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Let $n \geq 1 $ and let $\mathcal{D}_{2n}$ be the dihedral group of order $2n$. Find $\mathcal{Z}(\mathcal{D}_{2n})$ that is the centre of $\mathcal{D}_{2n}$.

All the distinct elements of the $n$-dihedral group $${\cal{D}}_{2n}=\left\langle{\rho,\,\tau \ | \ \rho^{n}=\tau^2={\rm{id}}, \ \tau\rho\tau=\rho^{n-1}}\right\rangle$$ are of the form $\rho^{k}\tau^{m}\,,\; k\in\{0,1,\ldots,n-1\},\; m\in\{0,1\}$.

For $n=1$, we have $Z({\cal{D}}_{2})=\{{\rm{id}},\tau\}={\cal{D}}_{2}$.

For $n>1$: To be the element $\rho^{k}\,,\; k\in\{1,\ldots,n-1\}$ in the center $Z({\cal{D}}_{2n})$ must hold: \begin{align*}
\rho^{k}\tau=\tau\rho^{k}\quad&\Rightarrow\quad \tau\rho^{n-k}=\tau\rho^{k}\\
&\Rightarrow\quad \rho^{n-k}=\rho^{k}\\
&\Rightarrow\quad n-k=k\\
&\Rightarrow\quad k=\frac{n}{2}\,.
\end{align*} So, if $n>1$ is odd, does not exists $k\in\{1,\ldots,n-1\}$ such that $\rho^{k}\in Z({\cal{D}}_{2n})$ and if $n$ is even, only the element $\rho^{\frac{n}{2}}$ may be in $Z({\cal{D}}_{2n})$. Because for every $s\in\{1,\ldots,n-1\}$: $\rho^{\frac{n}{2}}\rho^{s}=\rho^{s}\rho^{\frac{n}{2}}$, we conclude that if $n$ is even, then $\rho^{\frac{n}{2}}\in Z({\cal{D}}_{2n})$.
To be the element $\rho^{k}\tau\,,\; k\in\{1,\ldots,n-1\}$ in the center $Z({\cal{D}}_{2n})$, for every $s\in\{1,\ldots,n-1\}$ must hold:
\begin{align*}
\rho^{k}\tau\rho^{s}=\rho^{s}\rho^{k}\tau\quad&\Longrightarrow\quad \rho^{k}\rho^{n-s}\tau=\rho^{k+s}\tau\\
&\stackrel{s=1}{\Longrightarrow}\quad \rho^{k}\rho^{n-1}\tau=\rho^{k+1}\tau\\
&\Longrightarrow\quad \rho^{k+n-1}=\rho^{k+1}\\
&\Longrightarrow\quad n=2\,. \end{align*}
Finaly, if $n=1$, then $Z({\cal{D}}_{2})={\cal{D}}_{2}$, if $n>1$ is odd, then $Z({\cal{D}}_{2n})=\{{\rm{id}}\}$, if $n=2$, then $Z({\cal{D}}_{4})={\cal{D}}_{4}$ and if $n>2$ is even, then $Z({\cal{D}}_{2n})=\{\rho^{\frac{n}{2}},{\rm{id}}\}$.