Definition and Properties of Dihedral Group

Question

Definition: Here is equivalent definition of dihedral group $$\begin{align}D_{2n}&=\{\sigma \in S_n\mid \sigma(a+1)\equiv\sigma(a)+1\pmod n, \forall 1\leq a\leq n\} \bigcup \{\sigma \in S_n\mid \sigma(a+1)\equiv \sigma(a)-1\pmod n, \forall 1\leq a\leq n\} \\ &= \{T:\Bbb{R}^2\to \Bbb{R}^2\mid T(P)=P \text{ and } T \text{ is isometry}\}\\ &= \{\text{“symmetry” of $n$-gon}\} \end{align}$$

---

Fix a regular $n$-gon centered at the origin in an $x$, $y$ plane and label the vertices consecutively from $1$ to $n$ in a clockwise manner. Let $r$ be the rotation clockwise about the origin through $2\pi /n$ radian. Let $s$ be the reflection about the line of symmetry through vertex $1$ and the origin. We leave the details of the following calculations as an exercise

$(1)$ $1,r,r^2,…,r^{n-1}$ are all distinct and $r^n=1$, so $|r|=n$.

$(2)$ $|s|=2$.

$(3)$ $s\neq r^i$, for any $i$.

$(4)$ $sr^i \neq sr^j$, for all $0\leq i,j \leq n- 1$ with $i\neq j$, so $$D_{2n}=\{1,r,r^2,…,r^{n-1},s,sr,sr^2,…,sr^{n-1}\}$$ i.e. each element can be written uniquely in the form $s^kr^i$ for some $k=0$ or $1$ and $0\leq i\leq n-1$.

$(5)$ $rs=sr^{-1}$. [First work out what permutations effects on $\{1,2,...,n\}$ and then work out separately what each side in this equation does to vertices $1$ and $2$.] This shows in particular that $r$ and $s$ do not commute so that $D_{2n}$ is non-abelian.

$(6)$ $r^i s=sr^{-i}$, for all $0\leq i \leq n$. [Proceed by induction on $i$ and use the fact that $r^{i+1}s=r(r^is)$ together with the preceding calculation.] This indicates how to commute $s$ with powers of $r$.

My attempt: Let symmetry $r^p$ ($0\leq p\leq n-1$) and $s$ correspond to permutation $\sigma_p$ and $\mu$, respectively. Then $\sigma_p:J_n\to J_n$ such that $\sigma_p(a)=a+p$ (addition taken “modulo $n$”), $\forall a\in J_n$. It’s easy to check, $\sigma_p(b+1)=\sigma_p(b)+1$ for all $b$. By induction, $\sigma_p=(\sigma_1)^p$ for all $0\leq p\leq n-1$. Knowing image of two adjacent vertex, let say $1$ and $2$, completely determine the permutation (see third paragraph). Then $\mu:J_n\to J_n$ such that $\mu (1)=1$ and $\mu (2)=n$. It’s easy to check, $\mu (b+1)=\mu (b)-1$ for all $b$. Let $\sigma ,\tau\in D_{2n}$. Then $\sigma=\tau$ if and only if $\sigma(1)=\tau (1)$ and $\sigma (2)=\tau (2)$.

$(1)$ Let $i,j$ such that $0\leq i\lt j\leq n-1$. Then $1\leq 1+i\lt 1+j\leq n$. So $\sigma_i(1)=1+i\neq 1+j=\sigma_j(1)$. Thus $\sigma_i\neq \sigma_j$, if $i\neq j$. Hence $\sigma_0,\sigma_1,…,\sigma_{n-1}$ are all distinct. Let $\sigma_n=(\sigma_1)^n$. Since addition taken modulo $n$, we have $\sigma_n(a)=a+n=a$, $\forall a\in J_n$. So $\sigma_n=\sigma_0$. Thus $|\sigma_1|=n$.

$(2)$ We have $\mu ^2(1)=\mu (1)=1$ and $\mu^2(2)=\mu (n)=2$. Thus $\mu^2(1)=\sigma_0(1)$ and $\mu^2(2)=\sigma_0(2)$. Hence $\mu^2=\sigma_0$ and $|\mu|=2$.

$(3)$ We have $\sigma_i(1)=1+i$ and $\sigma_i(2)=2+i$. We have $\mu (1)=1$ and $\mu (2)=n$. Assume towards contradiction $\sigma_i=\mu$, for some $i$. Then $\sigma_i=\mu$ iff $\sigma_i(1)=\mu (1)$ and $\sigma_i(2)=\mu (2)$ iff $i=0$ and $i=n-2$. So $n=2\geq 3$. Thus we reach contradiction. Hence $\sigma_i\neq \mu$.

$(4)$ Assume towards contradiction $sr^i=sr^j$, for some $0\leq i\lt j\leq n-1$. Then $sr^i(r^i)^{-1}=sr^j(r^i)^{-1}=s=sr^{j-i}$. So $1=r^{j-i}$. Since $i\lt j$, we have $0\lt j-i\leq n-1$. By $(1)$, $1\neq r^{j-i}$. Thus we reach contradiction. Hence $sr^i\neq sr^j$ for all $0\leq i,j\leq n-1$ with $i\neq j$. That is $s,sr,…,sr^{n-1}$ are all distinct. I think we also need to show $r^i\neq sr^j$, for all $0\leq i,j\leq n-1$. Assume towards contradiction $r^i=sr^j$, for some $i,j$. Then $r^i(r^i)^{-1}=sr^j(r^i)^{-1}=1=sr^{j-i}$. By $(2)$, $s^2=1$. So $s=r^{j-i}$. Since $|r|=n$, we have $r^{j-i}\in \{1,r,…,r^{n-1}\}$. But by $(3)$, $s\neq r^{j-i}$. Thus we reach contradiction. Hence $1,r,…,r^{n-1},s,sr,…,sr^{n-1}$ are all distinct. Since $|D_{2n}|=2n$, we have $$D_{2n}=\{1,r,r^2,…,r^{n-1},s,sr,sr^2…,sr^{n-1}\}.$$

$(5)$ We have $\sigma_1\circ \mu(1)=\sigma_1 (1)=2$ and $\sigma_1 \circ \mu (2)=\sigma_1 (n)=1$. Since $\sigma_1(n)=1$ and $\sigma_1(1)=2$, we have $\sigma_1^{-1}(1)=n$ and $\sigma_1^{-1}(2)=1$. So $\mu \circ \sigma_1^{-1}(1)=\mu (n)=2$ and $\mu \circ \sigma_1^{-1}(2)=\mu (1)=1$. Thus $\sigma_1\circ \mu (1)=\mu \circ \sigma_1^{-1}(1)$ and $\sigma_1\circ \mu (2)=\mu \circ \sigma_1^{-1}(2)$. Hence $\sigma_1\circ \mu = \mu \circ \sigma_1^{-1}$, or equivalently $rs=sr^{-1}$. We claim $rs\neq sr$. Assume towards contradiction $rs=sr$. Then $rs=sr^{-1}=sr$. So $r^{-1}=r$. That is $|r|=2\geq 3$. Thus we reach contradiction. Hence $D_{2n}$ is non abelian.

$(6)$ $r^i s=sr^{-i}$, for all $0\leq i \leq n$. Base case: $i=1$. By $(5)$, $rs=sr^{-1}$. Inductive step: suppose $r^i s=sr^{-i}$, for some $0\leq i$. By inductive hypothesis, $r^{i+1}s=r(r^{i}s)=r(sr^{-i})$. By $(5)$, $r(sr^{-i})=(rs)r^{-i}=sr^{-1}r^{-i}=sr^{-(i+1)}$. So $r^{i+1}s=sr^{-(i+1)}$. By principle of mathematical induction, $r^i s=sr^{-i}$, for all $0\leq i \leq n$. Is my proof correct?

Que: Can we explicitly show permutation $\mu$, not just $\mu (1)$ and $\mu (2)$, like we did for $\sigma_p$? I think we have to consider two cases, first $n$ is even, second $n$ is odd.

Answer 1

Your very first definition $D_n=$ is false. Anyway, you don't have to translate Dummit's geometric definitions and propositions about $r,s,D_n$ into statements about permutations. It is much simpler to reason directly:

1. $r^k$ is the rotation of $k\frac{2\pi}n\pmod{2\pi}$ Since these angles are distinct for $k=0,1,\dots,n-1,$ the rotations $1,r,\dots,r^{n-1}$ are distinct. And since $n\frac{2\pi}n\equiv0\pmod{2\pi},$ $r^n=1.$

2. $s$ is a reflection.

3. a reflection reverses the orientation, whereas a rotation preserves it.

4. if $0\leq i\le j<n$ and $sr^i=sr^j,$ then $r^{j-i}=0$ hence by 1), $i=j.$ Similarly $sr^i\ne r^j$ by 3), hence $r^i\ne sr^j$ by 2). This proves that the $2n$ elements $sr^k,r^k(0\le k<n)$ are distinct. Let us now prove that any symmetry $t$ of the $n$-gon is one of them. Consider the symmetry $u$ of the $n$-gon equal to $st$ if $t$ is a reflection, and to $t$ if $t$ is a rotation. In both cases, $u$ is a rotation of the $n$-gon. Let $k$ be the number of the vertex to which $u$ maps the vertex number $1.$ Then $u=r^k,$ and we are done ($t=$ either $sr^k$ or $r^k$).

5. Denote by $v_k$ the vertex number $k.$ Then, $rs(v_1)=r(v_1)=v_2$ and $sr^{-1}(v_1)=s(v_n)=v_2.$ Since both $rs$ and $sr^{-1}$ are reflections and coincide on $0$ and $v_1,$ they are equal. Ok with your end of 5).

6. ok.

Re. $\mu(1)=1,\mu(2)=n,\mu(3)=n-1,\dots.$