Prove $rs=sr^{-1}$ in ${\rm Dih}(2n)$

Question

Let $r$ and $s$ be the rotation and reflection symmetries respectively in ${\rm Dih}(2n)$, the dihedral group of order $2n$. Show that $rs=sr^{-1}$.

I also need to show by induction that $r^js=sr^{-j}$ but I think that will follow easily.

Answer 1

I think you can find it in many Group theory book as a Theorem or as an exercise. Now, let $G_1,G_2$ are two groups and let $\phi:G_2\to Aut(G_1),~~\phi(b)=\phi_b$ (I assume you know what is $\phi_b$). We can prove that $G_1\times G_2$ with the following operation:

$$(a,b)(a',b')=(a\phi_b(a'),bb')$$

is a new group and you know that it is $G_1\times_{\phi} G_2$ called the semi-direct product of $G_1$ by $G_2$. In fact, assuming that $\phi_b$ where $b\in G_2$ is essential here. Now, let $G_1=\mathbb Z_n=\langle a\rangle$ and $G_2=\mathbb Z_2=\langle b\rangle$. Consider: $$\phi:G_2\to Aut(G_1)\\\\ \phi(b)=\phi_b,~~\phi(1)=\phi_1=id_{\mathbb Z_n}$$ $\phi$ is a group homomorphism and we denote $\mathbb Z_n\times_{\phi}\mathbb Z_2$ as $D_{2n}$. Here, you can find out why that relation is regarded. Just consider the way $\phi$ acts.

Answer 2

Hint: First you want to show that if $r$ is a rotation and $s$ is a reflection, then $rs$ must be a reflection. What is the square of any reflection? What is the inverse of any reflection?

Answer 3

You can see $\text{Dih}(2n)$ as a subgroup of $O(2,\mathbb{R})$. Thus $r= \left( \begin{matrix} \cos(\pi/n) & -\sin(\pi/n) \\ \sin(\pi/n) & \cos(\pi/n) \end{matrix} \right)$ and $s= \left( \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right)$. Then $srs= \left( \begin{matrix} \cos(\pi/n) & \sin(\pi/n) \\ -\sin(\pi/n) & \cos(\pi/n) \end{matrix} \right)=r^{-1}$.

Another possibility is to notice that $\det(srs)=\det(r)=1$ and $\text{Tr}(srs)=\text{Tr}(r)=2\cos(\pi/n)$. Therefore, $srs$ is a rotation of angle $\pm \pi/n$. If $srs=r$, then $r$ and $s$ commute and you deduce that $\text{Dih}(2n)$ is itself abelian, which is false. So $srs$ is a rotation of angle $-\pi/n$, ie. $srs=r^{-1}$.