dihedral group operation

Question

Let $D_{2n}$ be the dihedral group of order $2n$ with $r^n=1, s^2=1$. Then since $rs=sr^{-1}$, $r^as^b=s^br^{(-1)^ba}.$ Is this correct?

Answer 1

For $b$ even, your formula holds for every $a$. For $b$ odd, your formula reads:

$$r^as=sr^{-a} \tag 1$$

Induction on $a$: for $a=1$, $(1)$ holds; let $(1)$ hold for $a$; then: \begin{alignat}{1} r^{a+1}s &= r^ars \\ &= r^a(ss^{-1})rs \\ &= (r^as)s^{-1}rs \\ &= (sr^{-a})s^{-1}rs \\ &= (sr^{-a})s^{-1}rs^{-1} \\ &= sr^{-a}s^{-1}(rs^{-1}) \\ &= sr^{-a}s^{-1}(s^{-1}r^{-1}) \\ &= sr^{-a}(s^{-1}s^{-1})r^{-1} \\ &= sr^{-a}s^{-2}r^{-1} \\ &= sr^{-a}r^{-1} \\ &=sr^{-(a+1)} \end{alignat}

so that $(1)$ holds for every $a$.

In conclusion, yes, to me your formula looks OK.