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(a) Let $\sigma$ be the elements of the dihedral group $D_{2n}$ given as the rotation with $360/n$ and let $\rho \in D_{2n}$ be any reflection. Show that $\rho \sigma=\sigma^{-1}\rho$ and show that $\sigma \rho$ and $\rho$ are elements of order 2 which generate $D_{2n}$.

Approach: I understand how this works if I apply the permutation on any polyhedron. For example, I tried it for $D_4$ and the equality holds. Reflecting the polyhedron over any symmetric axes and then rotating it clockwise by 360/n is equivalent to rotatating it counterclockwise and reflecting it over the previous symmetric axes.

I don't know how to prove it for the general case. In this case 2n. How do you prove this?

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$D_{2n}$ acts on the Euclidean plane by rotating around the origin $O$/reflecting at axes through $O$. As $\rho$ is a reflection, let $A$ be a point $\ne O$ on its axis. With $B=\sigma A$, $C=\rho B$, and $D=\sigma C$, we have $\angle AOB = 360^\circ/n$, $\angle AOC=-\angle AOB=-360^\circ/n$, $\angle COD=\angle 360^\circ/n$, and therefore $\angle AOD=0^\circ$. As also $|OA|=|OB|=|OC|=|OD|$, we conclude $A=D$. Thus $A$ is a fixed point of $\sigma\rho\sigma$. This is only possible if $\sigma\rho\sigma$ is either the identity or the reflection at $OA$. The first case is impossible as it would imply $\rho=\sigma^2$, i.e., a reflection is a rotation (alternatively, you might track the effect of $\sigma\rho\sigma$ on any point outside $OA$). We conclude that $\sigma\rho\sigma=\rho$, or equivalently $\rho\sigma=\sigma^{-1}\rho$.

The order of $\rho$ is of course $2$. But also $(\sigma\rho)^2=\sigma\rho\sigma\cdot \rho=\rho^2=1$, i.e., $\sigma\rho$ has order two. As $\langle \rho,\sigma\rho\rangle$ contains $\sigma$, it contains at least the $n$ rotations obtainable from $\sigma$ and the non-rotation $\rho$, i.e., it is a subgroup of $D_{2n}$ of order $>n$. This implies that it is all of $D_{2n}$.

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At least in this case, I find it easier to think of the dihedral group in terms of numbers rather than shapes.

$D_{2n}$ acts on $Z/nZ$. The rotation $\sigma_i$ acts as

$\sigma_i(g) = (i + g) \mod n$.

The reflection $\rho_j$ acts as

$\rho_j(g) = (-g + j) \mod n$

(Try some examples to convince yourself of this). Now it's just plug and chug.

$\rho_j\sigma_i(g) = \rho_j(g + i) = -(g + i) + j = -g - i + j$

$\sigma_i^{-1}\rho_j(g) = \sigma_{-i}(-g + j) = -g - i + j$

Now to show that $D_{2n}$ is generated by the $\rho_j$, note that $\rho_i\rho_j(x) = x - j + i \mod n$, and you can choose $i$ and $j$ to make this match any rotation.

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