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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?

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2 Answers 2

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

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  • $\begingroup$ 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
    Nov 20, 2019 at 1:23
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    $\begingroup$ 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
    Nov 20, 2019 at 1:35
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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 $\theta_{T'}\equiv \theta_T \pmod {2\pi}\Leftrightarrow \theta_R\equiv 0\pmod {2\pi}$.

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