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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.

Chris
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    You're nearly there. If $Z(G) = G$, then $G$ is abelian. And simple. And infinite. – Andreas Caranti May 23 '18 at 14:23
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    @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
  • @Tortoise Thank you for your comment. There another way to prove it avoiding the proper subgroups or possibly the result with the center? – Chris May 23 '18 at 15:06
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    @Chris: Check out https://math.stackexchange.com/questions/322713/if-a-group-g-has-only-finitely-many-subgroups-then-show-that-g-is-a-finite or https://math.stackexchange.com/questions/1132946/show-that-a-group-that-has-only-a-finite-number-of-subgroups-must-be-a-finite-gr – Moritz May 23 '18 at 19:05
  • @Moritz Thank you for the useful link. – Chris May 23 '18 at 20:27
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    @Chris: If you follow the provided links above you should do fine. Good luck! – Tortoise May 24 '18 at 07:38
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    Slightly more involved exercise: in an infinite simple group, the only finite conjugacy class is ${1}$. – YCor May 24 '18 at 10:01

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