Prove Every Subgroup of Cyclic Group is Cyclic

In this page, we prove that every subgroup of cyclic group is Cyclic.

Every subgroup of a cyclic group is cyclic

Proof:

Let G be a cyclic group. By definition, there exists an element g ∈ G such that

G = <g> = {g^k : k∈$\mathbb{Z}$ }.

Let H be a subgroup of G. We need to show that H is cyclic. If H = {e}, then H = <e>, so H is cyclic. So assume that H≠{e}.

Since H ⊆ G, every element of H is of the form g^k for some k ∈$\mathbb{Z}$. Define the set

S = {k ∈ $\mathbb{Z}$ : g^k ∈ H}.

Since H ≠ {e}, the set S contains a nonzero integer. Again, as H is a subgroup we have g^k ∈ H ⇒ g^-k ∈ H. This shows that if k ∈ S, then -k∈S. Hence, we conclude that S must contain positive integers. Let us define the set

S⁺ = {k ∈ S : k > 0}.

By the Well-Ordering Principle, S⁺ has a least element. Let n = min S⁺. Then gⁿ ∈ H.

We claim that H = <gⁿ>.

Take h ∈ H.

Then we can write h = g^m for some m ∈ $\mathbb{Z}$.

By the division algorithm, there exist integers q and r such that m = nq + r, 0≤ r < n. Therefore,

g^m = g^nq+r = g^nq g^r = (gⁿ)^q g^r.

Now since H is a subgroup and gⁿ, g^m ∈ H, so we have g^r = (gⁿ)^-q g^m ∈ H.

But 0 ≤ r < n and g^r ∈ H implies that r=0; otherwise n fails to be the minimal positive integer such that gⁿ ∈ H.

Consequently, m=nq and so h = g^m = g^nq ∈ <gⁿ>.

This proves that H ⊆ <gⁿ>. As H is a subgroup, the converse inclusion is also true. It implies that H = <gⁿ>, a cyclic group. Thus, we have proved that every subgroup of a cyclic group is cyclic.

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