Suppose 𝐻 ⊆⟨𝑎⟩ is a subgroup. Clearly the trivial group ⟨𝑒⟩ is cyclic, so assume that 𝐻 contains at least one non-identity element. Since 𝐻 ⊆⟨𝑎⟩ all elements of 𝐻 are some power of 𝑎, and 𝐻 cannot contain only negative powers by closure.; Let 𝑚 ∈ℕ be the smallest positive integer such that 𝑎𝑚 ∈𝐻. By closure ⟨𝑎𝑚⟩ ⊆𝐻. We will prove ⟨𝑎𝑚⟩ =𝐻. Let some 𝑎𝑘 ∈𝐻. Then there exist 𝑞 ∈ℤ and 0 ≤𝑟 <𝑚 such that 𝑘 =𝑞𝑚 +𝑟 and thus 𝑎𝑘 =(𝑎𝑚)𝑞𝑎𝑟, whence 𝑎𝑟 =𝑎𝑘(𝑎𝑚)−𝑞. Since 𝑎𝑘,𝑎𝑚 ∈𝐻 by closure 𝑎𝑟 ∈𝐻 But 0 ≤𝑟 <𝑚 and 𝑚 is the least positive integer such that 𝑎𝑚 ∈𝐻. Therefore 𝑟 =0 and 𝑎𝑘 =(𝑎𝑚)𝑞 ∈⟨𝑎𝑚⟩. Hence 𝐻 ⊆⟨𝑎𝑚⟩ implying 𝐻 =⟨𝑎𝑚⟩.

Let |⟨𝑎⟩| =𝑛 and 𝐻 ⊆⟨𝑎⟩. From above, 𝐻 =⟨𝑎𝑚⟩ where 𝑚 is the smallest positive integer such that 𝑎𝑚 ∈𝐻. Letting 𝑘 =𝑛 as above, it follows 𝑛 =𝑚𝑞.

Let 𝑘 be some positive divisor of 𝑛. From the theorem on Order of powers of a group element, it follows that ∣⟨𝑎𝑘⟩∣ =𝑛/gcd(𝑛,𝑛𝑘) =𝑘. This proves existence, now we must show uniqueness. Let 𝐻 ⊆⟨𝑎⟩ be an arbitrary subgroup of order 𝑘. From above 𝐻 =⟨𝑎𝑚⟩ where 𝑚 divides 𝑛. Then 𝑚 =gcd(𝑛,𝑚), whence 𝑘 =|𝑎𝑚| =∣𝑎gcd(𝑛,𝑚)∣ =𝑛/gcd𝑛,𝑚 =𝑛/𝑚. Therefore 𝑚 =𝑛/𝑘 and 𝐻 =⟨𝑎𝑛/𝑘⟩.