sorry, but could you explain more? are you saying that if such a shape existed, that the composition of symmetries should be another symmetry yet rs cannot equal s but also cannot equal r, and therefore there is no such group (no closure)?
– tau
Commented Nov 20, 2019 at 1:23
• Yes, any group of symmetries of a 2D object that has a reflection and a non-trivial rotation would have at least one more reflection. Commented Nov 20, 2019 at 1:35

Consider the vertices in polar coordinates $$(r,\theta)$$, any rotation can be written as $$R:(r,\theta)\mapsto (r,\theta+\theta_R)$$, any reflection can be written as $$T:(r,\theta)\mapsto (r,\theta_T-\theta)$$. Now we have, $$R\circ T(r,\theta)=R(r,\theta_T-\theta)=(r,\theta_T+\theta_R-\theta)$$

Thus $$T'=R\circ T$$ is a reflection $$T':(r,\theta)\mapsto (r,\theta_T+\theta_R-\theta)=(r,\theta_{T'}-\theta)$$. $$T'$$ is equivalent to $$T$$ iff $$\theta_{T'}\equiv \theta_T \pmod {2\pi}\Leftrightarrow \theta_R\equiv 0\pmod {2\pi}$$.