Dihedral Group and Rotation of the Plane

Group Theory Problems and Solutions

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 $\theta=2 \pi/n$.


(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.


(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 $\GL_2(\R)$.


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

LoadingAdd to solve later

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 $\pm 1$.


LoadingAdd to solve later

Sponsored Links

More from my site

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

More in Group Theory
Normal Subgroups Problems and Solutions in Group Theory
Normal Subgroups Intersecting Trivially Commute in a Group

Let $A$ and $B$ be normal subgroups of a group $G$. Suppose $A\cap B=\{e\}$, where $e$ is the unit element...

Close