Exercise 1.2.4 (Center of dihedral groups (even case).)

Proof. We know that $r$ is an element of order $n$ in $D_{2n}$, thus $r^{n∕2}$ is an element of order $2$: for all $m\in$,

So $z=r^k$ is an element of order $2$.

Now we prove that $z$ commutes with all elements of $D_{2n}$.

Recall that $D_{2n}=\{r^j\mid 0\le j<n\}\cup \{s{r}^j\mid 0\le j<n\}$.

We know from Exercise 1.1.33 that $r^k=r^{-k}$, thus

This proves that $z=r^k$ commutes with all elements of $D_{2n}$.

Let $u$ be an element of $D_{2n}$ which commutes with all elements of $D_{2n}$.

If $u=s{r}^j$ for some $j=0,1,\dots ,n-1$, then $u$ commutes with $s{r}^{j+1}$. This gives

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 $r=r^{-1}$

Using anew Exercise 1.1.33, we know that $r^i=r^{-i}(1\le i<n)$ implies $i=n∕2$. So $r=r^{-1}$ implies $1=n∕2$ and $n=2$, in contradiction with the hypothesis $n\ge 4$. This show that $s{r}^j$ doesn’t commutes with all elements of $D_{2n}$.

Therefore $u=r^j$ for some $j=0,1,\dots ,n-1$. In particular $u$ commutes with $s$, so $r^{j}s=s{r}^j$. This gives $s{r}^{-j}=s{r}^j$, so $r^{-j}=r^j$. Exercise 1.1.33 shows that $j=0$ or $j=n∕2$. Therefore $u=r^{n∕2}$ is the only nonidentity element which commutes with all elements of $D_{2n}$ (if $n$ is even , $n\ge 4$). □