# Conjugacy class with $1$ element in an Infinite Simple Group

We would like to show that if $G$ is an infinite simple group, then the only conjugacy class of exactly one element is $\{1_G\}$.

My thoughts: We want to proove that if $|\mathrm{orb}(x)|=1\iff \mathrm{orb}(x)=\{x\}$ then $x=1_G$.

We know that $|\mathrm{orb}(x)|=1\iff x\in Z(G)$. But, $Z(G)\trianglelefteq G$ and $G$ is simple, so $Z(G)$ is $G$ or $\{1_G\}$. Also, from the Orbit-Stabilizer Theorem $|G|=|C_G(x)|=\infty$.

But, I'm afraid these don't help as. Any ideas please?

Thank you.

• You're nearly there. If $Z(G) = G$, then $G$ is abelian. And simple. And infinite. – Andreas Caranti May 23 '18 at 14:23
• @Chris: If $G$ is infinite, then it has an infinite number of proper subgroups. If $G$ is abelian, then all of these subgroups are normal which is impossible because $G$ is simple. So, $Z(G)=\{1_G\}$. – Tortoise May 23 '18 at 14:54
• – Moritz May 23 '18 at 19:05
• @Chris: If you follow the provided links above you should do fine. Good luck! – Tortoise May 24 '18 at 7:38
• Slightly more involved exercise: in an infinite simple group, the only finite conjugacy class is $\{1\}$. – YCor May 24 '18 at 10:01