A result on the number of cyclic subgroups of a finite group

Let $G$ be a finite group and $\alpha (G)=\frac{\mathrm{\mid }C(G)\mathrm{\mid }}{\mathrm{\mid }G\mathrm{\mid }}$,, where $C(G)$ denotes the set of cyclic subgroups of $G$. In this short note, we prove that $\alpha (G)\le \alpha (Z(G))$ and we describe the groups $G$ for which the equality occurs. This gives some sufficient conditions for a finite group to be $4$-abelian or abelian.