Centralizer, Normalizer, and Center of the Dihedral Group $D_{8}$

Group Theory Problems and Solutions in Mathematics

Problem 53

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 $r^2$.

LoadingAdd to solve later

Sponsored Links


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 $Z(D_8)=\{1, r^2\}$.


LoadingAdd to solve later

Sponsored Links

More from my site

You may also like...

1 Response

  1. 06/20/2017

    […] 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 […]

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
Group Theory Problems and Solutions
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...

Close