Question

If n is even, show that $\mathit{Z}\left(D_n\right)\mathbf{=}\left\{e,r^k\right\}$.

Short answer

It is proved that, $Z\left(D_n\right)=\left\{e,r^k\right\}$.

Step-by-step solution

Defining center of the group

Definition of Center of the Group

Let G be a group. The set of all elements which commutes with every element of group G , is called the center of the group.

It is denoted by $Z\left(G\right)$ and is defined as, $Z\left(G\right)=\left\{z\in G|zg=gz,\forall g\in G\right\}$.

Let $r^i{d}^j\in D_n$ and given that n is even.

Then, for any d we have:

$$\begin{gathered} \left(r^i{d}^j\right)d=d\left(r^i{d}^j\right) \\ \Rightarrow r^i{d}^{j+1}=r^{-i}{d}^{1+j} \\ \Rightarrow r^i=r^{-i} \end{gathered}$$

This is the case only if $i=0ori=k\mathrm{in}Z\left(D_n\right)$ in .

To prove Z(Dn)={e,rk}

Also, we have:

$$\begin{gathered} \left(r^k{d}^j\right)r=r\left(r^k{d}^j\right) \\ \Rightarrow r^{k+{\left(-1\right)}^J}{d}^j=r^{k+1}{d}^j \\ \Rightarrow r^k{r}^{{\left(-1\right)}^J}{d}^j=r^k{r}^1{d}^j \\ \Rightarrow r^{{\left(-1\right)}^J}=r^1 \end{gathered}$$

This is the case only if $j=0$.

Thus, the only nontrivial element in the center is $r^k$.

Hence, $Z\left(D_n\right)=\left\{e,r^k\right\}$.