Determine all the conjugacy classes of the dihedral group
\[D_{8}=\langle r,s \mid r^4=s^2=1, sr=r^{-1}s\rangle\] of order $8$.

Hint.

You may directly compute the conjugates of each element
but we are going to use the following theorem to simplify the computations.

Theorem.

The number of conjugates of an element $g$ in a group is the index $|G: C_G(s)|$ of the centralizer of $s$.

Solution.

Let us denote $G=D_8$.
Let $K_x$ be the conjugacy class in $G$ containing the element $x$.

Note that $\langle r \rangle < C_G(r) \lneq G$ and the order $|\langle r \rangle|=4$.
Hence we must have $C_G(r)=\langle r \rangle$.
Thus the element $r$ has $|G:C_G(r)|=2$ conjugates in $G$.

Since $srs^{-1}=r^3$, the conjugacy class $K_r$ containing $r$ is $\{r, r^3\}$.

Since $\langle s \rangle < C_G(s) \lneq G$ and $|\langle s \rangle|=2$, we have either $C_G(s)=\langle s\rangle$ or $|C_G(s)|=4$.
Since $r^2s=sr^2$, we must have $|C_G(s)|=4$ and hence the conjugacy class $K_s$ containing $s$ has $|G:C_G(s)|=2$ elements.

Since $rsr^{-1}=sr^2$, we have $K_s=\{s, sr^2\}$.

We know the center is $Z(G)=\{1,r^2\}$ by Problem Centralizer, normalizer, and center of the dihedral group D8.
Thus $K_1=\{1\}$ and $K_{r^2}=\{r^2\}$.

The remaining elements $sr$ and $sr^3$ should be in the same conjugacy class (otherwise these elements are in the center), thus $K_{sr}=\{sr, sr^3\}$.

In summary, the conjugacy classes of the dihedral group are
\[\{1\}, \{r^2\}, \{r, r^3\}, \{s, sr^2\}, \{sr, sr^3\}.\] Click here if solved 58

The post All the Conjugacy Classes of the Dihedral Group $D_8$ of Order 8 first appeared on Problems in Mathematics.

Let $D_8$ be the dihedral group of order $8$.
Using the generators and relations, we have
\[D_{8}=\langle r,s \mid r^4=s^2=1, sr=r^{-1}s\rangle.\]

(a) Let $A$ be the subgroup of $D_8$ generated by $r$, that is, $A=\{1,r,r^2,r^3\}$.
Prove that the centralizer $C_{D_8}(A)=A$.

(b) Show that the normalizer $N_{D_8}(A)=D_8$.

(c) Show that the center $Z(D_8)=\langle r^2 \rangle=\{1,r^2\}$, the subgroup generated by $r^2$.

Definitions (centralizer, normalizer, center).

Recall the definitions.

1.  The centralizer $C_{D_8}(A)$ is a subgroup of $D_8$ whose elements commute with $A$.
   That is $C_{D_8}(A)=\{ g\in D_8 \mid gxg^{-1}=x \text{ for all } x\in A\}$.
2. The normalizer $N_{D_8}(A)$ is a subgroup of $D_8$ defined as
   $N_{D_8}(A)=\{ g\in D_8 \mid gxg^{-1} \in A \text{ for any } x \in A\}$.
3. The center $Z(D_8)$ is a subgroup of $D_8$ whose elements commute with all elements of $D_8$.
   That is, $Z(D_8)=\{g \in D_8 \mid gxg^{-1}=x \text{ for all } x\in D_8\}$.

Proof.

(a) The centralizer $C_{D_8}(A)=A$

 Since any power of $r$ commutes with each other we have $A < C_{D_8}(A)$.
Since $sr=r^{-1}s$ and $r^{-1}s\neq rs$ (otherwise $r^2=1$), we see that $s\not \in C_{D_8}(A)$.

This also implies that any element of the form $r^as \in D_8$ is not in $C_{D_8}(A)$. In fact, if $r^as\in C_{D_8}(A)$, then $s=(r^{-a})\cdot (r^a s)$ is also in $C_{D_8}(A)$ because $r^{-a}\in C_{D_8}(A)$ and $C_{D_8}(A)$ is a group.

This is a contradiction. Therefore $C_{D_8}(A)=A$.

(b) The normalizer $N_{D_8}(A)=D_8$

 In general, the centralizer of a subset is contained in the normalizer of the subset. From this fact we have $A=C_{D_8}(A) < N_{D_8}(A)$.
Thus it suffices to show that the other generator $s \in D_8$ belongs to $N_{D_8}(A)$.

We have $sr^as^{-1}=r^{-1}ss^{-1}=r^{-1}\in A$ using the relation $sr=r^{-1}s$.
Thus $s \in N_{D_8}(A)$ as required.

(c) The center $Z(D_8)=\langle r^2 \rangle=\{1,r^2\}$

The center $Z(D_8)$ is contained in the centralizer $C_{D_8}(A)=A$.
Since $rsr^{-1}=sr^2 \neq s$ and $r^3s(r^3)^{-1}=r^{-1}sr=sr^2\neq s$, the elements $r$ and $r^3$ are not in the center $Z(D_8)$.

On the other hand, we have $sr^2=r^{-1}sr=r^{-2}s=r^2s$. Thus $r^2s(r^2)^{-1}=s$ and $r^2 \in Z(D_8)$.
Therefore we have $Z(D_8)=\{1, r^2\}$.

The post Centralizer, Normalizer, and Center of the Dihedral Group $D_{8}$ first appeared on Problems in Mathematics.

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.

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 $\pm 1$.

The post Dihedral Group and Rotation of the Plane first appeared on Problems in Mathematics.