Z(D_2n), Center of Dihedral Group D_2n

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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.

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}

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