Uniqueness of Subgroups in Cyclic Groups Given Any Order's Divisor

In any cyclic group $ G = \langle g \rangle $ of finite order $ n $, for every divisor $ d $ of $ n $, there exists precisely one subgroup of $ G $ with order $ d $, generated by $ g^{n/d} $.

This entails that $ g^n = e $, where $ e $ stands as the neutral element of the group, and $ n $ represents the group's order, $ G $.

The term 'order' of a group generally signifies the count of elements within that group.

In a cyclic group $ G $, it's not possible to have two distinct subgroups of the same order $ d $, as this would imply two different elements in $ G $ sharing the same order $ d $, which contradicts the nature of cyclic groups.

Example

Let's delve into a practical example by examining the cyclic group $ \mathbb{Z}_{12} $ under addition modulo 12.

The Z₁₂ group boasts an order of $ n=12 $, encompassing 12 elements, specifically the integers from 0 through 11.

$$ Z_{12} = \{ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 \} $$

Given $ n=12 $, its divisors – 1, 2, 3, 4, 6, and 12 – indicate the presence of corresponding subgroups for these divisors.

To pinpoint the generator of a subgroup with order $ d $, we compute $ g^{12/d} $ for each divisor of 12:

For $ d = 1 $, we have a subgroup of order 1.

This smallest divisor consistently divides $ n=12 $, with the generating element being $ g^{12/1} = g^{12} = 12 $, aligning with the neutral element $ 0 $ in $ \mathbb{Z}_{12} $.

In this context, $ g^{12} $ is not a multiplicative power but denotes the group operation of addition performed 12 times $$ g^{12} = 1+1+1+1+1+1+1+1+1+1+1+1=12 $$

Accordingly, this subgroup consists solely of the element $ \{0\} $ and possesses an order of 1.

$$ <12> = \{ 0 \} $$

For $ d = 2 $, the subgroup's order is 2.

Here, $ g^{12/2} = g^{6} = 1+1+1+1+1+1 = 6 $ serves as the subgroup's generator.

Thus, the subgroup comprising $ \{0, 6\} $ attains an order of 2.

$$ <6> = \{ 0, 6 \} $$

For $ d = 3 $, we observe a subgroup of order 3.

The subgroup's generator, $ g^{12/3} = g^{4} = 1+1+1+1 = 4 $, results in a cyclic subgroup with elements $ \{0, 4, 8\} $ and an order of 3.

$$ <4> = \{ 0, 4, 8 \} $$

Subgroup of order 4 for $ d = 4 $.

With $ d = 4 $, a divisor of 12, the corresponding subgroup is generated by $ g^{12/4} = g^3 = 1+1+1 = 3 $.

Hence, in $ \mathbb{Z}_{12} $, the subgroup $ \langle 3 \rangle $ comprises $ \{0, 3, 6, 9\} $ and holds an order of 4, as dictated by this property.

$$ <3> = \{ 0, 3, 6, 9 \} $$

For $ d = 6 $, the subgroup commands an order of 6.

The subgroup's generator, $ g^{12/6} = g^{2} = 1+1 = 2 $, ensures the subgroup includes 6 elements $ \{0, 2, 4, 6, 8, 10\} $ and secures an order of 6.

$$ <2> = \{ 0, 2, 4, 6, 8, 10 \} $$

A subgroup of order 12 for $ d = 12 $.

This order mirrors the group's own order, $ n=12 $, with the generator $ g^{12/12} = g^1 = 1 $ embodying the generating element of the entire group.

This discussion of "power" isn't in a multiplicative sense but rather signifies that the repeated addition of 1, the group operation, can engender all the group's elements.

Consequently, the subgroup aligns with the entire group $ \mathbb{Z}_{12} $, or $ \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\} $.

$$ <1> = \{ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 \} $$

In each scenario, the subgroup stands as the exclusive subgroup of its specific order,