How shall we show that the only possible orders of any element in dihedral group $D_n$ will be either a divisor of 2 or $n$?

Question

I am studying the dihedral group $D_n:=\{r_n, f_n: r_n^n=f_n^2=(r_nf_n)^2=e_n\}$.

I am willing to show that the possible orders of any element in it will be either a divisor of 2 or $n$. But don't know how shall I show it.

Can someone help me out?

Answer 1

As @Stefan Hamcke suggested, this is what I got.

Any element in $D_n$ is of the form $r^k$ or $r^kf$ where $k\in \{0, 1, 2, \cdots, n-1\}$. NOte that $D_n$ contains a cyclic subgroup of order $n$ which is generated by $r$ viz $\langle r\rangle=C_n\leq D_n$ and we know that for every divisor $d$ of $n$ the cyclic group $\langle r\rangle$ will contain a cyclic subgroup of order $d$.

On the other hand, any element of the form $r^kf$ will have order 2 because using the relation $(rf)^2=1$ i.e. $rf=fr^{-1}$ one see that $(r^kf)^2=r^kfr^kf=r^kr^{-k}ff=e$

hence any element of $D_n$ will have possible order as divisor of $2$ or $n$.

Did I make any mistake ?