Let $G$ be a group and $G'$ be the derived subgroup. Let $C=\langle c \rangle $ be a cyclic subgroup of $G$. Is it true that the intersection $H=G'\cap C$ is trivial?
I was thinking that if $g\in H$, then $g\in C$ and thus $g=c^{k}$, for some $k\in \mathbb{Z}$. Therefore, if $f$ is the homomorphism from $G$ to the abelianization $\bar{G}$ of $G$, then $f(g)=c^{k}$ and thus is not trivial. Hence since $\ker f=G'$ we have a contradiction.
Is this idea correct or is there a case where $f(g)$ can be trivial in the abelianization?