Subgroups

A subgroup $ S $ is a subset within a group $ G $ that, by adhering to the group's operation, itself constitutes a group. $$ S < G $$

This entity maintains the quintessential properties of closure, associativity, the presence of an identity element, and inverses, which are pivotal in defining a group.

Hence, a subgroup is not merely any subset of a group; it must also satisfy the specific requirements that classify it as a distinct group under the operation of its encompassing group.

An Illustrative Example
The Subgroup Criterion

An Illustrative Example

Consider the group of integers $\mathbb{Z}$ under addition.

The group $ ( \mathbb{Z}\, + ) $ serves as a prime example of an infinite group, wherein each element has an inverse (its negative) and the identity element is 0, for adding 0 to any number leaves it unchanged.

An intriguing subgroup of $ ( \mathbb{Z}\, + ) $ is the group $ ( 2\mathbb{Z} , + ) $, comprising all even integers $ 2\mathbb{Z} \subset \mathbb{Z} $.

$\mathbb{Z}$ qualifies as a subgroup under addition because it fulfills the essential conditions to be considered a group:

Closure
The sum of two even numbers invariably results in an even number. Taking any two elements $a, b \in 2\mathbb{Z}$, like 4 and 6, their sum $4 + 6 = 10$ remains an element of $2\mathbb{Z}$.
Identity Element
Zero stands as the identity element for the integers under addition and is also even, making it a member of $2\mathbb{Z}$.
Inverse
Every element $a \in 2\mathbb{Z}$ possesses an inverse $-a$ that likewise belongs to $2\mathbb{Z}$. For example, the inverse of 4 is -4, both of which are elements of $2\mathbb{Z}$.

Meeting these three criteria implicitly assures the associative property as well.

For instance, if the set $ 2\mathbb{Z} $ is closed under addition, it naturally inherits the associative property of addition $ a + (b+c) = (a+b) + c $ from its overarching group $ ( \mathbb{Z}, + ) $. This is because the operation of addition itself carries the associative property, and any subset forming a subgroup utilizes the same operation defined in the broader group, obviating the need for separate verification.

The $2\mathbb{Z}$ subgroup exemplifies a simpler construct within the intricate group $\mathbb{Z}$.

In essence, $ ( 2\mathbb{Z} , + ) $ is effectively a group in its own right.

Subgroups like $2\mathbb{Z}$ shed light on how groups can be assembled from smaller segments and the interaction of these segments with the larger group framework.

Broadly speaking, all subsets of $ \mathbb{Z} $ formed by the multiples of any integer $n$ are subgroups under addition. Thus, $ (n\mathbb{Z},+) $ also constitutes a subgroup of $\mathbb{Z} $. This highlights the diversity and abundance of potential subgroups within even seemingly straightforward mathematical scenarios.

The Subgroup Criterion

One particularly handy criterion for discerning whether a subset $S$ of a group $(G, *)$ qualifies as a subgroup is the one-step subgroup criterion, or the subgroup closure criterion.

This criterion posits that:

A non-empty subset $S$ of a group $(G, *)$ qualifies as a subgroup of $(G, *)$ if and only if, for any pair of elements within S, the outcome of operating the first element with the inverse of the second still falls within S $$ \forall \ a, b \in S \ , \ a * b^{-1} \in S $$

This criterion is exceptionally potent because it consolidates the necessary verifications to establish $S$ as a subgroup into a single, straightforward step.

Case Study

Let's apply the one-step subgroup criterion to the $2\mathbb{Z}$ subgroup under the addition operation in the group $\mathbb{Z}$.

Given that $2\mathbb{Z}$ encompasses all even integers, both $a$ and $b$ represent even numbers.

According to the criterion, to affirm $2\mathbb{Z}$ as a subgroup of $ \mathbb{Z} $, it's necessary to demonstrate that for any $a, b \in 2\mathbb{Z}$, the expression $a + (-b)$ remains within $2\mathbb{Z}$.

The expression $a + (-b)$ simplifies to the subtraction $a - b$, as our focus is on addition.

Selecting any two elements $a$ and $b$ from $2\mathbb{Z}$, say $a = 4$ and $b = 6$, and computing $a - b$, we get:

\[a - b = 4 - 6 = -2\]

The result, $-2$, is still even, hence remains within $2\mathbb{Z}$.

Yet, the criterion mandates this property to hold for every element pair in $2\mathbb{Z}$, beyond just the exemplar pair chosen.

Given that the subtraction of any two even numbers invariably yields an even number, it's clear that $2\mathbb{Z}$ indeed qualifies as a subgroup of $ \mathbb{Z} $ according to the one-step subgroup criterion.