Generating a Subgroup from a Set

Imagine you have a group \( (G,*) \) alongside a non-empty subset \( X \subseteq G \). The subgroup generated by \( X \), represented as \( \langle X \rangle \), encompasses all elements that can be produced through any combination of the elements \( x \in X \) using the group's * operation.

The subgroup formed by the set X is a subset within the group (G,*), encompassing all elements derived from any sequence of multiplications among the elements of X and their inverses.

$$ < X > = \{ u_1 * u_2 * ... * u_m \  |  \ u_i \in X \ or \ u_i^{-1} \in X, \ i,m \in \mathbb{N}  \} $$

This operation isn’t just regular multiplication, but the * operation as defined in the group G.

Let's take a group \( (G, *) \) and a set \( X = \{a, b\} \) comprising two elements of \( G \) as an example. The subgroup generated by \( X \), symbolized as \( \langle X \rangle \), encompasses all conceivable finite combinations of \( a \), \( b \), and their inverses \( a^{-1} \), \( b^{-1} \) utilizing the group's operation \( * \). $$  \langle X \rangle = \{ ab, ba, ab^{-1}, aba^{-1}, aba^{-1}b^{-1} ... \}$$ It's essential to remember that applying the operation \( * \) between an element and its inverse (for example, \( a*a^{-1} \) or \( b*b^{-1} \)), results in mutual cancellation because the outcome is the group's identity element \( e \). $$ aa^{-1} = a^{-1}a = bb^{-1} = b^{-1}b = e $$ To illustrate, the sequence $ abb^{-1}a $ simplifies to $ aa $ since the terms $ bb^{-1} $ cancel out.

The identity element of the group \( G \) is always included in any subgroup generated by a set $ X $ since elevating any element $ x \in X $ to the power of zero invariably yields the group's identity element $ e $.

$$ \forall \ x \in X \ , \ x^0 = e $$

The makeup of the subgroup generated from the set hinges on the specific * operation employed by the group.

For example, if \( G \) operates under addition, then \( \langle X \rangle \) will contain all possible finite sums of \( X \)'s elements and their negatives. Conversely, if \( G \) is based on multiplication, then \( \langle X \rangle \) will comprise all finite products of \( X \)'s elements and their reciprocals.

The generated subgroup \( \langle X \rangle \) stands as the most minimal subgroup of \( G \) that encapsulates every element of \( X \). It's so minimal that there's no other subgroup of \( G \) that includes \( X \) and is smaller than \( \langle X \rangle \). This phenomenon occurs because the generated subgroup \( \langle X \rangle \) is essentially the intersection of all subgroups S within $ G $ that encompass $ X $.

$$ < X >  =    \underset{X \subseteq S \le G}{ \cap } S   $$

If the set $ X $ is made up of just one element $ x \in X $, then we refer to the subgroup generated by $ x $ as a cyclic subgroup.

Illustrating with an Example

Consider the integers under addition, forming an additive group known as \((\mathbb{Z}, +)\).

Let's select the subset \( X = \{6\} \) from \(\mathbb{Z}\).

To uncover the subgroup generated by \( X \), it's necessary to explore all conceivable combinations of elements within \( X \) and their additive inverses.

With a singular element, this entails examining all powers of the element, be they positive, zero, or negative.

$ \vdots \\ x^{-2} = (-x)+(-x) \\ x^{-1} = -x \\ x^0 = e \\ x^1 = x \\ x^2 = x+x \\ x^3=x+x+x \\ \vdots $ 

In this instance, the subgroup generated by \( X \), known as \( \langle X \rangle \) or \( \langle 6 \rangle \), is the set of all integers that are multiples of 6. This is because by continuously adding or subtracting 6, we can derive every element within the subgroup.

This means, mathematically, the subgroup is:

\[ \langle 6 \rangle = \{ ... , -18, -12, -6, 0, 6, 12, 18, ... \} \]

Each number within this collection is a multiple of 6, and when you add or subtract any two numbers from this collection, the outcome remains a multiple of 6, confirming the set's closure under addition.

Moreover, it includes 0, the additive identity, ensuring that for every element, its inverse is also present (for example, the inverse of 6 is -6).

$$ x + x^{-1} = 6 + (-6) = 0 $$

Therefore, \( \langle 6 \rangle \) effectively qualifies as a subgroup of \((\mathbb{Z}, +)\).

This subgroup is cyclic because it is generated through the repeated application of the group operation (in this case, addition) on a single element, $x = 6$.

Example 2

Turning our attention back to the additive group $ ( Z,+) $ of integers under addition.

This time, let's consider a subset \( X = \{6, 7\} \) consisting of two elements.

In this scenario, the subgroup generated by \( X \), denoted \( \langle X \rangle \) or \( \langle 6, 7 \rangle \), will include all conceivable integer sums of 6 and 7, their negatives, and combinations thereof.

$$ \langle X \rangle = \{ 6 \cdot a + 7 \cdot b \ \ a ,b \in Z \}$$

Where $ a $ and $ b $ are any integers in a linear combination.

Given that 6 and 7 are consecutive integers and thus coprime (sharing no common divisors other than 1), any integer can be represented as a linear combination of 6 and 7.

In mathematical terms, for any integer \( n \), there exist integers \( a \) and \( b \) such that \( n = 6a + 7b \).

$$ \forall \ n \in Z , \ n = 6 \cdot a + 7 \cdot b $$

As such, the subgroup generated by \( \{6, 7\} \) essentially encompasses the entire group \( \mathbb{Z} \) itself.

$$ \langle 6,7 \rangle = \mathbb{Z} $$

Example 3

Using the group \((\mathbb{Z}, +)\), the set of integers with addition as the operation, let's consider the set \( X = \{2, 4\} \).

To identify the subgroup generated by \( X \), marked as \( \langle X \rangle \) or \( \langle 2, 4 \rangle \), we need to account for all possible integer sums of 2 and 4.

However, it quickly becomes apparent that 4 is a multiple of 2 (that is, \( 4 = 2 \cdot 2 \)), meaning any number achievable by adding 4 can also be reached by adding 2.

Therefore, the subgroup generated by \( X \) will mirror the subgroup that would be generated by just the smallest number in \( X \), which is 2.

This subgroup is the collection of all integer multiples of 2:

$$ \langle 2, 4 \rangle = \langle 2 \rangle = \{ ... , -8, -6, -4, -2, 0, 2, 4, 6, 8, ... \} $$

In summary

$$ \langle 2, 4 \rangle = \langle 2 \rangle = 2 \mathbb{Z} $$

This subgroup is cyclic because it can be fully generated by a single element, the number 2, through its repeated self-addition an arbitrary number of times (whether positive, negative, or zero).

In general, when a subgroup can be generated from a single element, such a subgroup is te