I'm struggling to answer this question,
Let $(G,\ast)$ be a group with $n$ elements and $a\in G$ with the following property $ab=ba \iff b\in \{ a^k\mid k\in \mathbb{Z} \}$, if $m$ is the order of $a$ in $G$, show that at least $\frac{n}{m}$ elements of order $m$ exist in $G$.
To me, it initially looks like something related to the centralizer and cyclic group, maybe $Z(x)=\langle x \rangle$? Some research landed me into thinking I might have to use something like if $\frac{G}{Z(G)}$ is cyclic $\implies G$ is abelian?
Any help would be appreciated.
The property in the claim means that $C_G(a)=\langle a\rangle$, where $C_G(a)$ is the centralizer of $a$ in $G$. Now, $C_G(a)=\operatorname{Stab}(a)$ with respect to the action by conjugation of $G$ on itself. Therefore (orbit-stabilizer theorem), all the $\frac{n}{m}$ elements of the orbit of $a$ are conjugate to $a$, and hence they have the same order of $a$ (see e.g. here), namely $m$.