Subgroups of Cyclic Groups

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=aⁿ for some integer n. In this article, we will study subgroups of a cyclic group.

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, a², …, 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, asd = aqkd+rd ⇒ e = ard as both <ak> and <as> 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

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