Generating Subgroups from a Set of Elements
A subgroup <X> generated by a set X⊆G of elements within a group (G,*) includes all elements that can be derived by repeatedly applying the group's * operation on the elements x∈X and their inverses.
When the set X consists of a single element X={x}, it is referred to as a cyclic subgroup.
This is the smallest subgroup that encompasses the elements of set X and all their possible combinations through the group's operations.
A Practical Example
Consider the additive group (Z,+) of integers with respect to addition
From this, we take the subset of elements X={2,3}.
The subgroup generated by X encompasses all linear combinations of the elements with respect to addition.
$$ <X> = \{ 2a+3b \ , \ a,b \in Z \} $$
Where "a" and "b" are coefficients, meaning any two integers that multiply the elements 2 and 3.
For instance
$$ 2 \cdot 1+3 \cdot 1=5 $$
$$ 2 \cdot 2+3 \cdot 1=4+3=7 $$
$$ 2 \cdot 1+3 \cdot 2=2+6=8 $$
$$ 2 \cdot 2+ 3 \cdot 2=4+6=10 $$
$$ -2+3=1 $$
$$ \vdots $$
In this scenario, the set X generates the entire group (Z,+) because the operation $ -2+3=1 $ allows us to multiply this product by a suitable coefficient to obtain any integer.
$$ <X> = (Z,+) $$
Note. It's interesting to observe that every subgroup always includes the group's identity element. For example, in the additive group (Z,+), the neutral element is zero. To achieve it, simply use two null coefficients a=0, b=0. $$0 \cdot 2 + 0 \cdot 3 = 0 $$
Example 2
Consider the additive group (Z,+) with respect to addition.
This time, let's look at a set X composed of the numbers 2 and 6.
$$ X= \{2,6 \} $$
Knowing that the sum of two even numbers can never yield an odd sum, it's clear that this subgroup can only generate even numbers.
$$ 2+(-2)=0 $$
$$ 2+6 \cdot 0 = 2 $$
$$ 2 \cdot 2+6 \cdot 0 = 4 $$
$$ 0 \cdot 2 + 1 \cdot 6 = 6 $$
$$ 1 \cdot 2 + 1 \cdot 6 = 8 $$
$$ 2 \cdot 2 + 1 \cdot 6 = 10 $$
Therefore, the subgroup generated by the set X={2,6} consists only of even numbers with respect to addition.
$$ <X>= (2Z,+) $$
Example 3
Consider the additive group $(\mathbb{Z}_6, +)$ of integers modulo 6. Here, addition serves as our operation.
$$ \mathbb{Z}_6 = \{0, 1, 2, 3, 4, 5\} $$
For this exercise, let's dive into the subset $X=\{2, 4\}$.
$$ X = \{2, 4\} $$
In the $(\mathbb{Z}_6, +)$ group, the subgroup generated by $X = \{2, 4\}$ captures all the possible sums of 2 and 4, considering modulo 6.
Let's break it down:
$$ 2 + 2 = 4 \mod 6 $$
$$ 2 + 4 = 6 \mod 6 = 0 $$
$$ 4 + 4 = 8 \mod 6 = 2 $$
$$ 2 + 2 + 2 = 6 \mod 6 = 0 $$
$$ 4 + 4 + 4 = 12 \mod 6 = 0 $$
$$ 2 + 4 + 4 = 10 \mod 6 = 4 $$
$$ \vdots $$
This demonstrates that any combination of 2 and 4 ultimately boils down to 0, 2, or 4 when viewed modulo 6.
Thus, the subgroup generated by $X$ is:
$$ \langle X \rangle = \{0, 2, 4\} $$
This subgroup not only includes the identity element 0, making it a true subgroup, but also ensures that each element within has its inverse in the group. This is clear as the inverse of 2 is 4 and vice versa (given that \(2 + 4 = 6 \mod 6 = 0\)), establishing that $\langle X \rangle$ is indeed closed under addition modulo 6.
And that's the gist of it.
