A cyclic group G is a group where there exists a ∈ G such that every element of G can expressed as an integral powers of a, that is, if x ∈ G then x=a^{n} for some integer n. In this article, we will study subgroups of a cyclic group.

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## Subgroups of Finite Cyclic Groups

The subgroups of a finite cyclic group have the following properties.

1. Each subgroup of a cyclic group is cyclic.

**Proof:** For a proof, see this page “Every Subgroup of a Cyclic Group is Cyclic“.

2. Let G be a cyclic group of order n. Then for every divisor d of n, there exists only one subgroup of order d.

**Proof:**

**(Existence)** Let G = <a>. Then o(a) = |G| = n. So we can write

G = {e, a, a^{2}, …, a^{n-1}}

where e denotes the identity element of G. If d = 1, n, then as there is only one subgroup {e} (trivial subgroup) of order 1 and only one subgroup G (improper subgroup) of order n, the result holds for d=1, n.

So assume 1 < d < n.

As d divides n, then n=dk for some integer k.

Note that a^{k} ∈ G and its order = $\dfrac{n}{\text{gcd}(k, n)} = \dfrac{n}{k}$ = d. Therefore, the subgroup <a^{k}> has order d.

**(Uniqueness)** Let H = <a^{k}>. Then H has order d.

If K is another subgroup of G of order d. As every subgroup of a cyclic group is cyclic, we have that K = <a^{s}> for some integer s. This integer s must the least such that K = <a^{s}>.

By division algorithm, s = qk+r for some integers q and r with 0 ≤ r < m. Now, a ^{sd} = a^{qkd+rd}⇒ e = a ^{rd} as both <a^{k}> and <a^{s}> have order d. |

Note that rd < kd=n.

This contradicts that the order of a is n. So we conclude that r=0.

Hence, s=qk and so <a^{s}> ⊂ <a^{k}> ⇒ K ⊂ H.

As both H and K have the same order, we conclude that H=K. This proves that H is a unique subgroup of order d.

So if G is a cyclic group of order n and d | n, then there is only one subgroup of order d.

**Corollary:**

A cyclic group of order n has φ(n) distinct generators and so it has φ(n) distinct subgroups, where φ denotes the Euler-phi function.

## Subgroups of Infinite Cyclic Groups

We know that any infinite cyclic group is isomorphic to the group (ℤ, +). As the subgroups of ℤ are mℤ where m is any integer, we conclude that each subgroup of an infinite cyclic group is isomorphic to mℤ for some integer m.

**Related Concepts:** Cyclic Group

Group of Prime Order is Cyclic: Proof

Every Subgroup of a Cyclic Group is Cyclic: Proof

Abelian Group | Quotient Group

Normal Subgroup | Simple Group

## FAQs

**Q1: How many subgroups of Z**

_{15}of order 5 have?Answer: As Z_{15} is a cyclic group of order 15, and 5 divides 15, we conclude that there is only subgroup of Z_{15} of order 5.

**Q2: How many generators of a cyclic group of order n has?**

Answer: There are φ(n) generators of a cyclic group of order n, where φ denotes Euler-phi function.

**Q3: How many generators of an infinite cyclic group has?**

Answer: As an infinite cyclic group is isomorphic to to the group (ℤ, +) which is generated by 1 and -1, so there are only two generators of a cyclic group.