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Disclaimer: I am still new to the subject of conjugacy classes, class equations and centralizers. Some guidance to solving this problem would be greatly appreciated on my part. I would like to know how to approach the following question:

Show that a finite group has exactly one conjugacy class if and only if it is trivial.

Thoughts

To some, the proof may be trivial (no pun intended), but to me, it's not. I want to understand what is going on here. I would imagine that if the question were to say "Show that a finite group has exactly $2$ conjugacy classes if anf only if it has order $2$", then we would apply a similar argument. So, if this were the case, then we would see some sort of pattern, but in order to move beyond, I need to understand how to approach my initial problem.

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Let $G$ be a finite group with exactly one conjugacy class.
This means that $1$ must be inside the conjugacy class.
Let $x\in G$. Then $x$ must be conjugate to $1$.
Hence $x=g1g^{-1}$ for some $g\in G$.
Thus $x=1$.
We conclude that $G=1$.

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  • $\begingroup$ Aren't we suppose to show that $G$ is abelian? $\endgroup$ – John Smith May 12 '17 at 4:14
  • $\begingroup$ We need to show that $G$ is trivial, that is $G=\{1\}$ where $1$ is the identity of $G$. @CartesianBread $\endgroup$ – Alan Wang May 12 '17 at 4:16
  • $\begingroup$ Ah yes. Sorry I didn't see your comment as I was typing it up. I will delete it. $\endgroup$ – John Smith May 12 '17 at 4:16
  • $\begingroup$ Since this is an if and only if statement, I would have to show the statement is true $\Leftarrow$ correct? $\endgroup$ – John Smith May 12 '17 at 4:21
  • $\begingroup$ @CartesianBread Yes but the converse is easy. If $G=1$, the conjugacy class of $G$ is 1 since $1$ is conjugate to itself. $\endgroup$ – Alan Wang May 12 '17 at 5:11

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