Algebra Solutions: Compute the order of each element in Dih(6), Dih(8), and Dih(10)

Recall that $D_{2n} = \langle r, s \ |\ r^n = s^2 = 1, rs = sr^{-1} \rangle$ . Compute the order of each element in the following groups.

$D_6$
$D_8$
$D_{10}$

Recall that every element of $D_{2n}$ can be represented uniquely as $s^i r^j$ for some $i = 0, 1$ and $0 \leq j < n$ . Moreover, $r^i s = s r^{-i}$ for all $0 \leq i \leq n$ . From this we deduce that $(sr^i)(sr^i) = ssr^{-i}r^i = 1$ , so that $sr^i$ has order 2 for $0 \leq i \leq n$ .

For $\alpha \in D_6$

$\alpha$ Reasoning $|\alpha|$

$1$ 1

$r$ $r \cdot r = r^2$
$r^2 \cdot r = r^3 = 1$ 3

$r^2$ $r^2 \cdot r^2 = r^4 = r$
$r \cdot r^2 = r^3 = 1$ 3

$s$ $s \cdot s = 1$ 2

$sr$ 2

$sr^2$ 2
For $\alpha \in D_8$

$\alpha$ Reasoning $|\alpha|$

$1$ 1

$r$ $r \cdot r = r^2$
$r^2 \cdot r = r^3$
$r^3 \cdot r = r^4 = 1$ 4

$r^2$ $r^2 \cdot r^2 = r^4 = 1$ 2

$r^3$ $r^3 \cdot r^3 = r^6 = r^2$
$r^2 \cdot r^3 = r^5 = r$
$r \cdot r^3 = r^4 = 1$ 4

$s$ $s \cdot s = 1$ 2

$sr$ 2

$sr^2$ 2

$sr^3$ 2
For $\alpha \in D_{10}$

$\alpha$ Reasoning $|\alpha|$

$1$ 1

$r$ $r \cdot r = r^2$
$r^2 \cdot r = r^3$
$r^3 \cdot r = r^4$
$r^4 \cdot r = r^5 = 1$ 5

$r^2$ $r^2 \cdot r^2 = r^4$
$r^4 \cdot r^2 = r^6 = r$
$r \cdot r^2 = r^3$
$r^3 \cdot r^2 = 1$ 5

$r^3$ $r^3 \cdot r^3 = r^6 = r$
$r \cdot r^3 = r^4$
$r^4 \cdot r^3 = r^7 = r^2$
$r^2 \cdot r^3 = r^5 = 1$ 5

$r^4$ $r^4 \cdot r^4 = r^8 = r^3$
$r^3 \cdot r^4 = r^7 = r^2$
$r^2 \cdot r^4 = r^6 = r$
$r \cdot r^4 = r^5 = 1$ 5

$s$ $s \cdot s = 1$ 2

$sr$ 2

$sr^2$ 2

$sr^3$ 2

$sr^4$ 2

Algebra Solutions

Compute the order of each element in Dih(6), Dih(8), and Dih(10)

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$\alpha$	Reasoning	$\|\alpha\|$
$1$		1
$r$	$r \cdot r = r^2$ $r^2 \cdot r = r^3 = 1$	3
$r^2$	$r^2 \cdot r^2 = r^4 = r$ $r \cdot r^2 = r^3 = 1$	3
$s$	$s \cdot s = 1$	2
$sr$		2
$sr^2$		2