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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?

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  • $\begingroup$ trivial conjugacy classes of G yes $\endgroup$
    – Rory
    Dec 16, 2014 at 8:08
  • $\begingroup$ This is an old question, but I decided to post an answer because I struggled with this proof a lot myself. I was only focusing on the fact that, if two elements are conjugate of one another, they are in the same conjugacy class. But I was missing the opposite direction: a conjugacy class contains of an element contains all its conjugates (obvious in hindsight). $\endgroup$ Jan 9, 2021 at 19:03

3 Answers 3

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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.

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  • $\begingroup$ Okay so with Hint 1, that gives me b is conjugate to b. Hint 2 gives me ba = ab -> $a^{-1}ba = b$ -> b is conjugate to b, but how does this help me say that Z(G) is the union of these conjugates. I am confused why an element being conjugate to itself helps me $\endgroup$
    – Rory
    Dec 16, 2014 at 21:44
  • $\begingroup$ edit: after changing the definition of Z(G) to show that its elements are conjugate to themselves by rearranging ba=ab it makes sense to say it is then trivial that the union of all of them comprise the group's center, is this correct thinking? $\endgroup$
    – Rory
    Dec 16, 2014 at 21:49
  • $\begingroup$ @Rory It shows that $b$ is in the center if and only if every conjugate of $b$ is $b$. $\endgroup$ Dec 17, 2014 at 2:39
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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}

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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, $b \in Z(G)$.

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