1. The problem statement, all variables and given/known data Show that Z has infinitely many subgroups isomorphic to Z. 2. Relevant equations Two groups are isomorphic iff there exists a bijective map between them. 3. The attempt at a solution I couldn't solve this before, I have a feeling that I have a stronger argument now so I'm hoping someone can help me. So the first thing we have to prove is that a group of infinite order has infinitely many subgroups. Abusing some logic, this is equivalent to proving a group of finite order has finitely many subgroups. So suppose that G is a finite group, we want to show G has finitely many subgroups. So there are two cases to consider. Case : |g| = ∞ for some g in G. If the order of g is ∞, then the cyclic subgroup generated by g, <g>, is isomorphic to Z. That is, <g> ≈ Z which has infinitely many subgroups. Thus G must have infinitely many subgroups contradicting the assumption. Case : |g| = n for all g in G. In this case, every element of g has finite order n. Since every group is a union of cyclic subgroups, G is also a union of the cyclic subgroups generated by its elements; each of which is finite. In simple terms, since G has only a finite amount of subgroups, G must be a union of these finitely many subgroups and thus G must be finite. Thus G is a finite group which has finitely many subgroups. Re-abusing our logic, we get that an infinite group has infinitely many subgroups. Hence Z must have infinitely many subgroups since |Z| = ∞. I thought about that one for awhile while I was studying, I hope it makes sense.