How do we prove that $\tau \rho = \rho^{-1} \tau$ in general for $D_{2n}$

Question

I found a problem similar to this one here: Prove that $\tau \rho = \rho^ {−1}\tau.$ except this is a particular example. I would think the best route to proving this is by a homomorphism from $\phi:D_{2n} \mapsto S_{n}$. But I have no idea how to translate $\tau,\rho$ as cycles. Any hints or tips would be appreciated. If this question has an answer somewhere else let me know.

Answer 1

Label the vertices of a regular $n$-gon by the first $n$ nonzero natural numbers. Then the symmetries of the $n$-gon correspond to particular elements of $S_n$. If any two of these permutations are the same, then clearly they are defined by the same symmetry; so the correspondence is injective. Fix a reflection $\tau$ and a rotation $\rho$. Use the fact that for any $\alpha,\beta\in S_n$, with $\alpha=(a_{i_1}\dots a_{i_{l_1}})\dots (a_{i_k}\dots a_{i_{l_k}})$ the cycle decomposition of $\alpha$, we have

$$\beta\alpha \beta^{-1}=(\beta(a_{i_1})\dots\beta (a_{i_{l_1}}))\dots (\beta(a_{i_k})\dots\beta( a_{i_{l_k}})).$$

Then note that $\rho $ corresponds to an $n$-cycle whose inverse is $(r_n\dots r_1)$, where $\rho=(r_1\dots r_n)$.