Question

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

Short answer

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

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 role="math" localid="1653636409337" $r^i{d}^j\in D_n$and given that n is odd.

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 $i=0\mathrm{in}Z\left(D_n\right)$ in .

To prove Z(Dn)={e}

Since $dr=r^{-1}d\ne rd$, we can say that $Z\left(D_n\right)$ contains only the identity element e .

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