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.

The above expression implies that $n^{t}=(m^s)^{-1}\in\langle m\rangle$. Thus $n^t\in\langle m\rangle\cap\langle n\rangle$. Hence $n^t=e$, and hence also $m^s=e$.
