Problem 11 - Algebra: Pure and Applied

Question

Show that if \(G\) is an Abelian group, then for all \(a, b \in G\) and for all integers \(n\), \((a b)^{n}=a^{n} b^{n}\)

Step-by-Step Solution

Answer

In an Abelian group, \((ab)^n = a^n b^n\) for all integers \(n\).

Step 1: Understand properties of Abelian groups

An Abelian group is a group where the group operation is commutative. This means that for all elements \(a, b \in G\), \(ab = ba\). We will use this property to show that \((ab)^n = a^n b^n\).

Step 2: Inductive Base Case

We need to show that the statement \((ab)^n = a^n b^n\) holds for \(n = 1\). For \(n = 1\), we have \((ab)^1 = ab\) and \(a^1 b^1 = ab\). Hence the statement is true for \(n = 1\).

Step 3: Inductive Hypothesis

Assume the statement is true for some integer \(k\), i.e., \((ab)^k = a^k b^k\). This is the inductive hypothesis we will use to prove the next step.

Step 4: Inductive Step

We need to show that if the hypothesis holds for \(k\), then it also holds for \(k + 1\). Consider \((ab)^{k+1}\): \((ab)^{k+1} = (ab)^k(ab) = a^k b^k ab\). Using the commutative property of Abelian groups, this equals \(a^k a b^k b = a^{k+1} b^{k+1}\). Thus, by induction, \((ab)^n = a^n b^n\) is true for all \(n\).

Step 5: Conclusion by Mathematical Induction

Having shown the base case and the inductive step, by mathematical induction, \((ab)^n = a^n b^n\) holds for all integers \(n\) in an Abelian group.

Key Concepts

Group Operation

A group operation is a mathematical function that combines two elements from a group to produce another element of the same group. Each element in the group remains within the group when the operation is applied. In the context of an Abelian group, this operation is specifically crucial because it is performed with elements such as \(a\) and \(b\) and results in a product like \(ab\). The operation must satisfy certain properties: closure, associativity, the existence of an identity element, and the existence of inverse elements.
For a better understanding, let's go over these properties quickly:

• Closure: For group elements \(a\) and \(b\), the result \(ab\) is still within the group.
• Associativity: For any elements \(a, b, c\), the equation \((ab)c = a(bc)\) holds.
• Identity: There exists a special element \(e\) such that for any element \(a\), \(ae = ea = a\).
• Inverse: For each element \(a\), there exists an element \(a^{-1}\) such that \(aa^{-1} = a^{-1}a = e\).

In Abelian groups, the group operation neatens calculations especially due to its commutative nature.

Commutative Property

The commutative property is a fundamental principle in mathematics that assures us that the order in which we apply an operation does not affect the outcome, specifically for addition and multiplication. For an Abelian group, this property means that for any two elements \(a\) and \(b\) in the group, the operation \(ab\) gives the same result as \(ba\). As a consequence, computations become more flexible and predictable.
This property is key when working with expressions like \((ab)^n\) because it allows us to rearrange and manipulate the elements freely to demonstrate equivalences like \((ab)^n = a^n b^n\). It's important to contrast this with non-Abelian groups, where such symmetrical manipulations would not be permitted. In everyday multiplication, whether multiplying numbers or elements, this glaring symmetry simplifies many algebraic manipulations.

Mathematical Induction

Mathematical induction is a powerful proof technique used to verify statements or propositions that are asserted to be true for all natural numbers. It is particularly useful when establishing the veracity of equations involving integer sequences or recursive definitions.
This technique involves two main steps:

• Base Case: You begin by proving the statement for the initial value, usually the smallest natural number, like \(n=1\).
• Inductive Step: You then assume that the statement is true for some integer \(k\), and you utilize this assumption, known as the inductive hypothesis, to show that the statement is also true for the next integer \(k+1\).

When both steps are rigorous, the principle of mathematical induction concludes that the statement is verified for all integers greater than or equal to the base case. Utilizing this method, we efficiently prove more complex expressions, such as the multiplication of group elements raised to powers in an Abelian group.

Inductive Hypothesis

The inductive hypothesis forms the backbone of the induction process in mathematical proofs. During mathematical induction, once the base case is established, an inductive hypothesis assumes that the statement holds for an arbitrary integer \(k\). This assumption is pivotal for deriving the truth of the statement for the subsequent integer \(k+1\).
In practice, with our example, the inductive hypothesis proposes that \((ab)^k = a^k b^k\) holds true. It is crucial because it serves as a foundational step for proving the statement for \((k+1)\), showing that if the hypothesis works for some integer \(k\), it must also be valid for \(k+1\).
The strength of the inductive hypothesis lies in its recursive nature, supporting the proof to leap from one true instance to another. Essentially, it functions as a ladder, allowing us to climb through an infinite sequence of cases without addressing each individually, which is especially powerful for confirming statements across all natural numbers.