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 definition of abelian group:
- $\map Z {D_n} = D_n$
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:
\(\ds x \alpha\) | \(=\) | \(\ds \alpha x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \alpha^i \beta^j \alpha\) | \(=\) | \(\ds \alpha^{i + 1} \beta^j\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \beta^j \alpha\) | \(=\) | \(\ds \alpha \beta^j\) |
But for $j = 1$ this means:
\(\ds \alpha \beta\) | \(=\) | \(\ds \beta \alpha\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \alpha^{-1} \beta\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \alpha\) | \(=\) | \(\ds \alpha^{-1}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \alpha^2\) | \(=\) | \(\ds e\) |
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:
\(\ds x \beta\) | \(=\) | \(\ds \beta x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \alpha^i \beta\) | \(=\) | \(\ds \beta \alpha^i\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \alpha^{n - i} \beta\) | Product of Generating Elements of Dihedral Group | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \alpha^i\) | \(=\) | \(\ds \alpha^{n - i}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \alpha^n \alpha^{-i}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \alpha^{-i}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \alpha^{2 i}\) | \(=\) | \(\ds e\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds n\) | \(\divides\) | \(\ds 2 i\) | 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: $\S 50 \beta$