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.
Table of Contents
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, …, an-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 ak ∈ G and its order = $\dfrac{n}{\text{gcd}(k, n)} = \dfrac{n}{k}$ = d. Therefore, the subgroup <ak> has order d.
(Uniqueness) Let H = <ak>. 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 = <as> for some integer s. This integer s must the least such that K = <as>.
| 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 <as> ⊂ <ak> ⇒ 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
Answer: As Z15 is a cyclic group of order 15, and 5 divides 15, we conclude that there is only subgroup of Z15 of order 5.
Answer: There are φ(n) generators of a cyclic group of order n, where φ denotes Euler-phi function.
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.
