1
$\begingroup$

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?

$\endgroup$

2 Answers 2

1
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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)? $\endgroup$
    – tau
    Commented Nov 20, 2019 at 1:23
  • 1
    $\begingroup$ 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. $\endgroup$
    – lanskey
    Commented Nov 20, 2019 at 1:35
1
$\begingroup$

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 $\theta_{T'}\equiv \theta_T \pmod {2\pi}\Leftrightarrow \theta_R\equiv 0\pmod {2\pi}$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .