Recall that the centralizer of $H$ in $G$ is
\[C_G(H)=\{g \in G \mid gh=hg \text{ for any } h\in H\}.\]

The normalizer of $H$ in $G$ is
\[N_G(H)=\{g\in G \mid gH=Hg \}.\]

Proof.

(a) Prove $N_G(H)=C_G(H)$

In general, we have $C_G(H) \subset N_G(H)$. We show that $N_G(H) \subset C_G(H)$.

Take any $g \in N_G(H)$. We have $gH=Hg$. Since $|H|=2$, let $H=\{1,h\}$.
Then $gH=\{g,gh\}$ and $Hg=\{g, hg\}$. Since $gH=Hg$, we have $gh=hg$. Namely $g\in C_G(H)$.

This proves that $N_G(H) \subset C_G(H)$, hence $N_G(H)= C_G(H)$.

(b) If $H$ is normal, then $H$ is a subgroup of $Z(G)$

Suppose that $H$ is a normal subgroup of $G$, that is $G=N_G(H)$.
By part (a), this implies that $G=C_G(H)$. Hence $H < Z(G)$.

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

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 […]

Equivalent Definitions of Characteristic Subgroups. Center is Characteristic.
Let $H$ be a subgroup of a group $G$. We call $H$ characteristic in $G$ if for any automorphism $\sigma\in \Aut(G)$ of $G$, we have $\sigma(H)=H$.
(a) Prove that if $\sigma(H) \subset H$ for all $\sigma \in \Aut(G)$, then $H$ is characteristic in $G$.
(b) Prove that the center […]

The Index of the Center of a Non-Abelian $p$-Group is Divisible by $p^2$
Let $p$ be a prime number.
Let $G$ be a non-abelian $p$-group.
Show that the index of the center of $G$ is divisible by $p^2$.
Proof.
Suppose the order of the group $G$ is $p^a$, for some $a \in \Z$.
Let $Z(G)$ be the center of $G$. Since $Z(G)$ is a subgroup of $G$, the order […]

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 Order of a Conjugacy Class Divides the Order of the Group
Let $G$ be a finite group.
The centralizer of an element $a$ of $G$ is defined to be
\[C_G(a)=\{g\in G \mid ga=ag\}.\]
A conjugacy class is a set of the form
\[\Cl(a)=\{bab^{-1} \mid b\in G\}\]
for some $a\in G$.
(a) Prove that the centralizer of an element of $a$ […]

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 […]

The Normalizer of a Proper Subgroup of a Nilpotent Group is Strictly Bigger
Let $G$ be a nilpotent group and let $H$ be a proper subgroup of $G$.
Then prove that $H \subsetneq N_G(H)$, where $N_G(H)$ is the normalizer of $H$ in $G$.
Proof.
Note that we always have $H \subset N_G(H)$.
Hence our goal is to find an element in […]