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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 $ | $ ba= ab$ for all $a \in G$}. 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 '14 at 8:08
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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 '14 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 '14 at 21:49
  • $\begingroup$ @Rory It shows that $b$ is in the center if and only if every conjugate of $b$ is $b$. $\endgroup$ – zibadawa timmy Dec 17 '14 at 2:39

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