Every Subgroup of a Cyclic Group is Cyclic: Proof

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 aⁿ 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.

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 eⁿ = 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=ak for some non-zero k ∈ ℤ. Now since H is a subgroup, x-1 = a-k ∈ H. So we obtain that ak, 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

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