The centralizer $C_{D_8}(A)$ is a subgroup of $D_8$ whose elements commute with $A$.
That is $C_{D_8}(A)=\{ g\in D_8 \mid gxg^{-1}=x \text{ for all } x\in A\}$.
The normalizer $N_{D_8}(A)$ is a subgroup of $D_8$ defined as
$N_{D_8}(A)=\{ g\in D_8 \mid gxg^{-1} \in A \text{ for any } x \in A\}$.
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 $Z(D_8)=\{1, r^2\}$.
Normalizer and Centralizer of a Subgroup of Order 2
Let $H$ be a subgroup of order $2$. Let $N_G(H)$ be the normalizer of $H$ in $G$ and $C_G(H)$ be the centralizer of $H$ in $G$.
(a) Show that $N_G(H)=C_G(H)$.
(b) If $H$ is a normal subgroup of $G$, then show that $H$ is a subgroup of the center $Z(G)$ of […]
Dihedral Group and Rotation of the Plane
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 […]
All the Conjugacy Classes of the Dihedral Group $D_8$ of Order 8
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 […]
The Center of a p-Group is Not Trivial
Let $G$ be a group of order $|G|=p^n$ for some $n \in \N$.
(Such a group is called a $p$-group.)
Show that the center $Z(G)$ of the group $G$ is not trivial.
Hint.
Use the class equation.
Proof.
If $G=Z(G)$, then the statement is true. So suppose that $G\neq […]
Conjugate of the Centralizer of a Set is the Centralizer of the Conjugate of the Set
Let $X$ be a subset of a group $G$. Let $C_G(X)$ be the centralizer subgroup of $X$ in $G$.
For any $g \in G$, show that $gC_G(X)g^{-1}=C_G(gXg^{-1})$.
Proof.
$(\subset)$ We first show that $gC_G(X)g^{-1} \subset C_G(gXg^{-1})$.
Take any $h\in C_G(X)$. Then for […]
Abelian Normal subgroup, Quotient Group, and Automorphism Group
Let $G$ be a finite group and let $N$ be a normal abelian subgroup of $G$.
Let $\Aut(N)$ be the group of automorphisms of $G$.
Suppose that the orders of groups $G/N$ and $\Aut(N)$ are relatively prime.
Then prove that $N$ is contained in the center of […]
1 Response
[…] 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 […]