Problem

Let \(n \in \mathbb{Z}\) with \(n \ge 3\). Prove the following:

• \(Z(D_{2n}) = \{1\}\) if \(n\) is odd
• \(Z(D_{2n}) = \{1,r^k\}\) if \(n = 2k\)

Solution

By definition of center,

$$ Z(G) = \{g \in G \mid gx = xg \text{ for all } x \in G\}. $$

\(x \in D_{2n}\) is reperesented as either one of the following.

• \(1\)
• \(sr^i\) for some \(0 \le i \lt n\)
• \(r^i\) for some \(0 < i < n\)

First, consider \(1\). By definition of group, \(1x = x1\) for all \(x \in G\). Therefore, \(1 \in Z(D_{2n})\). — (A)

Next, consider \(sr^i\) where \(0 \le i <n\). Note $$ sr^i\cdot r = r\cdot sr^i \iff sr^{i+1} = sr^{i-1} \iff s^{-1}sr^{i+1}r^{-i+1} = s^{-1}sr^{i-1}r^{-i+1} \iff r^2 = 1. $$

Since \(n \ge 3\), \(r^2 \ne 1\). Therefore, \(sr^i \notin Z(D_{2n})\). — (B)

Finally, consider \(r^i\) where \(0 < i < n\). \(r^{i}\) commutes with \(1\) and \(r^m\) for any \(m\). It remains to show if it commutes with \(sr^m\). We have

$$ sr^m \cdot r^i = sr^{m+i} \quad\land\quad r^i \cdot sr^m = sr^{m-i}. $$

Since \(0 < i < n\), $$ sr^{m+i} = sr^{m-i} \iff s^{-1}sr^{m+i}r^{-m+i} = s^{-1}sr^{m-i}r^{-m+i} \iff r^{2i} = 1 \iff 2i = n. $$

Therefore, \(r^i\) commutes with \(sr^m\) if and only if \(n = 2i\). In other words, \(r^k \in Z(D_{2n})\) if and only if \(n = 2k\). — (C)

Hence, from (A), (B), and (C) above,

$$ Z(D_{2n}) = \begin{cases} \{1\}\quad\quad\text{if } n \text{ is odd}\\ \{1,r^k\}\quad\text{if } n = 2k \end{cases} $$

$$\tag*{$\blacksquare$}$$