Showing that $f(r^k)^{-1}=f(r^k)$, where $f$ is a representation of the dihedral group $D_n$

Question

Let $D_n$ be the dihedral group, which contains elements of the form $\{r^k\}_{k\in K}$ and $\{sr^k\}_{k\in K}$ where $K=\{1,\dots,n-1\}$ and $r$ and $s$ satisfy the equations $$ r^n=s^2=1,\quad sr^ks=r^{-k}. $$ I know that if $f$ is a one-dimensional matrix representation of $D_n$, then $f(r^k)\in \{(-1),(+1)\}$. This would suggest that $$\tag{1} f(r^k)^{-1}=f(r^k), $$ but I cannot see why that is true. Can someone help me show equation $(1)$?

Answer 1

Since your representation is one-dimensional, $f(ab) = f(a) f(b) = f(b) f(a) = f(ba)$, thus $$f(r^k) = f(r^k s^2) = f(sr^k s) = f(r^{-k}).$$

If you add $f(r^{k})^{-1} = f(r^{-k})$, you get $f(r^k)^{-1} = f(r^k)$.