Exercise 1.6.25 (Matrix representation of $D_{2n}$)

Proof. Let $A=\left(\begin{array}{cc}\hfill \cos <!--FUNCTION\; APPLICATION-->𝜃\hfill & \hfill -\sin <!--FUNCTION\; APPLICATION-->𝜃\hfill \\ \hfill \sin <!--FUNCTION\; APPLICATION-->𝜃\hfill & \hfill \cos <!--FUNCTION\; APPLICATION-->𝜃\hfill \end{array}\right)$ and let $\rho$ be the endomorphism of $E={ℝ}^2$ such that ${ℳ}_{ℬ}(\rho )=B$, where $ℬ=(e_1,e_2)=((1,0),(0,1))$ is the canonical orthonormal basis of $E$.

Put $S=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)$.

(If $n=1$, then $A=I$ and $D_2=⟨s⟩\simeq \mathbb{Z}∕2\mathbb{Z}$. We suppose in the following that $n\ge 2$.)

(a)

Since $A{A}^{}$

T = I_2 $and$ $det(A)=1$, $A$ is a rotation matrix and $\rho$ is a rotation. The measure of the angle of the rotation is the measure of the angle $\widehat{(e_1,\rho (e_1))}=\widehat{(e_1,\cos <!--FUNCTION\; APPLICATION-->𝜃e_1+\sin <!--FUNCTION\; APPLICATION-->𝜃e_2)}$, so is $𝜃$. We may write $\rho =Rot(𝜃)$.

(b)

Note that, since $𝜃=2\pi ∕n$ $$A^n={\left(\begin{array}{cc}\hfill \cos <!--FUNCTION\; APPLICATION-->𝜃\hfill & \hfill -\sin <!--FUNCTION\; APPLICATION-->𝜃\hfill \\ \hfill \sin <!--FUNCTION\; APPLICATION-->𝜃\hfill & \hfill \cos <!--FUNCTION\; APPLICATION-->𝜃\hfill \end{array}\right)}^n=\left(\begin{array}{cc}\hfill \cos <!--FUNCTION\; APPLICATION-->n𝜃\hfill & \hfill -\sin <!--FUNCTION\; APPLICATION-->n𝜃\hfill \\ \hfill \sin <!--FUNCTION\; APPLICATION-->n𝜃\hfill & \hfill \cos <!--FUNCTION\; APPLICATION-->n𝜃\hfill \end{array}\right)=I_2,$$

$$S^2={\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)}^2=I_2.$$

(More precisely , for all integers $k\in \mathbb{Z}$,

$$\begin{array}{llll}\hfill A^k=I_2& ⟺{\rho }^k={Id}_E& \hfill & \\ \hfill & ⟺Rot(k𝜃)={Id}_E& \hfill & \\ \hfill & ⟺k\left(\frac{2\pi }{n}\right)\equiv 0(\mathrm{mod}<!--FUNCTION\; APPLICATION-->2\pi )& \hfill & \\ \hfill & ⟺n\mid k.& \hfill & \end{array}$$

Therefore the order of $A$ in the group ${GL}_2(ℝ)$ is $|A|=n$.)

Moreover,

$$\begin{array}{llll}\hfill AS& =\left(\begin{array}{cc}\hfill \cos <!--FUNCTION\; APPLICATION-->𝜃\hfill & \hfill -\sin <!--FUNCTION\; APPLICATION-->𝜃\hfill \\ \hfill \sin <!--FUNCTION\; APPLICATION-->𝜃\hfill & \hfill \cos <!--FUNCTION\; APPLICATION-->𝜃\hfill \end{array}\right)\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill -\sin <!--FUNCTION\; APPLICATION-->𝜃\hfill & \hfill \cos <!--FUNCTION\; APPLICATION-->𝜃\hfill \\ \hfill \cos <!--FUNCTION\; APPLICATION-->𝜃\hfill & \hfill \sin <!--FUNCTION\; APPLICATION-->𝜃\hfill \end{array}\right),& \hfill & \\ \hfill S{A}^{-1}& =\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)\left(\begin{array}{cc}\hfill \cos <!--FUNCTION\; APPLICATION-->𝜃\hfill & \hfill \sin <!--FUNCTION\; APPLICATION-->𝜃\hfill \\ \hfill -\sin <!--FUNCTION\; APPLICATION-->𝜃\hfill & \hfill \cos <!--FUNCTION\; APPLICATION-->𝜃\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill -\sin <!--FUNCTION\; APPLICATION-->𝜃\hfill & \hfill \cos <!--FUNCTION\; APPLICATION-->𝜃\hfill \\ \hfill \cos <!--FUNCTION\; APPLICATION-->𝜃\hfill & \hfill \sin <!--FUNCTION\; APPLICATION-->𝜃\hfill \end{array}\right),& \hfill & \end{array}$$

so $SA=S{A}^{-1}$.

Since

$$D_{2n}=⟨r,s\mid r^n=s^2=1,rs=s{r}^{-1}⟩,$$

and $A^n=S^2=I_2,AS=S{A}^{-1}$, where $A,S\in {GL}_2(ℝ)$, there exists an homomorphism $\phi :D_{2n}\to {GL}_2(ℝ)$ such that $\phi (r)=A,\phi (s)=S$ (see Section 6.3).

Since every element $g\in D_{2n}$ is of the form $g=r^h{s}^k,0\le h<n,0\le k<2$, we obtain

$$\phi (g)=\phi (r^h{s}^k)=A^h{S}^k.$$

(Alternatively, we may verify as in Exercise 24, that $\phi$ defined by $\phi (r^h{s}^k)=A^h{S}^k$ is an isomorphism, knowing that the order of $A$ is $n$, the order of $S$ is $2$, and $AS=S{A}^{-1}$.)

(c)

We show that $S\notin ⟨A⟩$.

Assume that $S\in ⟨A⟩$. Then $S=A^i$ for some integer $i$. Then

$$-1=\det <!--FUNCTION\; APPLICATION-->(S)={(\det <!--FUNCTION\; APPLICATION-->(A))}^i=1.$$

The contradiction $1=-1$ (since $1\ne -1$ in $ℝ$) shows that $S\notin ⟨A⟩$.

Now let $g=r^h{s}^k\in \ker <!--FUNCTION\; APPLICATION-->(\phi )$, where $0\le h<n,0\le k<2$. Then $I_2=\phi (g)=A^h{S}^k$.

• If $k=1$, then $S=A^{-h}\in ⟨A⟩$: this is impossible.

• If $k=0$, then $A^h=I_2$, thus $n\mid h$, where $0\le h<n$, so $h=k=0$, and $g=r^h{s}^k=e$.

This shows that $\ker <!--FUNCTION\; APPLICATION-->(\phi )=\{e\}$, so $\phi$ is injective.