I'm trying to prove that any group $G$ of infinite order has an infinite number of subgroups.
I think that if the group has an element of infinite order, then it's easy because I can take the groups generated by the powers of this element.
What if it doesn't? Every element generates a cyclic subgroup. Every element belongs to at least one cyclic subgroup (that generated by itself). So the group is the union of its cyclic subgroups. If all these are finite, we would have to have an infinite collection of subgroups anyway.
Is that correct?