# Conjugate class in the dihedral group

List all the conjugate classes in the dihedral group of order $2n$ and verify the class equation.

The dihedral group is generated by two elements $r$ and $s$.

The order of $r$ is two since $r^2=e$ and $s$ is $n$ since $s^n = e$. And I know all elements can be produced as either $s^k$ or $rs^k$. How can I list all the conjugacy class?

• Use the fact that $sr=rs^{-1}$, you can express the conjugacy classes easily. – Henry Nov 8 '13 at 4:48
• The way you write the first line, as a declarative sentence, makes this sound like homework. Is it? – KCd Nov 8 '13 at 5:00
• @KCd Nope, just practice problems I am doing. Homework was due yesterday lol. – user104235 Nov 8 '13 at 5:03
• @KCd, I disagree. The way the OP writes the first line as an imperative sentences makes it sound like homework. – PrimeRibeyeDeal May 8 '15 at 0:48

First we are going to deal with $\rm{D}_{2n}$ where $n$ is an odd number. Then $\rm{D}_{2n} = <a,b>$ where $a^2=1$ is the reflection and $b^n$ is the rotation. Claim if $n=2m+1$ then the conjugacy classes, are the $\{ b^{i}, b^{-i} \}$ for $1 \leq i \leq m$ and $\{a b^{k} | 0\leq k < n \}$. $\\ \\$ We note that $\rm {D} _{2n}= \{a^{i}b^{j}| 0\leq i \leq 1 , 0 \leq j \leq n\}$. Take $1 \leq l <m$ and $0\leq k <n$ an element which is only rotation $b^l$, then we see that $b^{k} b^l b^{- k}= b^l$ and $a b^{k}b^l ( a b^{k} )^{-1}= a b^{k}b^l b^{-k}a = b^{-l}$ which proves that the orbit of $b^l$ is just $\{ b^l , b^{-l} \}$. $\\ \\$ For the other type of conjugacy classes again we do the same computation for an element $a$ then its orbit is $\{a b^{k} | 0\leq k < n \}$ using the relation $ab^{k}a= b^{-k}$ we get that $b^k a b^{-k} =a b^{-2k}$ and because $\gcd(2,n) = 1$, $<b^{-2k}>= <b>$ and by taking conjugation with $a b^k$ we get $a b^k a b^{-k} a =a b^{2k}$. Therefore we have prove the orbit always stays in the elements of the form $a b^{p}$ and that it covers all the elements which completes the analysis for $n$ odd. We get the conjugacy equation (and of course the identity element has orbit only itself) $2 n=1 + \underbrace {2 +2+ \cdots +2} _{m \rm{\,times}} +n.$
For the even case the analysis is the same but we get some different type of orbits, which is expected since the even dihedral has center whereas the odd does not. Take $n = 2 m$ for $b^{k}$ for $1 \leq k < m$ the same computation gives that the orbit of $b^{k}$ is the same namely $\{ b^k , b^{-k}\}$. We have an incosistency with $r^m$ because tihs element with the same computation we get it goes to $r^{-m}$ but $r^{-m} = r^{m}$ which proves it is only a one element orbit. Now again when we try to find the orbit of the reflection element $a$, again with the same computation as before we take elements of the form $a b^{2k}$ but since $\gcd(2, n) =2$, $b^2$ has order $m$ and so its orbit take only the elements of the form $a b^l$ where $l$ is even. We are left with one more orbit namely $a b^{p}$ where $p$ is odd and by doing again the same computation this is indeed the orbit of $a b$. The conjugacy equation is now $$n = 1+1+ \underbrace {2 +2 +\cdots +2 }_{m-1 \rm{times}} +m +m.$$
• Thank you Clark!! I have a quick question, why is $b^{k} b^l b^{- k}= b^l$ and $a b^{k}b^l ( a b^{k} )^{-1}= a b^{k}b^l b^{-k}a = b^{-l}$? – user104235 Nov 8 '13 at 5:11
• Any time! for the first we have $b^k b^l b^{-l}$ since they have the same base we see that this is equal to $b^{k+l-k}=b^l$. In the dihedral we have the relation $a^{-1}b a=b^{-1} \Rightarrow (a^{-1}b a)^l= b^{-l} \Rightarrow a^{-1} b^{l} a =b^{-l}$. So firstly we used $(a b^l)^{-1} =b^{-l} a^{-1}$, and then we applied $a^{-1} b^{l} a =b^{-l}$ remember $a=a^{-1}$. – clark Nov 8 '13 at 5:22
• The thing is that when we have the odd case when we take a conjugations of $a$ with different elements we end up with elements of the form $ab^{2k}$ and $k$ goes from $1$ to $n$, so the orbit would be how many different element of $<b>$ do we hit with the $b^{2k}$. This is the same as asking the order of $<b^{2}>$ but this has order order $\frac{n}{\gcd (2,n)}=n$ (since $n$ is odd) so it gives all the elements. In the even case though when we do the same we need to answer the same question $<b^2>$ how many elements this things spans, we have the same formula but this time it gives – clark Nov 10 '13 at 17:01
• $\frac{n}{\gcd (2,n)}=\frac{n}{2}$ so there are some elements left out from the orbit of $a$. But this elements are picked by the orbit of $ab$ which gives the element of the form $a b^k$ where $k$ is odd – clark Nov 10 '13 at 17:03