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I was trying to do the following from a past exam of my Rings and Groups' professor

Classify all conjugacy classes of the elements in the dihedral group $D_n$ = $\{ 1,r,r^2, ... , r^{n-1} ,s ,rs ,r^2s , ... ,r^{n-1} \}$ = $\mathopen{<} r,s \ | \ r^n=1; s^2=1, (rs)^2=1 \mathopen{>}$

A brief reminder that $D_n$ can be understood from a geometrical point of view, $D_n$ is the group of "symmetries" of a regular $n$-gon.

Given two elements $x,a$ in a group G, we define the conjugate of $a$ through $x$ as the element $x^{-1}ax$

We define the conjugacy class of an element $a$ in a group G as $[a]=\{ x^{-1}ax \ | \ x \in G \}$

More specifically, I'm trying to calculate the following :

  1. The conjugate of a rotation $r^k$ through a rotation $r^j$ and through a symmetry $r^js$.
  2. Determine the conjugacy class of $r^k$ in $D_n$. How many elements does it have?
  3. The conjugate of a symmetry $s$ through a rotation $r^j$ and through a symmetry $r^js$.
  4. Determine the conjugacy class of $s$ in $D_n$. How many elements does it have?

  1. Evidently through $r^j$ we have, $r^{-j}r^kr^j = r^{-j+k+j}=r^k$.

    Now through $r^js$ we get, $(r^js)^{-1}r^k(r^js) = s^{-1}r^{-j}r^kr^js = s^{-1}r^ks$.

I don't know how to classify this last one, and the ones from 3.

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It should be sufficient to solve this to know that $$s^{-1}=s$$ which you can see from $s^2=1$, so $s^2s^{-1}=s^{-1}$, and since $$rsrs = 1$$ we have $$srs = r^{-1}$$ so $$sr = r^{-1}s$$ I'm not going to answer all of the questions, but this should be enough to do it yourself.

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  • $\begingroup$ Could you please explain why those equalities are true? $\endgroup$ – M. Navarro Apr 28 at 14:17
  • $\begingroup$ @M.Navarro Edited, take a look. $\endgroup$ – Matt Samuel Apr 28 at 14:21
  • $\begingroup$ Obviously, thank you so much. I’ll update my answer when I make some progress. Otherwise I’ll come back to ask for help $\endgroup$ – M. Navarro Apr 28 at 14:23

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