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?

  • $\begingroup$ trivial conjugacy classes of G yes $\endgroup$ – Rory Dec 16 '14 at 8:08

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.