If every proper subgroup of an infinite group $G$ is cyclic, then $G$ is abelian ?
I know there is an infinite abelian with this property(every proper subgroup is cyclic). But I was unable to find any example of a non abelian group with this property. So I try to prove it.
If there exist $a,b \in G$ such that $ab \neq ba$ then we take $<a,b>$(Group generated by a and b).
If $<a,b>$ is a proper subgroup of $G$, then $<a,b> = <c>$ for some $c \in G$.
So $ab=ba$. So $<a,b> = G$.
Again I was not able to complete the proof.
It is clear to me if such non abelian group exist then it must be generated by exactly two elements. So how can I proceed ?
Thanks in advance.