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

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.

• Mar 17 '19 at 19:52
• Why did someone vote to close this as off-topic? It has a very clear attempt! Mar 17 '19 at 19:59
• @Shaun thanks bro :) Mar 17 '19 at 20:03
• You're welcome, @Rivaldo. Does my answer make sense to you? :) Mar 17 '19 at 20:04
• @Shaun think I see it now thanks Mar 17 '19 at 20:27

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 $$g\in Z(G)$$.