$(\subset)$ We first show that $gC_G(X)g^{-1} \subset C_G(gXg^{-1})$.
Take any $h\in C_G(X)$. Then for any element $x \in X$, we have
\begin{align*}
ghg^{-1}(gxg^{-1}) &=ghxg^{-1}\\
&=gxhg^{-1} \,\,\,\, \text{ since } h\in C_G(X)\\
&=gxg^{-1}(ghg^{-1}).
\end{align*}

This computation shows that $ghg^{-1} \in C_G(gXg^{-1})$ for any $h \in C_G(X)$.
Thus this proves $g C_G(X) g^{-1} \subset C_G(gXg^{-1})$.

$(\supset)$ Conversely, take any element $k \in C_G(gXg^{-1})$.
Then for any $x \in X$, we have
\begin{align*}
g^{-1}kg(x)&=g^{-1}k(gxg^{-1})g\\
&=g^{-1}(gxg^{-1})kg \,\,\,\, \text{ since } k\in C_G(gXg^{-1})\\
&=x(g^{-1}kg).
\end{align*}
This implies that $g^{-1}kg \in C_G(X)$, hence $k\in gC_G(X)g^{-1}$ for any $k \in C_G(gXg^{-1})$. This proves $C_G(gXg^{-1}) \subset gC_G(X)g^{-1}$.

Two Normal Subgroups Intersecting Trivially Commute Each Other
Let $G$ be a group. Assume that $H$ and $K$ are both normal subgroups of $G$ and $H \cap K=1$. Then for any elements $h \in H$ and $k\in K$, show that $hk=kh$.
Proof.
It suffices to show that $h^{-1}k^{-1}hk \in H \cap K$.
In fact, if this it true then we have […]

If a Sylow Subgroup is Normal in a Normal Subgroup, it is a Normal Subgroup
Let $G$ be a finite group. Suppose that $p$ is a prime number that divides the order of $G$.
Let $N$ be a normal subgroup of $G$ and let $P$ be a $p$-Sylow subgroup of $G$.
Show that if $P$ is normal in $N$, then $P$ is a normal subgroup of $G$.
Hint.
It follows from […]

Abelian Normal Subgroup, Intersection, and Product of Groups
Let $G$ be a group and let $A$ be an abelian subgroup of $G$ with $A \triangleleft G$.
(That is, $A$ is a normal subgroup of $G$.)
If $B$ is any subgroup of $G$, then show that
\[A \cap B \triangleleft AB.\]
Proof.
First of all, since $A \triangleleft G$, the […]

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 […]

If a Subgroup Contains a Sylow Subgroup, then the Normalizer is the Subgroup itself
Let $G$ be a finite group and $P$ be a nontrivial Sylow subgroup of $G$.
Let $H$ be a subgroup of $G$ containing the normalizer $N_G(P)$ of $P$ in $G$.
Then show that $N_G(H)=H$.
Hint.
Use the conjugate part of the Sylow theorem.
See the second statement of 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 […]

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 […]

If a Group is of Odd Order, then Any Nonidentity Element is Not Conjugate to its Inverse
Let $G$ be a finite group of odd order. Assume that $x \in G$ is not the identity element.
Show that $x$ is not conjugate to $x^{-1}$.
Proof.
Assume the contrary, that is, assume that there exists $g \in G$ such that $gx^{-1}g^{-1}=x$.
Then we have
\[xg=gx^{-1}. […]