Show That The Union Of A Nonempty Directed Family Of Subgroups Of A Group G Is A Subgroup Of G?
Understanding how subgroups combine within a larger group is a fundamental aspect of group theory, a core area of abstract algebra. One intriguing question is under what conditions the union of multiple subgroups itself forms a subgroup. Specifically, if we have a nonempty family of subgroups of a group \( G \), and this family is directed, then the union of these subgroups is guaranteed to be a subgroup of \( G \). This result is not only elegant but also essential in understanding structures like directed systems and direct limits in algebra.
In this article, we will explore this concept thoroughly. We will define what a directed family of subgroups is, and then carefully demonstrate why their union under these conditions is a subgroup. Along the way, examples and key properties will be discussed to solidify understanding.
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Preliminaries and Definitions
Before diving into the main proof, it’s important to establish clear definitions and notations.
Subgroups of a Group \( G \)
A subset \( H \subseteq G \) is called a subgroup of \( G \) if it satisfies the following conditions:
- Closure: For all \( a, b \in H \), the product \(ab \in H \).
- Identity: The identity element \( e \in H \).
- Inverses: For all \( a \in H \), the inverse \( a^{-1} \in H \).
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Family of Subgroups and Their Union
Suppose \( \mathcal{F} = \{ H_\alpha \mid \alpha \in A \} \) is a family of subgroups of \( G \), indexed by some set \( A \). The union of this family is:
\[
\bigcup{\alpha \in A} H\alpha = \{ x \in G \mid x \in H_\alpha \text{ for some } \alpha \in A \}.
\]
A natural question arises: Is this union always a subgroup? The answer, in general, is no. The union of subgroups need not be a subgroup unless additional conditions are satisfied.
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Directed Families of Subgroups
The key condition that ensures the union of subgroups is itself a subgroup is the directedness of the family.
Definition of a Directed Family
A family \( \mathcal{F} = \{ H\alpha \}{\alpha \in A} \) of subgroups of \( G \) is called directed if:
- \(\mathcal{F}\) is nonempty; that is, there exists at least one \( H_\alpha \in \mathcal{F} \).
- For any two subgroups \( H\alpha, H\beta \in \mathcal{F} \), there exists another subgroup \( H_\gamma \in \mathcal{F} \) such that:
\[
H\alpha \subseteq H\gamma \quad \text{and} \quad H\beta \subseteq H\gamma.
\]
In other words, given any two subgroups in the family, there is a third subgroup that contains both.
This directedness condition ensures the family is "upward-directed" or "coherent" in the sense of inclusion, which is crucial for the union to inherit subgroup properties.
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The Main Theorem and Its Proof
Theorem:
Let \( G \) be a group, and \( \mathcal{F} = \{ H\alpha \}{\alpha \in A} \) be a nonempty, directed family of subgroups of \( G \). Then, the union \( H = \bigcup{\alpha \in A} H\alpha \) is a subgroup of \( G \).
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Step 1: Show that \( H \) contains the identity element \( e \).
Since each \( H\alpha \) is a subgroup, it contains \( e \). Given that \( \mathcal{F} \) is nonempty, pick any \( H{\alpha_0} \in \mathcal{F} \). Then:
\[
e \in H{\alpha0} \subseteq H,
\]
so the union \( H \) contains \( e \).
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Step 2: Show that \( H \) is closed under inverses.
Take any element \( x \in H \). By the definition of union, there exists some \( H_\alpha \in \mathcal{F} \) such that:
\[
x \in H_\alpha.
\]
Since \( H\alpha \) is a subgroup, it contains the inverse \( x^{-1} \). Because \( x^{-1} \in H\alpha \), it follows that:
\[
x^{-1} \in H.
\]
Thus, \( H \) is closed under inverses.
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Step 3: Show that \( H \) is closed under multiplication.
This is the critical part. Take any \( x, y \in H \). Then:
- There exist \( H\alpha, H\beta \in \mathcal{F} \) such that
x \in H\alpha, \quad y \in H\beta.
\]
Because \( \mathcal{F} \) is directed, there exists some \( H_\gamma \in \mathcal{F} \) such that:
\[
H\alpha \subseteq H\gamma \quad \text{and} \quad H\beta \subseteq H\gamma.
\]
Since both \( x, y \in H\gamma \) (by inclusion), and \( H\gamma \) is a subgroup, it contains the product:
\[
xy \in H_\gamma.
\]
Therefore,
\[
xy \in H,
\]
since \( H_\gamma \subseteq H \). This confirms that \( H \) is closed under multiplication.
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Conclusion: \( H \) is a Subgroup of \( G \)
Having verified the three subgroup criteria—containing the identity, closed under inverses, and closed under multiplication—it's clear that:
\[
H = \bigcup{\alpha \in A} H\alpha
\]
is a subgroup of \( G \).
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Implications and Applications
This result has several important implications in algebra and related fields.
Construction of Direct Limits
In category theory and algebra, the concept of a direct limit involves taking a directed system of groups (or modules, rings, etc.) and forming their union in a way that preserves structure. The theorem ensures that the union of a directed family of subgroups is itself a subgroup, serving as a building block for direct limits.
Building Larger Subgroups from Smaller Ones
Often, complex groups are constructed as unions of smaller subgroups, especially in infinite groups. The theorem assures us that when these smaller subgroups are arranged in a directed manner, their union can be treated as a new subgroup, facilitating analysis and further constructions.
Understanding the Structure of Infinite Groups
Many infinite groups are studied via chains or directed systems of subgroups. Recognizing when unions are subgroups simplifies the analysis of their structure, particularly in the context of solvable groups, nilpotent groups, and other classes where ascending chains are common.
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Examples to Illustrate the Theorem
Example 1:
Let \( G = \mathbb{Q} \) (the additive group of rational numbers). Define the family:
\[
H_n = \left\{ \frac{m}{2^n} \mid m \in \mathbb{Z} \right\}
\]
for each positive integer \( n \). Each \( H_n \) is a subgroup of \( \mathbb{Q} \), and the family is directed because:
\[
H1 \subseteq H2 \subseteq H_3 \subseteq \cdots
\]
The union:
\[
H = \bigcup{n=1}^\infty Hn
\]
is the set of all rational numbers with denominators powers of 2, which is a subgroup of \( \mathbb{Q} \).
Example 2:
Suppose \( G \) is a group, and \( \{ H\alpha \} \) is a family of subgroups such that for any two \( H\alpha, H\beta \), there exists \( H\gamma \) containing both. Their union is then a subgroup. Conversely, if the family is not directed, the union may fail to be a subgroup (for example, the union of two subgroups that are not contained within a common larger subgroup may not be a subgroup).
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Summary
To summarize, the key points are:
- The union of a nonempty family of subgroups of a group \( G \) is not necessarily a subgroup.
- When the family is directed, meaning for any two subgroups there exists a larger subgroup containing both, the union is a subgroup.
- The proof hinges
