**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.6 Exercise 1.6.25**

Let $n \in \mathbb{Z}^+$, let $r$ and $s$ be the usual generators of $D_{2n}$, and let $\theta = 2 \pi / n$.

(1) Prove that the matrix $T = \left[ {\cos \theta \atop \sin \theta} {-\sin \theta \atop \cos \theta} \right]$ is the matrix of the linear transformation which rotates the $xy$-plane about the origin in a counterclockwise direction by $\theta$ radians.

(2) Prove that the map $\varphi : D_{2n} \rightarrow GL_2(\mathbb{R})$ defined on the generators by $\varphi(r) = \left[ {\cos \theta \atop \sin \theta} {-\sin \theta \atop \cos \theta} \right]$ and $\varphi(s) = \left[ {0 \atop 1} {1 \atop 0} \right]$ extends to a homomorphism of $D_{2n}$ to $GL_2(\mathbb{R})$.

(3) Prove that the homomorphism in Part (2) is injective.

Solution:

(1) Let $v = \langle x, y \rangle$ be a vector in $\mathbb{R}^2$. In polar coordinates, this point is $\langle \sqrt{x^2 + y^2}, \tan^{-1} \frac{y}{x} \rangle$. Consider the action of the given matrix on $v$: we have $$\left[ {\cos \theta \atop \sin \theta} {-\sin \theta \atop \cos \theta} \right] \cdot \left[ x \atop y \right] = \left[ {x \cos \theta -y \sin \theta} \atop {x \sin \theta + y \cos \theta} \right],$$ Which in polar coordinates has radius $$\sqrt{(x \cos \theta - y \sin \theta)^2 + (x \sin \theta + y \cos \theta)^2}$$ and angle $$\tan^{-1} \left( \frac{x \sin \theta + y \cos \theta}{x \cos \theta - y \sin \theta} \right).$$ Note, however, that \begin{align*} &\ (x \cos \theta - y \sin \theta)^2 + (x \sin \theta + y \cos \theta)^2\\ = &\ x^2(\cos^2 \theta + \sin^2 \theta) + y^2(\cos^2 \theta + \sin^2 \theta)\\ = &\ x^2 + y^2, \end{align*} so the radius of the image of $v$ is in fact $\sqrt{x^2 + y^2}$. Moreover, note that \begin{align*} &\ \tan \left( \tan^{-1} \left( \frac{y}{x} \right) + \theta \right)\\ =&\ \frac{\sin(\tan^{-1}(y/x) + \theta)}{\cos(\tan^{-1}(y/x) + \theta)}\\ =&\ \frac{x \sin \theta + y \cos \theta}{x \cos \theta - y \sin \theta} \end{align*} by the angle sum formulas for sine and cosine. Thus we have $$ \tan^{-1} \left( \frac{x \sin \theta + y \cos \theta}{x \cos \theta - y \sin \theta} \right) = \tan^{-1}\left( \frac{y}{x} + \theta \right).$$ Hence in polar coordinates, $$ T \cdot \langle \sqrt{x^2 + y^2}, \tan^{-1}(y/x) \rangle = \langle \sqrt{x^2 + y^2}, \tan^{-1}(y/x) + \theta \rangle.$$ So $T$ is in fact the transformation which rotates the xy-plane by \theta radians counterclockwise about the origin.

(2) First we prove a lemma.

**Lemma 1.** Let $S = \{ r,s \}$ and $G$ be a group. $If \overline{\varphi} : S \rightarrow G$ is a mapping such that $\overline{\varphi}(r)^n = 1$, $\overline{\varphi}(s)^2 = 1$, and $\overline{\varphi}(r) \overline{\varphi}(s) = \overline{\varphi}(s) \overline{\varphi}(r)^{-1}$, then $\overline{\varphi}$ extends to a homomorphism $\varphi : D_{2n} \rightarrow G$. Moreover, if no power of $\overline{\varphi}(r)$ is equal to $\overline{\varphi}(s)$ and the powers $\overline{\varphi}(r)^k$ are distinct for $0 \leq k < n$, then $\varphi$ is injective.

Proof: We have that $\overline{\varphi}(r)$ and $\overline{\varphi}(s)$ satisfy the relations of $D_{2n}$. Now every element of $D_{2n}$ can be written uniquely as $s^ar^b$ for some $0 \leq a < 2$ and $0 \leq b < n$; define $\varphi$ by $$\varphi(s^ar^b) = \overline{\varphi}(s)^a \overline{\varphi}(r)^b.$$ Now if $s^ar^b$ and $s^cr^d$ are elements of $D_{2n}$, we have $$\varphi(s^ar^bs^cr^d) = \varphi(s^{a+c} r^{d-b}) = \overline{\varphi}(s)^{a+c} \overline{\varphi}(r)^{d-b} = \overline{\varphi}(s)^a \overline{\varphi}(r)^b \overline{\varphi}(s)^c \overline{\varphi}(r)^d = \varphi(s^ar^b) \varphi(s^cr^d),$$ hence $\varphi$ is a homomorphism. To see injectivity, suppose that $\varphi(s^ar^b) = \varphi(s^cr^d)$ but that $s^ar^b \neq s^cr^d$; then either $a \neq c$ or $b \neq d$. Then we have $$\overline{\varphi}(s)^a \overline{\varphi}(r)^b = \overline{\varphi}(s)^c \overline{\varphi}(r)^d.$$ If $a \neq c$ then without loss of generality, $a = 1$ and $c = 0$. Then we have $\overline{\varphi}(r)^{b-d} = \overline{\varphi}(s)$, a contradiction. On the other hand, if $a = c$ and $b \neq d$ then we have $\overline{\varphi}(r)^b = \overline{\varphi}(r)^d$, a contradiction. Thus $\varphi$ is injective.

Now by the lemma, it suffices to show that $\overline{\varphi}(r)^n = 1$, $\overline{\varphi}(s)^2 = 1$, and $$\overline{\varphi}(r) \overline{\varphi}(s) = \overline{\varphi}(s) \overline{\varphi}(r)^{-1}.$$ To this end, we prove another lemma.

**Lemma 2.** For all $k \in \mathbb{Z}^+$, $\overline{\varphi}^k = \left[ {\cos k \theta \atop \sin k \theta} {-\sin k \theta \atop \cos k \theta} \right]$.

Proof: We proceed by induction on $k$. The base case is trivial. Now suppose the conclusion holds for some k; then \begin{align*} &\ \overline{\varphi}(r)^{k+1} = \overline{\varphi}(r)^k \cdot \overline{\varphi}\\ =&\ \left[ {\cos k\theta \atop \sin k\theta} {-\sin k\theta \atop \cos k\theta} \right] \cdot \left[ {\cos \theta \atop \sin \theta} {-\sin \theta \atop \cos \theta} \right]\\ =&\ \left[ {\cos \theta \cos k\theta - \sin \theta \sin k \theta \atop \cos \theta \sin k\theta + \sin \theta \cos k\theta} {-(\sin \theta \cos k\theta + \cos \theta \sin k\theta) \atop \cos \theta \cos k\theta - \sin \theta \sin k\theta} \right]\\ =&\ \left[ {\cos (k+1)\theta \atop \sin (k+1)\theta} {-\sin (k+1)\theta \atop \cos (k+1)\theta} \right] \end{align*} using the angle sum formulas for sine and cosine. By induction, the lemma is proved. $\blacksquare$

We now show that $\varphi$ is a homomorphism. We have $$\overline{\varphi}(r)^n = \left[ {\cos 2\pi \atop \sin 2\pi} {-\sin 2\pi \atop \cos 2\pi} \right] = I_2.$$ Also, clearly $\overline{\varphi}(s)^2 = I$. Finally it is easy to check that $(\overline{\varphi}(s) \overline{\varphi}(r))^2 = I_2$, and hence $$\overline{\varphi}(r)\overline{\varphi}(s) = \overline{\varphi}(s)\overline{\varphi}(r)^{-1}.$$ By Lemma 1, then, $\varphi$ is a homomorphism.

(3) By Lemma 1, it suffices to show that the powers $\overline{\varphi}(r)^k$ are distinct for $0 \leq k < n$ and that $\overline{\varphi}(r)^k \neq \overline{\varphi}(s)$. These both follow from Lemma 2.