Exercise 2.2.7 (Center of $D_{2n}$)

Proof. By definition $D_{2n}=⟨r,s\mid r^n=s^2=1,\mathit{rs}=s{r}^{-1}⟩$.

We recall that $D_{2n}=\{1,r,r^2,\dots ,r^{n-1},s,\mathit{rs},r^{2}s,\dots ,r^{n-1}\}$, and the law is given by

In particular, $s{r}^h=r^{-h}s$ for all integers $h$.

Let $a$ be any element in $Z(D_{2n})$, where $n\ge 3$. Then $a=r^h{s}^k$, where $0\le h<n,0\le s<2$.

Then $\mathit{as}=\mathit{sa}$, thus $(r^h{s}^k)r=r(r^h{s}^k)$. Simplifying by $r^h$, this gives

If $k=1$, we obtain $\mathit{sr}=\mathit{rs}=s{r}^{-1}$, thus $r=r^{-1}$, i.e. $r^2=1$. This is impossible, because the order of $r$ is $n\ge 3$. Hence $k=0$ and $a=r^h$.

Then $\mathit{as}=\mathit{sa}$ gives $r^{h}s=s{r}^h=r^{-h}s$, so $r^h=r^{-h}$ and

Since the order of $r$ is $n$, we have $n\mid 2h$.

(a)

If $n$ is odd, then $g.c.d(n,2)=1$, hence $n\mid h$, where $0\le h<n$. Therefore $h=0$, and $a=1$. This shows that $Z(D_{2n})\subseteq \{1\}$. Of course, since $Z(D_{2n})$ is a subgroup, $\{1\}\subseteq Z(D_{2n})$, so $$Z(D_{2n})=\{1\}.$$

(b)

If $n$ is even, then $n\mid 2h$ implies $n∕2\mid h$, where $0\le h<n$, so $h=0$ or $h=n∕2$. This shows that $Z(D_{2n})\subseteq \{1,r^{n∕2}\}$. Conversely $r^{n∕2}$ commute with $r$, and since $r^n=1$, $$s{r}^{n∕2}=r^{-n∕2}s=r^{n∕2}s,$$

so $r^{n∕2}$ commute with $s$. Therefore $r^{n∕2}$ commute with every element $r^h{s}^k$, so $r^{n∕2}\in Z(D_{2n})$. This shows that $\{1,r^{n∕2}\}\subseteq Z(D_{2n})$, so

$$Z(D_{2n})=\{1,r^{n∕2}\}.$$

In conclusion, 

• $Z(D_{2n})=\{1\}$ if $n$ is odd.
• $Z(D_{2n})=\{1,r^k\}$ if $n=2k$.