Group Theory symmetry questions 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?
 A: 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.
A: 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
A: 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.
A: 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.
