Question

If $\mathit{n}\mathbf{=}\mathbf{2}\mathit{k}$, show that ${\mathit{r}}^{\mathbf{k}}$is in the center of ${\mathit{D}}_{\mathbf{n}}$.

Short answer

It is proved that, $r^k\in Z\left(D_n\right)$.

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, localid="1653635090512" $Z\left(G\right)=\left\{z\in G|zg=gz,\forall g\in G\right\}$.

Given that n=2k.

Since $r^k{r}^k=r^{2k}=e$, we have:

$\begin{array}{rcl}{\left(r^k\right)}^{-1}& =& r^{-k}\\ & =& r^k\end{array}$

$\therefore {\left(r^k\right)}^{-1}=r^k$

To prove rk  is in the center of Dn  

We have to prove that $r^k\in Z\left(D_n\right)$.

For any $r^i{d}^j\in D_n$, we have:

$\begin{array}{rcl}{r}^i{d}^j{r}^k& =& r^{i+{\left(-1\right)}^{J}k}{d}^j\\ & =& r^i{r}^{{\left(-1\right)}^{J}k}{d}^j\\ & =& r^k{r}^i{d}^j\end{array}$

Hence, $r^k\in Z\left(D_n\right)$.