Subgroups within Cyclic Groups Are Themselves Cyclic

A principle holds true within the realm of cyclic groups: each of their subgroups is also cyclic.

To bring this concept to life, let’s use the integers modulo $ n $ as an example. These integers constitute a cyclic group when combined with the operation of addition modulo $ n $.

Take, for instance, the cyclic group $ \mathbb{Z}_{12} $, or the integers modulo 12. The number 1 serves as its generator because adding 1 modulo 12 repeatedly brings forth every other group element.

The makeup of $ \mathbb{Z}_{12} $ is $ \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\} $.

Let’s examine the subgroup $ H $ generated by the element 4 in $ \mathbb{Z}_{12} $. To identify all the elements of $ H $, we calculate:

$ 4 \cdot 1 = 4 \mod 12 $
$ 4 \cdot 2 = 8 \mod 12 $
$ 4 \cdot 3 = 12 \mod 12 = 0 $ (circling back to the group's identity element)
$ 4 \cdot 4 = 16 \mod 12 = 4 $ (where the elements start repeating)

This reveals $ H $'s composition to be $ \{0, 4, 8\} $.

It’s evident that $ H $ maintains a cyclic structure, being generated by the element 4.

Each element of $ H $ is effectively a multiple of 4, modulo 12.

Furthermore, 4 emerges as the smallest positive integer capable of generating all elements of $ H $ through its power (in this context, via multiplication modulo 12).

Another Example

What happens when we explore the element 5 in $ \mathbb{Z}_{12} $ for its subgroup generating potential? Let’s delve into whether a subgroup formed by 5 would be cyclic.

Calculating the elements generated by 5 in $ \mathbb{Z}_{12} $ gives us:

$ 5 \cdot 1 = 5 \mod 12 $
$ 5 \cdot 2 = 10 \mod 12 $
$ 5 \cdot 3 = 15 \mod 12 = 3 $
$ 5 \cdot 4 = 20 \mod 12 = 8 $
$ 5 \cdot 5 = 25 \mod 12 = 1 $ (looping back to the identity and completing the cycle)
$ 5 \cdot 6 = 30 \mod 12 = 6 $
$ 5 \cdot 7 = 35 \mod 12 = 11 $
$ 5 \cdot 8 = 40 \mod 12 = 4 $
$ 5 \cdot 9 = 45 \mod 12 = 9 $
$ 5 \cdot 10 = 50 \mod 12 = 2 $
$ 5 \cdot 11 = 55 \mod 12 = 7 $
$ 5 \cdot 12 = 60 \mod 12 = 0 $

This process illustrates that through the powers of 5, every element of $ \mathbb{Z}_{12} $ is generated.

In this scenario, the subgroup created by 5 indeed encompasses the entire group $ \mathbb{Z}_{12} $, and given $ \mathbb{Z}_{12} $'s cyclic nature, the subgroup (encompassing the entire group here) is categorically cyclic.

The Underlying Principle

This theorem posits a straightforward notion: every subgroup nestled within a cyclic group retains a cyclic structure.

The proof unfolds through the following logical steps:

Identify a subgroup $ S $ within the cyclic group $ G $, brought into existence by an element $ g $, meaning $ G = \langle g \rangle $.
If $ S $ is composed solely of the neutral element $ e $, it naturally qualifies as cyclic.
In cases where $ S $ encompasses an element $ g^t $ beyond $ e $, it invariably contains powers of $ g $, expressed as natural numbers $ h $.
Take, for example, the element g^t=4²=8 with t=2 within the group $ \mathbb{Z}_{12}, + $, where G spans the finite set of integers from 0 to 11, that is, $ \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\} $. This element gives rise to the subgroup $ S = \{ 0,4,8 \} $
Within this subset of elements, the integer $ m $ represents the smallest positive figure such that $ g^m $ falls within $ S $, leading to the hypothesis that $ S $ is propelled by $ g^m $, symbolized by $ S = \langle g^m \rangle $.
For instance, the element g^m=4¹=4 with m=1 also fits snugly within the subgroup $ S = \{ 0, 4, 8 \} $.
It stands to reason that every power of $ g^m $ finds a place in $ S $ (given that $ g^m $ is part of $ S $ and the powers of any element within $ S $ remain confined to $ S $).
To assert that each element of $ S $ manifests as a power of $ g^m $, we consider a generic element $ g^k $ from $ S $, dividing $ k $ by $ m $ to yield a quotient $ q $ and a remainder $ r $, with $ 0 \leq r < m $.
For instance, the element g^k=4²=8 with k=2 still falls within the subgroup $ S = \{ 0, 4, 8 \} $. Dividing k=2 by m=1 produces a quotient q=2 and a remainder r=0.
We then demonstrate that $ g^k $ can be articulated as $ g^{mq+r} $, thereby ensuring $ g^r $ is part of $ S $. However, since $ m $ represents the minimal integer whose power aligns with $ S $, the remainder $ r $ must zero out to dodge any logical inconsistency.
This concludes that $ k $ is an integer multiple of $ m $, allowing $ g^k $ to be depicted as $ (g^m)^q $, a distinct power of $ g^m $.

In essence, every component of $ S $ embodies a power of $ g^m $, underscoring that $ S $ is spawned by $ g^m $ and is, by all accounts, cyclic.