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I know that if $z\in Z(G)$, the centre of group $G$ then it is true that $cl(z)=\{z\}$ where $cl(g)$ is the conjugacy class that contains element $g\in G$.

But what if $cl(g)=\{g\}$ can we imply that $g\in Z(G)?$

My attempt: $$cl(g)=\{h\in G|\exists k\in G\text{ such that }h=k^{-1}gk\}=\{g\}$$ so there exists a $k$ in $G$ such that $g=k^{-1}gk$, but might not be the case for all $k$ in $G$ hence the above statement is false.

Thanks.

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    $\begingroup$ Here's a MathJax tutorial :) $\endgroup$
    – Shaun
    Mar 17 '19 at 19:52
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    $\begingroup$ Why did someone vote to close this as off-topic? It has a very clear attempt! $\endgroup$
    – Shaun
    Mar 17 '19 at 19:59
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    $\begingroup$ @Shaun thanks bro :) $\endgroup$
    – Rivaldo
    Mar 17 '19 at 20:03
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    $\begingroup$ You're welcome, @Rivaldo. Does my answer make sense to you? :) $\endgroup$
    – Shaun
    Mar 17 '19 at 20:04
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    $\begingroup$ @Shaun think I see it now thanks $\endgroup$
    – Rivaldo
    Mar 17 '19 at 20:27
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We have $\operatorname{cl}(g)=\{g\}$, so if $h\in G$, then $h^{-1}gh\in \operatorname{cl}(g)$ implies $h^{-1}gh=g$, i.e., $$gh=hg.$$ But $h$ was arbitrary in $G$. Hence $g\in Z(G)$.

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