Exercise 1.2.5 (Center of Dihedral groups (odd case).)

Proof. Reasoning by contradiction, assume there is some element $u\ne e$ in $D_{2n}$ which commutes with all elements of $D_{2n}=\{r^j\mid 0\le j<n\}\cup \{s{r}^j\mid 0\le j<n\}$.

If $u=s{r}^j$ for some $j$, $0\le j<n$, then $u$ commute with $s{r}^{j+1}$, thus

But $s{r}^{j}s{r}^{j+1}=\mathit{ss}{r}^{-j}{r}^{j+1}=r$, and $s{r}^{j+1}s{r}^j=\mathit{ss}{r}^{-j-1}{r}^j=r^{-1}$. This shows that $r=r^{-1}$. But exercise 1.1.33 shows that there is no $i\in \{1,\dots ,n-1\}$ such that $r^i=r^{-i}$ if $n$ is odd, and here $1=i<n$ since $n\ge 3$. Therefore no element $s{r}^j$ of $D_{2n}$ commutes with all element of $D_{2n}$.

If $u=r^j$ for some $j$, $0\le j<n$, then $r^{j}s=s{r}^j$. Therefore $s{r}^{-j}=s{r}^j$, so $r^{-j}=r^j$. By Exercise 1.1.33, this is impossible when $n$ is odd and $1\le j<n$. Therefore $j=0$, and $u=e$.

The identity is the only element of $D_{2n}$ which commutes with all elements of $D_{2n}$ when $n\ge 3$ is odd. □