Q:

Centraliser of a cyclic subgroup

Suppose $a$ belongs to a group $G$ and order of $a$ is $5$. Prove that $C(a)=C(a^3)$, where $C(a)$ is the centralizer of $a$ in $G$.  
We can show that $C(a)=C(a^4)$ as we always have the identity $C(a)=C(a^{-1})$, but what about what is asked above?   

A:

For the counterexample when the order of $a$ is $6$, consider $a= (123)(45) \in S_5$.
Then $a^3 = (45)$, which commutes with any permutation that doesn't touch $4$ and $5$, e.g. $(12)$. However, $(12)$ does not commute with $a$.
Concerning the question, note that if $x\in C(a^3)$, then $xa = x(a^3)^2 = (a^3)x(a^3) = (a^3)^2 x = ax$, hence $x\in C(a)$. The other inclusion is trivial.

