Subgroup of any group always exists. If the group is cyclic, then its subgroup is also cyclic. In this article, we prove that each subgroup of a cyclic group is also cyclic.

Let us first understand what are subgroups and cyclic groups.

**What are groups and subgroups?**

A set G forms a group with respect to the binary operation ∗, denoted by (G, ∗), if it satisfies the following properties.

- ∗ is closed, that is, a∗b ∈ G ∀ a, b ∈ G.
- ∗ is associative on G, i.e., (a∗b)∗c = a∗(b∗c) ∀ a, b, c ∈ G.
- The identity element exists, that is, there exists an element e in G such that a∗e = e∗a = a ∀ a ∈ G.
- For any a ∈ G, ∃ a
^{-1}∈ G such that a∗a^{-1}= a^{-1}∗a = e.

If a subset H of G forms a group under the operation ∗, then the pair (H, ∗) is called a subgroup of (G, ∗).

**Cyclic Groups:**

A group G is called a cyclic group, if all of its elements can be expressed as a^{n} for some fixed a ∈ G and integer n. The element a is called a generator of the group and it is written as follows: G = <a>.

Note that the order of a in G is equal to the cardinality of the group G.

Table of Contents

## Prove that every subgroup of a cyclic group is cyclic

**Proof:**

Let G be a cyclic group generated by a, that is, G = <a>. Let H be a subgroup of G. We need to prove that H is a cyclic group.

If H=G, then H is cyclic.

So assume that H ≠ G.

**Case 1:** Let H = {e} be trivial group where e is the identity element. As e^{n }= e, for all integers n, the group H is cyclic.

**Case 2:** Let H ≠ {e} be a proper subgroup of G.

Consider an element x in H other than e.

As x ∈ G, we have x=a^{k} for some non-zero k ∈ ℤ. Now since H is a subgroup, x^{-1} = a^{-k} ∈ H. So we obtain thata ^{k}, a^{-k} ∈ H for some integer k ≠ 0. |

Thus H contains some positive integral powers of a. Let m be the least positive integer such that a^{m} ∈ H. The existence of such an integer by the well ordering property of natural numbers.

We claim that H = <a^{m}>.

Let h ∈ H.

Then h = a^{s} for some integer s.

By the division algorithm of integers, there exist integers q and r such that s = qm+r with 0 ≤ r < m.

Note a^{m} ∈ H ⇒ a^{-qm} ∈ H, because H is a subgroup.

Also, a^{s} ∈ H.

Combining, we deduce that a^{s-qm} ∈ H

⇒ a^{r} ∈ H as s = qm+r.

Now since 0 ≤ r < m, this contradicts that m is the least positive integer such that a^{m} ∈ H. So r must be 0. This implies that

a^{s} = a^{qm} = <a^{m}>.

As x is an arbitrary element, and it is expressed as an integral power of a^{m}, we conclude that H is generated by a^{m}, that is, H = <a^{m}>. This proves that H is cyclic.

So each subgroup of a cyclic group is also cyclic.

**Related Concept:** Cyclic Group

Group of Prime Order is Cyclic: Proof

Abelian Group | Quotient Group

Normal Subgroup | Simple Group

## FAQs

**Each subgroup of a cyclic group is cyclic, true or false?**

Answer: Yes, every subgroup of a cyclic group is cyclic.

This article is written by Dr. T. Mandal, Ph.D in Mathematics. On Mathstoon.com you will find Maths from very basic level to advanced level. Thanks for visiting.