Exercise 1.2.2 (Elements of $D_{2n}$ which are not powers of $r$)

Question

Use the generators and relations above to show that if $x$ is any element of $D_{2n}$ which is not a power of $r$, then $rx =xr^{-1}$.

Richard Ganaye · Accepted Answer

\begin{proof} 
We know from the text that
$$D_{2n} = \langle r,s \mid r^n = s^2 = 1, rs = sr^{-1}\rangle = \{s^i r^j \mid 0\leq i <2, 0 \leq j < n\}.$$

Let $x$ be any element of $D_{2n}$ which is not a power of $r$. Then $x = sr^j$ for some $j=0,1,\ldots,n-1$. Then
$$rx = r(sr^j)=  (rs) r^j = s r^{-1} r^j  = sr^{j-1}.$$
(If $1\leq j <n$, then $rx = s r^{j-1}, \ 0\leq j <n-1$, and if $j = 0$, then $rx = sr^{-1} = sr^{n-1}$.)
Moreover, 
$$xr^{-1} = sr^jr^{-1} = sr^{j-1}.$$
This shows that 
$$rx = xr^{-1}.$$
\end{proof}

Exercise 1.2.2 (Elements of $D_{2n}$ which are not powers of $r$)

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Exercise 1.2.2 (Elements of $D_{2n}$ which are not powers of $r$)

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