Let $G$ be a group. We fix an element $x$ of $G$ and define a map
\[ \Psi_x: G\to G\]
by mapping $g\in G$ to $xgx^{-1} \in G$.
Then prove the followings. (a) The map $\Psi_x$ is a group homomorphism.

(b) The map $\Psi_x=\id$ if and only if $x\in Z(G)$, where $Z(G)$ is the center of the group $G$.

(c) The map $\Psi_y=\id$ for all $y\in G$ if and only if $G$ is an abelian group.

For any elements $g, h\in G$, we have
\begin{align*}
\Psi_x(gh)=x(gh)x^{-1}\stackrel{(*)}{=} xgx^{-1}xhx^{-1}=\Psi_x(g) \Psi_x(h),
\end{align*}
where we inserted the identity element $e=x^{-1}x$ between $g$ and $h$ to obtain (*).
Hence $\Psi_x$ is a group homomorphism.

(b) The map $\Psi_x=\id$ if and only if $x\in Z(G)$

$(\implies)$ Suppose that $\Psi_x=\id$. Then for any $g\in G$, we have
\begin{align*}
\Psi_x(g)=\id(g)
\end{align*}
and thus we have
\begin{align*}
xgx^{-1}=g.
\end{align*}
This implies that we have $xg=gx$ for all $g \in G$, and hence $x\in Z(G)$.

$(\impliedby)$ On the other hand, if $x$ is in the center $Z(G)$, then we have
\[\Psi_x(g)=xgx^{-1}=xx^{-1}g=g\]
for any $g\in G$, where the second equality follows since $x \in Z(G)$.
This yields that $\Psi_x=\id$.

(c) The map $\Psi_y=\id$ for all $y\in G$ if and only if $G$ is an abelian group

$(\implies)$ Suppose that the map $\Psi_y=\id$ for all $y\in G$. Then by part (b), we have $y\in Z(G)$ for all $y\in G$. This means that we have $G=Z(G)$, and hence $G$ is an abelian group.

$(\impliedby)$ Now suppose that $G$ is an abelian group. Then for any $y\in G$ we have
\[\Psi_y(g)=ygy^{-1}=yy^{-1}g=g=\id(g)\]
for any $g\in G$, where the second equality follows since $G$ is an abelian group.
Thus we have $\Psi_y=\id$ for any $y \in G$.

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