Exercise 4.3.31 (Conjugacy classes in $D_{2n}$, even case)

Question

Using the usual generators and relations for the dihedral group $D_{2n}$ (cf. Section 1.2) show that for $n = 2k$ an even integer the conjugacy classes in $D_{2n}$ are the following: $\{1\}, \{r^k\}, \{r^{\pm1}\},  \{r^{\pm2}\},\ldots, \{r^{\pm(k-1)}\},\{sr^{2b} \mid b = 1,\ldots,k\}$ and ${\{sr^{2b-1} \mid b = 1,\ldots, k\}}$. Give the class equation for $D_{2n}$.

Richard Ganaye · Accepted Answer

\begin{proof} Here $n = 2k$ is even.

From Section 1.2, we know that
\begin{align*}
D_{2n} &= \langle r,s \mid r^n = s^2 = 1,\  rs = sr^{-1} angle\
&= \{1,r,r^2,\ldots,r^{n-1}, s, sr,sr^2,\ldots,sr^{n-1}\}.
\end{align*}

By Exercise 1.2.4, the center of $D_{2n}$ is
$$Z(D_{2n}) = \{1, r^k\}\qquad (n = 2k).$$
Therefore the conjugacy class of $1$ is $\{1\}$, and the conjugacy class of $r^k = r^{n/2}$ is $\{r^k\}$.
$${\mathcal O}_1 = \{1\},\qquad {\mathcal O}_{r^k} = \{r^k\}.$$
For $1\leq i \leq k - 1$
\begin{align*}
&r^j r^ir^{-j} = r^{i} ,\
&(sr^j) r^i (sr^j)^{-1} = sr^is = r^{-i},
\end{align*}
 for all integers $j$, so the conjugacy class or $r^i$ is $\{r^i, r^{-i}\}$, where $r^i 
e r^{-i}$:
$${\mathcal O}_{r^i} = \{r^i, r^{-i}\}\qquad( 1\leq i \leq  k-1).$$
For all $b \in [\![1, k]\!]$,
$$ (sr^b) s (sr^b)^{-1} = sr^b s r^{-b} s = r^{-2b} s = sr^{2b},$$
so
\begin{align}
{\mathcal O}_{s} \supseteq \{sr^{2b} \mid b = 1,\ldots, k\}.
\end{align}
Similarly,
$$(sr^b) sr (sr^b)^{-1} = sr^b s r r^{-b} s = r^{-2b+1} s = s r^{2b-1}.$$
Therefore
\begin{align}
{\mathcal O}_{sr} \supseteq \{sr^{2b-1} \mid b = 1,\ldots, k\}.
\end{align}
Since 
$$D_{2n} = \{1\} \cup \{r^k\} \cup \bigcup_{i=1}^{k-1} \{r^i, r^{-i}\} \cup \{sr^{2b} \mid b = 1,\ldots, k\} \cup \{sr^{2b-1} \mid b = 1,\ldots, k\},$$
 the two inclusions in (1) and (2) are equalities: if some element $ho \in {\mathcal O}_{s}$ is not of the form $sr^{2b}\ (1\leq b \leq k)$, then it is in another conjugacy class: this is impossible because the conjugacy classes form a partition of $G$. Hence
 \begin{align*}
 {\mathcal O}_{s} &= \{sr^{2b} \mid b = 1,\ldots, k\},\
 {\mathcal O}_{sr}&=\{sr^{2b-1} \mid b = 1,\ldots, k\}.
 \end{align*}
In conclusion, the conjugacy classes in $D_{2n}$, where $n = 2k$, are the following:
$$\{1\},\  \{r^k\},\ \{r,r^{-1}\},\  \{r^2,r^{-2}\}, \ldots, \{r^{k-1},\ r^{-(k-1)}\},\ \{sr^{2b} \mid b = 1,\ldots, k\},\ \{sr^{2b-1} \mid b = 1,\ldots, k\}.$$
Since $|Z(D_{2n})| = 2$, and $k = n/2$, the class equation is
$$2n = 2 + \left(\frac{n}{2} - 1ight) 2 + \frac{n}{2} + \frac{n}{2},$$
which seems true.

\end{proof}

Exercise 4.3.31 (Conjugacy classes in $D_{2n}$, even case)

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Exercise 4.3.31 (Conjugacy classes in $D_{2n}$, even case)

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