Exercise 1.2.7 (Another presentation of dihedral groups)

Question

Show that $\langle a,b \mid a^2 = b^2 = (ab)^n = 1\}$ gives a presentation for $D_{2n}$ in terms of the two generators $a = s$ and $b =sr$ of order $2$ computed in Exercise 3 above. [Show that the relations for $r$ and $s$ follow from the relations for $a$ and $b$ and, conversely, the relations for $a$ and $b$ follow from those for $r$ and $s$.]

Richard Ganaye · Accepted Answer

\begin{proof} Consider $G = \langle a,b \mid a^2 = b^2 = (ab)^n = 1 angle$. Define $s = a, r = a b \in G$. Then $a = s , b = s r$. Therefore $G =\langle a, bangle= \langle r,s angle$.

Then $s^2 = a^2 = 1$, and $r^n = (ab)^n = 1$. Moreover, 
\begin{align*}
rs &= aba\
sr^{-1} &= a(ab)^{-1}= ab^{-1} a^{-1} = aba,
\end{align*}
so $rs = sr^{-1}$.

\medskip
Conversely, consider $D_{2n}= \langle r,s \mid r^n= s^2 = 1, rs = sr^{-1} angle$. Define $a =s, b =sr$. Then $\langle a, bangle= \langle r,s angle$, and
\begin{align*}
a^2 &=s^2 = 1,\
b^2 &= (sr)(sr) =s(rs)r = s(sr^{-1}) r= s^2 = 1\
(ab)^n = (s(sr))^n r^n =1.
\end{align*}
This shows that
 $$\langle a,b \mid a^2 = b^2 = (ab)^n = 1 angle \simeq \langle r,s \mid r^n= s^2 = 1, rs = sr^{-1} angle$$
 are two presentations of the diehedral group.
 
 (For a complete proof of this isomorphism, we need a more formal definition of  presentations of groups.)
\end{proof}

Exercise 1.2.7 (Another presentation of dihedral groups)

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Exercise 1.2.7 (Another presentation of dihedral groups)

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