Dihedral Group and Rotation of the Plane

Problem 52

Let $n$ be a positive integer. Let $D_{2n}$ be the dihedral group of order $2n$. Using the generators and the relations, the dihedral group $D_{2n}$ is given by
\[D_{2n}=\langle r,s \mid r^n=s^2=1, sr=r^{-1}s\rangle.\] Put $\theta=2 \pi/n$.

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(a) Prove that the matrix $\begin{bmatrix}
\cos \theta & -\sin \theta\\
\sin \theta& \cos \theta
\end{bmatrix}$ is the matrix representation of the linear transformation $T$ which rotates the $x$-$y$ plane about the origin in a counterclockwise direction by $\theta$ radians.

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(b) Let $\GL_2(\R)$ be the group of all $2 \times 2$ invertible matrices with real entries. Show that the map $\rho: D_{2n} \to \GL_2(\R)$ defined on the generators by
\[ \rho(r)=\begin{bmatrix}
\cos \theta & -\sin \theta\\
\sin \theta& \cos \theta
\end{bmatrix} \text{ and }
\rho(s)=\begin{bmatrix}
0 & 1\\
1& 0
\end{bmatrix}\] extends to a homomorphism of $D_{2n}$ into $\GL_2(\R)$.

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(c) Determine whether the homomorphism $\rho$ in part (b) is injective and/or surjective.

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Hint.

1. For (a), consider the unit vectors of the plane and consider where do the unit vector go by the linear transformation $T$.
2. Show that $\rho(r)$ and $\rho(s)$ satisfy the same relations as $D_{2n}.
3. Consider the determinant.

Proof.

(a) The matrix representation of the linear transformation $T$

Let $\mathbf{e}_1, \mathbf{e}_2$ be the standard basis of the plane $\R^2$. That is
\[\mathbf{e}_1=\begin{bmatrix}
1 \\
0
\end{bmatrix} \text{ and }
\mathbf{e}_2=\begin{bmatrix}
0 \\
1
\end{bmatrix}.\] Then by the $\theta$ rotation $\mathbf{e}_1$ moves to the point $\begin{bmatrix}
\cos \theta \\
\sin \theta
\end{bmatrix}$ and $\mathbf{e}_2$ moves to the point $\begin{bmatrix}
-\sin \theta \\
\cos \theta
\end{bmatrix}$.
Therefore the matrix representation of $T$ is the matrix $\begin{bmatrix}
\cos \theta & -\sin \theta\\
\sin \theta& \cos \theta
\end{bmatrix}$.

(Recall that if $T$ is a linear transformation from a vector space $V$ to itself with a basis $\{\mathbf{e}_1, \dots, \mathbf{e}_n\}$, its representation matrix is given by the matrix $[T(\mathbf{e}_1) \cdots T(\mathbf{e}_n)]$ whose $i$-th column is the vector $T(\mathbf{e}_i)$.)

(b) $\rho$ is a homomorphism of $D_{2n}$ into $\GL_2(\R)$

Any element $x \in D_{2n}$ can be written as $x=r^as^b$ using the relations.
Then we define the value of $\rho$ on $x$ by
\[\rho(x):=\rho(r)^a\rho(s)^b=\begin{bmatrix}
\cos \theta & -\sin \theta\\
\sin \theta& \cos \theta
\end{bmatrix}^a
\begin{bmatrix}
0 & 1\\
1& 0
\end{bmatrix}^b.\]

We need to show that this is well defined.
To do this, we show that $\rho(r)$ and $\rho(s)$ satisfy the same relation as $D_{2n}$.

We have
\[\rho(r)^n=\begin{bmatrix}
\cos \theta & -\sin \theta\\
\sin \theta& \cos \theta
\end{bmatrix}^n=
\begin{bmatrix}
\cos (n\theta) & -\sin (n\theta)\\
\sin (n\theta)& \cos (n\theta)
\end{bmatrix}
=I_2,\] where $I_2$ is the $2\times 2$ identity matrix. Also we have $\rho(s)^=I_2$.

Finally, we compute
\begin{align*}
\rho(r)\rho(s)\rho(r)&=\begin{bmatrix}
\cos \theta & -\sin \theta\\
\sin \theta& \cos \theta
\end{bmatrix}
\begin{bmatrix}
0 & 1\\
1& 0
\end{bmatrix}
\begin{bmatrix}
\cos \theta & -\sin \theta\\
\sin \theta& \cos \theta
\end{bmatrix}
\\[6pt] &=\begin{bmatrix}
– \sin \theta & \cos \theta\\
\cos \theta& \sin \theta
\end{bmatrix}
\begin{bmatrix}
\cos \theta & -\sin \theta\\
\sin \theta& \cos \theta
\end{bmatrix} \\[6pt] &=\begin{bmatrix}
0 & \sin^2 \theta + \cos^2 \theta\\
\cos^2+\sin^2 \theta& 0
\end{bmatrix}=I_2
\end{align*}
Therefore, the extension of $\rho$ does not depend on the expression of $x=r^as^b$.

(c) Determine whether the homomorphism $\rho$ in part (b) is injective and/or surjective.

We first show that $\rho$ is injective.
Suppose that we have $\rho(x)=I_2$ for $x \in D_{2n}$. Write $x=r^as^b$.
Then we have $\rho(r)^a\rho(s)^b=I_2$.

We compute the determinant of both sides and get
\[\det(\rho(r))^a \det(\rho(s))^b=1.\] Since $\det(\rho(r))=1$ and $\det(\rho(s))=-1$ we have $(-1)^b=1$, thus $b$ must be even.
Then $x=r^a$ since the order of $s$ is two.
Then $\rho(r)^a=I_2$ implies that $r\theta=2\pi m$ for some $m\in \Z$.

Hence $r=nm$ and we obtain $x=r^{nm}=1$ since the order of $r$ is $n$. Therefore the kernel of $\rho$ is trivial, hence the homomorphism $\rho$ is injective.

As the argument shows, the determinant of $\rho(x)$ is either $\pm 1$. The homomorphism $\rho$ is not surjective since $\GL_2(\R)$ contains elements with determinants not equal to $\pm 1$.

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