dihedral – Mathtuition88

Question: What is $Z(D_{2n})$ , the center of the dihedral group $D_{2n}$ ?

Algebraically, the dihedral group may be viewed as a group with two generators $a$ and $b$ , i.e. $\boxed{D_{2n}=\{1,a,a^2,\dots,a^{n-1},b,ab,a^2b,\dots,a^{n-1}b\}}$ with $a^n=b^2=1$ , $bab=a^{-1}$ .

Answer: $Z(D_2)=D_2$

$Z(D_4)=D_4$ .

For $n\geq 3$ , $Z(D_{2n})=\begin{cases}1&,\ n\ \text{is odd}\\ \{1,a^{n/2}\}&,\ n\ \text{is even} \end{cases}$

Proof: For $n=1$ , $D_2=\{1, b\}\cong\mathbb{Z}_2$ which is abelian. Thus, $Z(D_2)=D_2$ .

For $n=2$ , $D_4=\{1,a,b,ab\}\cong V$ , the Klein four-group, which is also abelian. Thus, $Z(D_4)=D_4$ .

Let $A=\{1,a,a^2,\dots,a^{n-1}\}$ , $B=\{b,ab,a^2b,\dots,a^{n-1}b\}$ . Clearly elements in $A$ commute with each other.

Let $a^k$ be an element in $A$ . ( $0\leq k\leq n-1$ ). Let $a^lb$ be an element in $B$ . ( $0\leq l\leq n-1$ )

$\begin{aligned}a^k(a^lb)=(a^lb)a^k&\iff a^kb=ba^k\\ &\iff a^kba^{-k}b^{-1}=1\\ &\iff a^kb(bab)^kb=1\ (\text{here we used}\ bab=a^{-1})\\ &\iff a^kb(ba^kb)b=1\\ &\iff a^{2k}=1\\ &\iff k=0\ \text{or}\ n/2 \end{aligned}$

I.e. the only element in $A$ (other than 1) that is in the center is $a^{n/2}$ , which is only possible if $n$ is even.

Let $a^kb$ , $a^lb$ be two distinct elements in $B$ . ( $0\leq k< l\leq n-1$ )

$\begin{aligned}(a^kb)(a^lb)=(a^lb)(a^kb)&\iff ba^l=a^{l-k}ba^k\\ &\iff ba^{l-k}=a^{l-k}b \end{aligned}$

By earlier analysis, this is true iff $l-k=n/2$ . Each $a^kb\ (0\leq k\leq n-2)$ is not in the center since we may consider $l=k+1$ , i.e. $a^{k+1}b$ . Then $l-k=1<n/2$ . (since $n\geq 3$ ). $a^{n-1}b$ also does not commute with $a^{n-2}b$ for the same reason.