Determine the number of subgroups of a cyclic group

Question

Can we actually tell the number of subgroups of a particular cyclic group without actually listing them?

Answer 1

I assume you are talking of a finite cyclic group, of order $n$, say. So it's the number of divisors of $n$. If you are given the factorization $$ n = p_1^{e_1} p_2^{e_2} \dots p_{k}^{e_k}, $$ where the $p_i$ are distinct primes, and the $e_i > 0$, then the number is $$ (e_1 + 1) (e_2 + 1) \dots (e_k + 1). $$

Answer 2

If $G=\langle g \rangle $ is a finite cyclic group of order $n$, then any subgroup of $G$ has the form $S=\langle g^d\rangle$ for a divisor $d\mid n$. Different values of $d$ give different sizes so there is just one subgroup of G having a given size. Hence there are $\tau(n)$ different subgroups.
Each non-trivial subgroup of an infinite cyclic group has two generators, which are inverses of each other. Fixing one generator $g$ of the whole group, we can write each subgroup in the form $⟨g^n⟩$ for a unique $n ≥ 0$. So the number of subgroups is countably infinite.