Exercise 1.2.1 (Order of the elements of $D_6,\ D_8,\ D_{10}$)

Proof. By definition

with the relations $r^n=s^2=e,\mathit{sr}{s}^{-1}=r^{-1}$. This implies that

In particular,

so the $n$ elements $s,\mathit{rs},r{s}^2,\dots ,r^{n-1}s$ are of order $2$:

For the elements of the cyclic subgroup $⟨r⟩=\{e,r,r^2,\dots ,r^{n-1}\}$, the order of $r^h$ is

(We write $g.c.d.(n,h)=n\wedge h$.)

Indeed, for all $i\in \mathbb{Z}$,

because $\left(\frac{n}{n\wedge h}\right)\wedge \left(\frac{h}{n\wedge h}\right)=1$. So (2) is proven.

In particular, this gives the order of the elements of $D_6,D_8,D_{10}$

(a)

Order of the elements of $D_6$:

$x$	$e$	$r$	$r^2$	$s$	$\mathit{rs}$	$r^{2}s$
$|x|$	1	3	3	2	2	2

(b)

Order of the elements of $D_8$:

$x$	$e$	$r$	$r^2$	$r^3$	$s$	$\mathit{rs}$	$r^{2}s$	$r^{3}s$
$|x|$	1	4	2	4	2	2	2	2

(c)

Order of the elements of $D_{10}$:

$x$	$e$	$r$	$r^2$	$r^3$	$r^4$	$s$	$\mathit{rs}$	$r^{2}s$	$r^{3}s$	$r^{4}s$
$|x|$	1	5	5	5	5	2	2	2	2	2