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.
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 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 am ∈ H. The existence of such an integer by the well ordering property of natural numbers.
We claim that H = <am>.
Let h ∈ H.
Then h = as 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 am ∈ H ⇒ a-qm ∈ H, because H is a subgroup.
Also, as ∈ H.
Combining, we deduce that as-qm ∈ H
⇒ ar ∈ H as s = qm+r.
Now since 0 ≤ r < m, this contradicts that m is the least positive integer such that am ∈ H. So r must be 0. This implies that
as = aqm = <am>.
As x is an arbitrary element, and it is expressed as an integral power of am, we conclude that H is generated by am, that is, H = <am>. 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
Answer: Yes, every subgroup of a cyclic group is cyclic.
