Let $H$ be commutator subgroup of a free group $G$. This free group $G$ is generated by $a$ and $b$ such that both have infinite order. We know that $H$ is not finitely generated. However, can we infer that $H$ too has at least two generators with infinite order? Although $a$ and $b$ are not in $H$ as $a$ and $b$ cannot be written in the form of commutator(s) but still every element of $H$ is some combination of $a$ and $b$. Consequently, the answer to the above question should be yes.
I was thinking that since $a$ and $b$ have infinite order, therefore $[a,b]$ should have infinite order and it is also in the generating set of $H$. Similarly, I can take $[ab,ba]$ be another element in the generating set that too has infinite order.