Is the group $(G,*)$ abelian?

Question

Let $(G,*)$ be a finite group in which the sets $C_a$={$x\epsilon G$|$ax=xa$} have the same cardinality, for all $a \epsilon G$ \{e}. My question is: is the group abelian?

Answer 1

Yes, it is true for finite groups and there are counter examples in infinite groups.

case$1$: Let $G$ be a finite group;

Assume that $G$ is nonabelian.

Let $g\in Z(G)$ and $g\neq1$ then $C_g=G$ then it is true for all element which is a contradiction since we assumed that $G$ is nonabelain. So, We must also assume that $Z(G)$ is a trivial group.

Let's write class equation and fix $k=[G:C_G(a)]$

$$|G|=|Z(G)|+ {nk}$$ for some $n\in \mathbb Z$ and since $|Z(G)|=1 $ we have

$$|G|=1+ nk$$ Then we have, $$|G|\equiv1 \text{ } mod(k)$$ Which is a contradiction since $k$ is a divisior of $|G|$.

Note that:Class equation is $|G|=|Z(G)|+\sum [G:C_a] $

case$2$: For infinite group it is not true.

Let $G=S_3\times \mathbb Z$ then clearly $G$ is not abelian but for every element $a$ in $G$

$C_G(a) $ has countably infinite elements.

Conclusion: We see that it is true for finite groups and found counter example for infinite groups. Even if it is not true for infinite groups, I wonder whether it is true or not for locally finite infinite groups.