Suppose that a group $G$ contains two elements $m$ and $n$ such that $mn=nm$ and $\langle m\rangle\cap\langle n\rangle=\{e\}$. Suppose that a group $G$ contains two elements $m$ and $n$ such that $mn=nm$ and $\langle m\rangle\cap\langle n\rangle=\{e\}$.
Show that if $m^s\cdot n^t = e$, that $m^s=e$ and $n^t = e$.
Does $\langle m\rangle\cap\langle n\rangle=\{e\}$ mean that the two subgroups are relatively prime? To be honest I am having trouble on where to go with this.

Hint Show that $m^s \in \langle m \rangle \cap \langle n \rangle$.
