Conjugacy classes of dihedral groups

Posted: March 8, 2022 in Basic Algebra, Groups
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For a group G and x \in G, we denote by \text{Cl}(x) the conjugacy class of x, i.e. \text{Cl}(x):=\{gxg^{-1}: \ g \in G\}.

Recall that D_{2n}, the dihedral group of order 2n, is defined as follows

D_{2n}=\langle a,b: \ a^2=b^n=1, \ ab=b^{-1}a\rangle.

So D_{2n}=\{1,a,b,b^2, \cdots , b^{n-1},ab,ab^2, \cdots ,ab^{n-1}\}, where multiplication is done using the relations a^2=b^n=1, \ ab=b^{-1}a. An easy induction shows that in D_{2n} we have

ab^j=b^{-j}a, \ \ \ \ \ \forall j \in \mathbb{Z}. \ \ \ \ \ \ \ \ \ \ (*)

We now use (*) to find all the conjugacy classes of D_{2n}.

Theorem. i) If n is even, then the conjugacy classes of D_{2n} are

\displaystyle \{1\}, \{b,b^{-1}\}, \{b^2,b^{-2}\}, \cdots , \{b^{\frac{n}{2}-1},b^{1-\frac{n}{2}}\},\{b^{\frac{n}{2}}\}, \{a,ab^2, ab^4, \cdots , ab^{n-2}\}, \{ab,ab^3, ab^5, \cdots ,ab^{n-1}\}.

ii) If n is odd, then the conjugacy classes of D_{2n} are

\displaystyle \{1\},\{b,b^{-1}\}, \{b^2,b^{-2}\}, \cdots , \{b^{\frac{n-1}{2}},b^{\frac{1-n}{2}}\}, \{a,ab, ab^2, \cdots, ab^{n-1}\}.

Proof. An element of D_{2n} is in the form a^ib^j, \ 0 \le i \le 1, \ 0 \le j \le n-1. So if x=a^kb^{\ell}, \ g=a^ib^j are two elements of D_{2n}, then

gxg^{-1}=a^ib^ja^kb^{\ell}b^{-j}a^i=a^ib^ja^kb^{\ell-j}a^i. \ \ \ \ \ \ \ \ \ \ (\dagger)

We now consider two cases.

Case 1: k=0. In this case, x=b^{\ell} and, by (\dagger), \ gxg^{-1}=a^ib^{\ell}a^i. So if i=0, then gxg^{-1}=b^{\ell} and if i=1, then, by (*), \ gxg^{-1}=ab^{\ell}a=b^{-\ell}. Thus we have shown that \text{Cl}(b^{\ell})=\{b^{\ell},b^{-\ell}\} for all 0 \le \ell \le n-1.

Case 2: k=1. In this case, we have x=ab^{\ell} and, by (\dagger), \ gxg^{-1}=a^ib^jab^{\ell-j}a^i. Thus if i=0, then gxg^{-1}=b^jab^{\ell-j}=ab^{\ell-2j}, by (*), and if i=1, then gxg^{-1}=ab^jab^{\ell-j}a=b^{\ell-2j}a=ab^{2j-\ell}, by (*). So we have shown that \text{Cl}(ab^{\ell})=\{ab^{\ell-2j}, ab^{2j-\ell}: \ \ 0 \le j \le n-1\}, for all 0 \le \ell \le n-1.

We’re now ready to complete the proof.

i) If n is even, then, by Case 1,

\displaystyle \text{Cl}(1)=\{1\}, \ \text{Cl}(b)=\{b,b^{-1}\}, \ \cdots , \text{Cl}(b^{\frac{n}{2}-1})=\{b^{\frac{n}{2}-1},b^{1-\frac{n}{2}}\}, \ \text{Cl}(b^{\frac{n}{2}})=\{b^{\frac{n}{2}}\}.

The above conjugacy classes cover all the elements of \langle b\rangle. Also, by Case 2,

\displaystyle \text{Cl}(a)=\{a,ab^2, ab^4, \cdots , ab^{n-2}\}, \ \ \text{Cl}(ab)=\{ab,ab^3, ab^5, \cdots ,ab^{n-1}\},

which cover the rest of the elements of D_{2n}.

ii) Suppose now that n is odd. We get from Case 1 that

\displaystyle \text{Cl}(1)=\{1\}, \ \text{Cl}(b)=\{b,b^{-1}\}, \ \cdots , \text{Cl}(b^{\frac{n-1}{2}})=\{b^{\frac{n-1}{2}},b^{\frac{1-n}{2}}\}.

The above conjugacy classes cover all the elements of \langle b\rangle. Now, since n is odd, for any integers m, there exists an integer j such that 2j \equiv m \mod n and hence, by Case 2,

\displaystyle \text{Cl}(a)=\{a,ab, ab^2, \cdots, ab^{n-1}\},

and that covers the rest of the elements of D_{2n}. \ \Box

Remark 1. So, by the Theorem, the number of conjugacy classes of D_{2n} is \displaystyle \begin{cases}\frac{n+6}{2} & \text{if} \ 2 \mid n \\ \frac{n+3}{2} & \text{if} \ 2\nmid n\end{cases}, which can also be written as \displaystyle \frac{2n+9+(-1)^n3}{4}.

Remark 2. The center of a group G is the set of all the elements x \in G such that |\text{Cl}(x)|=1. The Theorem now gives the center of D_{2n}, a result we already proved here.

Example. Let p_n be the probability that two randomly selected elements of D_{2n} commute. Find p_n and show that \displaystyle \lim_{n\to\infty}p_n=\frac{1}{4}.

Solution. By the Remark in this post and Remark 1,

\displaystyle p_n=\text{Pr}(D_{2n})=\frac{\text{number of conjugacy classes of} \ D_{2n}}{|D_{2n}|}=\frac{\frac{2n+9+(-1)^n3}{4}}{2n}=\frac{2n+9+(-1)^n3}{8n},

which gives \displaystyle \lim_{n\to\infty}p_n=\frac{1}{4}. \ \Box

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