Center of Group and Conjugacy Classes I am trying to prove that the center of a group $G$ is the union of the trivial conjugacy classes of $G$.
So far what I have:
We know the center $Z(G)$ of group $G$ is defined by $\{ b \in G\mid ba= ab\,\forall a \in G\}$. We also know if there is an element $ x \in G $ such that $ b = x^{-1}ax$, then $a$ is conjugate to $b$.
Since we know the elements $b$ of $G$ commute with every element of $G$ in the center, we may say every element in $Z(G)$ has a conjugate. Therefore, the center is the union of all these conjugacy classes.
Is this correct, or where did I run off course?
 A: Hint 1: If $b\in Z(G)$ and $x\in G$, what is $xbx^{-1}$?
Hint 2: In the definition of $Z(G)$, write the condition "$ba=ab$" in a way that makes one side a conjugate of $b$.
Otherwise, what you say isn't meaningful.  Every element of a group has a conjugate (namely itself), regardless of whether it is central or not.  A subgroup being a union of conjugacy classes is a generally non-trivial thing (though here it is relatively simple). They're called normal subgroups, and not all subgroups are normal in general.
A: Since $\operatorname{cl}(b):=\{c\in G\mid c=aba^{-1}, a\in G\}$, then:
\begin{alignat}{1}
Z(G) &:= \{b\in G\mid ba=ab, \forall a\in G\} \\
&= \{b\in G\mid b=aba^{-1}, \forall a\in G\} \\
&= \{b\in G\mid c\in\operatorname{cl}(b)\Longrightarrow c=b\} \\
&= \{b\in G\mid c\in\operatorname{cl}(b)\Longrightarrow c\in\{b\}\} \\
&= \{b\in G\mid \operatorname{cl}(b)\subseteq\{b\}\} \\
&= \{b\in G\mid \operatorname{cl}(b)=\{b\}\} \\
&= \bigcup_{\{b\}=\operatorname{cl}(b)}\{b\} \\
&= \bigcup_{\left|\operatorname{cl}(b)\right|=1}\operatorname{cl}(b) \\
\end{alignat}
A: As @zibadawa timmy pointed out, this is equivalent to saying that $b$ is in the center if and only if every conjugate of $b$ is $b$.
In the forward direction, for all $a \in G$, there is $x \in G$ such that:
$$
\begin{equation}
\begin{split}
a \sim b & \Rightarrow ax = xb \\
 & \Rightarrow  ax = bx &\ \text{(because $b \in Z(G))$} \\
 & \Rightarrow a = b
\end{split}
\end{equation}
$$
In the backward direction, consider the conjugacy class of $b$. It contains all elements of the form $xbx^{-1}$. By the premise, the conjugacy class of $b$ contains only itself: $[b] = \{b\}$. So, for any $x \in G$, $b = xbx^{-1}$ and, therefore, $bx = xb$. In other words, $b$ commutes with all elements of $G$ and, as such, $b \in Z(G)$.
