4
$\begingroup$

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.

$\endgroup$
  • 5
    $\begingroup$ You're nearly there. If $Z(G) = G$, then $G$ is abelian. And simple. And infinite. $\endgroup$ – Andreas Caranti May 23 '18 at 14:23
  • 1
    $\begingroup$ @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, $Z(G)=\{1_G\}$. $\endgroup$ – Tortoise May 23 '18 at 14:54
  • 1
    $\begingroup$ @Chris: Check out math.stackexchange.com/questions/322713/… or math.stackexchange.com/questions/1132946/… $\endgroup$ – Moritz May 23 '18 at 19:05
  • 1
    $\begingroup$ @Chris: If you follow the provided links above you should do fine. Good luck! $\endgroup$ – Tortoise May 24 '18 at 7:38
  • 1
    $\begingroup$ Slightly more involved exercise: in an infinite simple group, the only finite conjugacy class is $\{1\}$. $\endgroup$ – YCor May 24 '18 at 10:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.