# 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?

• trivial conjugacy classes of G yes
– Rory
Dec 16, 2014 at 8:08
• 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). Jan 9, 2021 at 19:03

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.

• 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
– Rory
Dec 16, 2014 at 21:44
• 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?
– Rory
Dec 16, 2014 at 21:49
• @Rory It shows that $b$ is in the center if and only if every conjugate of $b$ is $b$. Dec 17, 2014 at 2:39

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}

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:

$$$$\begin{split} a \sim b & \Rightarrow ax = xb \\ & \Rightarrow ax = bx &\ \text{(because b \in Z(G))} \\ & \Rightarrow a = b \end{split}$$$$

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, $$b \in Z(G)$$.