# Z(D_2n), Center of Dihedral Group D_2n

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. (since $n\geq 3$). $a^{n-1}b$ also does not commute with $a^{n-2}b$ for the same reason.

Therefore,

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

## Author: mathtuition88

https://mathtuition88.com/

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