Exercise 4.3.32 (Conjugacy classes in $D_{2n}$, odd case)

Question

For $n = 2k + 1$ an odd odd integer show that the conjugacy classes in $D_{2n}$ are $\{1\}, \{r^{\pm 1}\}, \{r^{\pm 2}\},\ldots, \{r^{\pm k}\}, \{sr^b \mid b = 1,\ldots,n\}$. Give the class equation for $D_{2n}$.

Richard Ganaye · Accepted Answer

\begin{proof} (By exercise 1.2.5, since $n = 2k+1$ is odd, $Z(D_{2n}) = \{1\}$.)

The conjugacy class of $1$ is $\{1\}$.

For $0 \leq l \leq k$, and for all integers $j$,
\begin{align*}
&r^j r^ir^{-j} = r^{i} ,\
&(sr^j) r^i (sr^j)^{-1} = sr^is = r^{-i},
\end{align*}
 so the conjugacy class or $r^i$ is $\{r^i, r^{-i}\}$, where $r^i 
e r^{-i}$:
$${\mathcal O}_{r^i} = \{r^i, r^{-i}\}\qquad( 1\leq i \leq  k).$$
Moreover, for all integers $a$, since $r^{2k+1} = 1$,
\begin{align*}
&(sr^a) s (sr^a)^{-1} = sr^{2a},\
&r^{k-a} s (r^{k-a})^{-1} = sr^{-2(k-a)} = s r^{2k + 1 - 2(k-a)}= s r^{2a+1}.
\end{align*}
Thus ${\mathcal O}_s$ contains all elements of the form $sr^b$ ($b$ odd or even), so
$${\mathcal O}_s \supseteq \{sr^b \mid b = 1,\ldots n\}.$$
Since
$$D_{2n} = \{1\} \cup \bigcup_{i = 1}^k \{r^i, r^{-i}\} \cup \{sr^b \mid b = 1,\ldots n\},$$
the preceding inclusion is an equality:
$${\mathcal O}_s = \{sr^b \mid b = 1,\ldots n\}.$$
For $n = 2k + 1$ an odd odd integer, the conjugacy classes in $D_{2n}$ are 
$$\{1\}, \{r,r^{-1}\}, \{r^2,r^{-2}\},\ldots,\{r^k,r^{-k}\}, \{sr^b \mid b = 1,\ldots,n\}.$$
Since $k = (n-1)/2$, the class equation is
$$ 2n = 1 + \frac{n-1}{2} \cdot 2 + n.$$
\end{proof}

Exercise 4.3.32 (Conjugacy classes in $D_{2n}$, odd case)

Answers

Comments

Exercise 4.3.32 (Conjugacy classes in $D_{2n}$, odd case)

Answers

Comments

Add answer