Exercise 1.2.3 (Generators of order 2)

Question

Use the generators and relations above to show that every element of $D_n$ which is not a power of $r$ has order $2$. Deduce that $D_{2n}$ is generated by the two elements $s$ and $sr$, both of which have order $2$.

Richard Ganaye · Accepted Answer

\begin{proof} 
Every element $x$ of $D_n$ which is not a power of $r$ satisfies $x = s r^j$, where $0\leq j <n$.
Using $r^j s = sr^{-j}$ (see 1.2 eq.(6)),
$$x^2 = (sr^j)(sr^j) = s(r^j s )r^j = s(sr^{-j}) r^j= s^2 =e.$$
Since $x 
e e$, $x$ has order $2$: $|x| = 2$.

\bigskip
Therefore $|s| = |sr| = 2$.

Since $r,s \in D_{2n}$, $  \langle s, sr angle \subseteq D_{2n}$. Moreover $r = s(sr) \in \langle s, sr angle$, and $s \in  \langle s, sr angle$, thus $D_{2n} = \langle r,s angle \subset  \langle s, sr angle$. This proves
$$D_{2n} =  \langle r,s angle=\langle s, sr angle.$$
 $D_{2n}$ is generated by the two elements $s$ and $sr$, both of which have order $2$.
\end{proof}

Exercise 1.2.3 (Generators of order 2)

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Exercise 1.2.3 (Generators of order 2)

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