Center of Dihedral Group

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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.

$\blacksquare$


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