Problem 21 - A Computational Introduction to Number Theory and Algebra

Question

Let \(H\) be a subgroup of an abelian group \(G,\) and let \(a, b \in G\) Show that \([a+b]_{H}=\left\\{x+y: x \in[a]_{H}, y \in[b]_{H}\right\\}\)

Step-by-Step Solution

Answer

Question: Prove that for an abelian group \(G\) and its subgroup \(H\), the sum of the cosets \([a+b]_{H}\) can be expressed as the set of sums of elements from \([a]_{H}\) and \([b]_{H}\), where \(a, b\) are elements of \(G\). Answer: To prove this, we first define the sets \(A = \left\\{x+y: x \in[a]_{H}, y \in[b]_{H}\right\\}\) and \(B = [a+b]_{H}\) and then show that \(A = B\). We do this by showing that \(A \subseteq B\) and \(B \subseteq A\). Since the group \(G\) is abelian, we can rearrange the terms and use the closure property of the subgroup \(H\) to simplify our proof. After showing that both \(A \subseteq B\) and \(B \subseteq A\), we can conclude that \(A = B\) and \([a+b]_{H}=\left\\{x+y: x \in[a]_{H}, y \in[b]_{H}\right\\}\).

Step 1: Define the sets that we want to show are equal

Let \(A = \left\\{x+y: x \in[a]_{H}, y \in[b]_{H}\right\\}\) and \(B = [a+b]_{H}\). Our goal is to prove that \(A = B\).

Step 2: Prove that A is a subset of B

To prove that \(A \subseteq B\), we need to show that for any element \(z \in A\), \(z \in B\) as well. Let's take an arbitrary \(z \in A\). Then, by the definition of \(A\), we know that \(z = x + y\), where \(x \in [a]_H\) and \(y \in [b]_H\). Since \(x \in [a]_H\), it means that \(x = a + h_1\) for some \(h_1 \in H\). Similarly, we have \(y = b + h_2\) for some \(h_2 \in H\). Now, let's add these two expressions: $$z = x + y = (a + h_1) + (b + h_2) \enspace (1)$$ Since \(G\) is abelian, we can rearrange the terms to get: $$z = a + b + (h_1 + h_2) \enspace (2)$$ Since \(H\) is a subgroup, it's closed under the operation, so \(h_1 + h_2 \in H\). Thus, we have \(z \in [a+b]_H\). This proves that \(A \subseteq B\).

Step 3: Prove that B is a subset of A

To prove that \(B \subseteq A\), we need to show that for any element \(z \in B\), \(z \in A\) as well. Let's take an arbitrary \(z \in B\). Then, by the definition of \(B\), we know that \(z \in [a+b]_H\), which means \(z = (a+b)+h\) for some \(h \in H\). Now, let's rewrite this expression as: $$z = (a + h_1) + (b + h_2) \enspace (3)$$ Where we define \(h_1 = h\) and \(h_2 = 0\), noting that \(h_2 = 0\) because the identity element is always in \(H\). Since \(a + h_1 \in [a]_H\) and \(b + h_2 \in [b]_H\) and using the definition of the set A, we see that \(z \in A\). Thus, we conclude that \(B \subseteq A\).

Step 4: Conclude that A equals B

Since we have shown that \(A \subseteq B\) and \(B \subseteq A\), we can conclude that \(A = B\). Therefore, \([a+b]_{H}=\left\\{x+y: x \in[a]_{H}, y \in[b]_{H}\right\\}\).

Key Concepts

Subgroup

In the realm of group theory, a subgroup is a subset of a group that also qualifies as a group under the same operation as the parent group. A core feature of a subgroup is that it contains the group's identity element. Additionally, it is closed under the group operation and includes the inverse of each of its elements.

Consider a group \( G \) and a subgroup \( H \). For \( H \) to be a subgroup of \( G \), it must satisfy:

• Identity: The identity element \( e \) of \( G \) must be in \( H \).
• Closure: If \( a, b \in H \), then \( a \cdot b \in H \).
• Inverses: For every \( a \in H \), the inverse \( a^{-1} \in H \).

By these properties, a subgroup behaves like a mini-group within \( G \). In this exercise, knowing \( H \) is a subgroup is critical as it allows the use of closure and identity properties to manipulate elements effectively, proving statements about cosets.

Coset

Cosets are a fascinating concept in group theory, especially crucial when dealing with subgroups. When a subgroup \( H \) is given, the left coset of \( H \) in \( G \) for any element \( a \in G \) is defined as:

• \([a]_H = \{ a \, h \mid h \in H \} \).

This essentially means adding the element \( a \) to every element in \( H \).

Similarly, a right coset is obtained by multiplying each \( h \in H \) on the right:

• \( H a = \{ h \, a \mid h \in H \} \).

In the context of an Abelian group, left and right cosets are the same, as the group operation is commutative.

The significance of cosets lies in how they partition the group \( G \). Each element of \( G \) belongs to exactly one coset. In the exercise at hand, understanding how cosets relate through subgroup properties is pivotal for proving the equality \([a+b]_H = \{ x+y \mid x \in [a]_H, y \in [b]_H \}\).

Group Theory

Group theory is a rich field of mathematics that studies the symmetry of structures via groups. A group consists of a set, combined with an operation that satisfies four key properties:

• Closure: Performing the operation on two elements of the set results in another element within the same set.
• Associativity: The group operation is associative.
• Identity element: There exists an identity element which leaves other elements unchanged when combined.
• Inverses: Every element has an inverse such that the combination of an element and its inverse yields the identity.

When a group is abelian, it adds the layer of commutativity to the operation, meaning the order of combining elements does not matter. This property is crucial in the provided exercise because it allows the rearrangement of terms, which in turn facilitates equality aspects of cosets.

By utilizing these properties, group theory allows for deep analysis and simplification of complex algebraic structures, making it a powerful mathematical tool.

Proof Techniques

Proof techniques are fundamental in mathematics, providing clarity and validation to mathematical statements. One common technique in group theory proofs is showing set equality by demonstrating two subset relations: \( A \subseteq B \) and \( B \subseteq A \).

In the given exercise, this method is employed to demonstrate the equality of two sets. The proof is twofold:

• Subset Relationship: Prove \( A \subseteq B \) by picking any element from \( A \) and showing it belongs to \( B \).
• Subset Reversal: Prove \( B \subseteq A \) by picking any element from \( B \) and showing it belongs to \( A \).

Once both directions of inclusion are covered, you can conclude \( A = B \).

Such proofs heavily rely on logical reasoning, definitions, and underlying properties (e.g., subgroup properties). This well-established method ensures a solid foundation for comparing mathematical structures, supporting deeper understanding and exploration in group theory.