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I was wondering if there is a special technique to find the conjugacy classes of $\mathcal D_{10}=\left<a,b\mid a^5=b^2=1,bab^{-1}=a^{-1}\right>$, and of $\mathcal D_{2n}=\left<a,b\mid a^n=b^2=1,bab^{-1}=a^{-1}\right> $ ? I can do it by hand (but it's a little bit long), and there is probably a simple technic.

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  • $\begingroup$ Do you want to know how to calculate the conjugacy class or simply how many of them are there? $\endgroup$ Commented Jun 14, 2016 at 10:13
  • $\begingroup$ Yes I know, but my method is quite complicate (since I do all computation). $\endgroup$
    – MSE
    Commented Jun 14, 2016 at 10:26
  • $\begingroup$ MSE A consequence of those defining relations is $ba^{-j}=a^{-j}b$, for all $j$. With that extra relation the calculation is quite straightforward. $\endgroup$ Commented Jun 14, 2016 at 11:49

1 Answer 1

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In general, we have the following techniques for finding the conjugacy classes of dihedral group $D_{2n}=\langle a,b\vert a^n=b^2=1,b^{-1}ab=a^{-1}\rangle$:

If $n$ is odd:
The number of the conjugacy classes of dihedral group $D_{2n}$ is $\frac{n+3}{2}$ and these classes are as follows
$\{1\},\{a,a^{-1}\},\{a^2,a^{-2}\},\ldots,\{a^{\frac{n-1}{2}},a^{-\frac{n-1}{2}}\},\{b,ab,\ldots,a^{n-1}b\}$.

If $n=2m$ is even:
The number of the conjugacy classes of dihedral group $D_{2n}$ is $m+3$ and these classes are as follows
$\{1\},\{a^m\},\{a,a^{-1}\},\{a^2,a^{-2}\},\ldots,\{a^{m-1},a^{-m+1}\},\{a^{2j}b:0\le j\le m-1\}, \{a^{2j+1}b:0\le j\le m-1\}$

In the following, I will explain the first case in details. The other case can be proven similarly.

When $n$ is odd, for finding the conjugacy classes containing $a^i$, $1\le i \le n-1$, we know that $\langle a \rangle \le C_G(a^i)$. So, $\vert G:C_G(a^i) \vert\le \vert G:\langle a \rangle \vert=2$. This means that $\vert CL_G(a^i) \vert\le 2$. Since $b^{-1}ab=a^{-1}$, $a^i$ and $a^{-i}$ are conjugate. Therefore, $\{a^i,a^{-i}\}\subseteq CL_G(a^i)$.
If $a^i=a^{-i}$, then $n \vert 2i$. Since $n$ is odd, $n\vert i$, that is a contradiction with $1\le i \le n-1$. So, $\vert CL_G(a^i) \vert= 2$ and $CL_G(a^i)=\{a^i,a^{-i}\}$ for $1\le i \le n-1$.
Obviousely, $\{1,b\}\subseteq C_G(b)$. By the second relation of the group we have $b^{-1}a^ib=a^{-i}\ne a^i$. Hence, $b(a^ib)\ne (a^ib)b$ and $ba^i\ne a^ib$ for $1\le i \le n-1$. So, $C_G(b)=\{1,b\}$ and we have $\vert CL_G(b) \vert= n$. Therefore, $CL_G(b)=\{b,ab,\ldots,a^{n-1}b\}$.

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    $\begingroup$ Maybe you could explain how you found this result. $\endgroup$
    – Watson
    Commented Jun 14, 2016 at 10:33
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    $\begingroup$ @Watson: I added some details to the answer. $\endgroup$ Commented Jun 14, 2016 at 11:11
  • $\begingroup$ What is $\langle a \rangle$ and $\mathrm{CL}_G$? $\endgroup$
    – Ziyuan
    Commented Mar 13, 2022 at 22:30

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