# Center of Dihedral Group

## Theorem

Let $n \in \N$ be a natural number such that $n \ge 3$.

Let $D_n$ be the dihedral group of order $2 n$, given by:

- $D_n = \gen {\alpha, \beta: \alpha^n = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$

Let $\map Z {D_n}$ denote the center of $D_n$.

Then:

- $\map Z {D_n} = \begin{cases} e & : n \text { odd} \\ \set {e, \alpha^{n / 2} } & : n \text { even} \end{cases}$

## Proof

By definition, the center of $D_n$ is:

- $\map Z {D_n} = \set {g \in D_n: g x = x g, \forall x \in D_n}$

For $n \le 2$ we have that $\order {D_n} \le 4$ and so by Group of Order less than 6 is Abelian $D_n$ is abelian for $n < 3$.

Hence by Group equals Center iff Abelian $\map Z {D_n} = D_n$ for $n < 3$.

So, let $n \ge 3$.

By Group Presentation of Dihedral Group:

- $D_n = \gen {\alpha, \beta: \alpha^n = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$

From Product of Generating Elements of Dihedral Group:

- $\beta \alpha^k = \alpha^{n - k} \beta$

for all $k \in \Z_{\ge 0}$.

We have that $D_n$ is generated by $\alpha$ and $\beta$.

Thus:

- $x \in \map Z {D_n} \iff x \alpha = \alpha x \land x \beta = \beta x$

Let $x \in \map Z {D_n}$.

We have that $x$ can be expressed in the form:

- $x = \alpha^i \beta^j$

As $x \in \map Z {D_n}$, we have:

\(\displaystyle x \alpha\) | \(=\) | \(\displaystyle \alpha x\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \alpha^i \beta^j \alpha\) | \(=\) | \(\displaystyle \alpha^{i + 1} \beta^j\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \beta^j \alpha\) | \(=\) | \(\displaystyle \alpha \beta^j\) |

But for $j = 1$ this means:

\(\displaystyle \alpha \beta\) | \(=\) | \(\displaystyle \beta \alpha\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \alpha^{-1} \beta\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \alpha\) | \(=\) | \(\displaystyle \alpha^{-1}\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \alpha^2\) | \(=\) | \(\displaystyle e\) |

But the order of $\alpha$ is $n$ and $n > 2$, and hence:

- $\alpha^2 \ne e$

So if $x \in \map Z {D_n}$ it follows that $x$ has to be in the form:

- $x = \alpha^i$

for some $i \in \Z_{\ge 0}$.

Again, as $x \in \map Z {D_n}$, we have:

\(\displaystyle x \beta\) | \(=\) | \(\displaystyle \beta x\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \alpha^i \beta\) | \(=\) | \(\displaystyle \beta \alpha^i\) | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \alpha^{n - i} \beta\) | Product of Generating Elements of Dihedral Group | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \alpha^i\) | \(=\) | \(\displaystyle \alpha^{n - i}\) | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \alpha^n \alpha^{-i}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \alpha^{-i}\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \alpha^{2 i}\) | \(=\) | \(\displaystyle e\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle n\) | \(\divides\) | \(\displaystyle 2 i\) | Definition of Order of Group Element |

So either $i = 0$ or $n = 2 i$, as $0 \le i \le n$.

If $i = 0$ then $x = \alpha^0 = e$.

If $2 i = n$ then $n$ is even and so:

- $x = \alpha^{n / 2}$

Hence the result.

$\blacksquare$

## Sources

- 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 50 \beta$