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Is it possible to draw a figure that has exactly one reflection symmetry (flip) and one (or more) non-trivial rotational symmetry? (Note: The trivial symmetry is the 0 degrees or 360 degreesrotation).

I am doing this problem for a homework assignment, and was assuming that it wasn't possible, but my proof may not be rigorous enough or even proof at all. This is what i think.

if you assume that something has one reflectional symmetry and one non-trivial rotational symmetry you can rotate it and another reflectional symmetry will become present which contradicts the original assumption of a single reflective symmetry. furthermore any figure with a single reflective symmetry will have atleast 2 rotational non-trivial symmetries.

can someone help me word this so it makes more sense? or steer me on the right path?

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  • $\begingroup$ Are you assuming that it is a bounded figure? $\endgroup$ – José Carlos Santos Jul 8 '17 at 18:46
  • $\begingroup$ the question is stated exactly that way on my assignment sheet. yes i am assuming it is bounded $\endgroup$ – August Haze Jul 8 '17 at 18:56
  • $\begingroup$ @AugustHaze Please do not vandalize posted question, yours own included. $\endgroup$ – dxiv Jul 11 '17 at 0:50
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You can use orientation of symmetries here: Per definition a symmetry $\sigma$ is a reflection if and only if $\det\sigma$ is negative. So if there's a rotation $\sigma$ and a reflection $\rho$ in some symmetry group, then $\det(\sigma\rho)=\det\sigma\cdot\det\rho<0$ and $\sigma\rho$ is a reflection as well.

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Hint: Consider combining the operations to create new symmetries. I will use the notation from the Wikipedia article on co-ordinate rotations and reflections: let a rotation about the origin $O$ by an angle $\theta$ be denoted as $Rot(\theta)$. Let a reflection about a line $L$ through the origin which makes an angle $\theta$ with the $x$-axis be denoted as $Ref(\theta)$.

Then we have the following identities: $$Rot(\theta)Ref(\phi) = Ref\left(\phi+\frac12\theta\right)\\ Ref(\phi)Rot(\theta) = Ref\left(\phi-\frac12\theta\right)$$

Without loss of generality, suppose your (unique) reflection symmetry is about $\phi = 0$. (Note that this is equivalent to $\phi = \pi$.) Now use the above equations to arrive at a contradiction.

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Suppose that figure exists. Let $G$ be its symmetry group. Let $\sigma$ be the only symmetry and $\rho$ the only non-trivial rotation. Let $e$ be the identity. Let $\operatorname{ord}(\rho)=n$, that is, the smallest number greater than zero such that $\rho^n=e$. Then, it's clear than $n$ must be $2$, otherwhise exists another rotation $\kappa\equiv\rho^{n-1}\ne e,\rho$ with $\kappa^n=(\rho^{n-1})^n=(\rho^n)^{n-1}=e^{n-1}=e$. So $\rho$ has to be a rotation of $\pi$ radians.

So, we have, $\rho,\sigma,e\in G$ with $\rho\ne\sigma$ and $\rho^2=\sigma^2=e$

Think about what is $\sigma\rho$ and what is $\operatorname{ord}(\sigma\rho)$ and you will get your answer

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My proof is going to be slightly different from the one given by @kremerd.

Suppose the symmetry group $G_F$ of a figure $F$ contains a reflection $s$ and a rotation $r\ne e$, where $e$ denotes the neutral element of the group, i.e. the identity. Since $G_F$ is a group, then $sr=s\circ r$ is also members of $G_F$. But the composition of a reflection and a rotation is also a reflection, therefore $$ sr=s $$ because $s$ is the only reflection in $G_F$. It follows that $$ r=er=s^2r=s(sr)=ss=s^2=e, $$ contracting the fact that $r$ is different from $e$.

Thus you cannot have exactly one reflection.

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