Problem 1 - A Book of Abstract Algebra

Question

left(b a b^{-1}\right)^{n}=b a^{n} b^{-1}\(, for every positive integer \)n .\( Prove by induction. (Remember that to prove a formula such as this one by induction, you first prove it for \)n=1\(; next you prove that if it is true for \)n=k\(, then it must be true for \)n=k+1\(. You may conclude that it is true for every positive integer \)n$.)

Step-by-Step Solution

Answer

Proven by induction for all positive integers.

Step 1: Base Case (n = 1)

Start by proving the base case. For \(n = 1\), we need to verify that \(\left(b a b^{-1}\right)^{1} = b a^{1} b^{-1}\). This is trivially true because \(\left(b a b^{-1}\right)^{1} = b a b^{-1}\) and \(b a b^{-1} = b a^{1} b^{-1}\). Thus, the statement holds for \(n = 1\).

Step 2: Inductive Hypothesis (n = k)

Assume that the statement is true for \(n = k\), which means \(\left(b a b^{-1}\right)^{k} = b a^{k} b^{-1}\). This assumption is known as the inductive hypothesis and will be used to prove the case for \(n = k+1\).

Step 3: Inductive Step (n = k+1)

We need to show that if \(\left(b a b^{-1}\right)^{k} = b a^{k} b^{-1}\) is true, then \(\left(b a b^{-1}\right)^{k+1} = b a^{k+1} b^{-1}\) is also true. Begin by expressing \(\left(b a b^{-1}\right)^{k+1}\) as follows: \[\left(b a b^{-1}\right)^{k+1} = \left(b a b^{-1}\right)^{k} \cdot \left(b a b^{-1}\right)\]Using the inductive hypothesis, substitute \(\left(b a b^{-1}\right)^{k}\) with \(b a^{k} b^{-1}\):\[\left(b a b^{-1}\right)^{k+1} = (b a^{k} b^{-1})(b a b^{-1})\]Perform multiplication:\[= b a^{k} (b^{-1} b) a b^{-1}\]Since \(b^{-1} b = e\) (where \(e\) is the identity element), simplify:\[= b a^{k} e a b^{-1} = b a^{k} a b^{-1} = b a^{k+1} b^{-1}\]Thus, the statement holds for \(n = k+1\).

Step 4: Conclusion

Since we have shown the base case is true and that if the statement is true for \(n = k\), then it is also true for \(n = k+1\), we conclude by the principle of mathematical induction that \(\left(b a b^{-1}\right)^{n} = b a^{n} b^{-1}\) is true for every positive integer \(n\).

Key Concepts

Inductive Hypothesis

The inductive hypothesis is a key part of proving statements using mathematical induction. To understand this concept, think of it like making a prediction based on an assumption. You assume that your statement is true for some arbitrary case, often called \( n = k \). For example, in our exercise, you assume that the equation \( (b a b^{-1})^k = b a^k b^{-1} \) holds true.

This assumption becomes a building block to prove that if it works for \( n = k \), it should also work for the next step, which is \( n = k+1 \).

• The inductive hypothesis provides the basis upon which you demonstrate that your statement is valid for the subsequent step.
• If you establish that \((b a b^{-1})^k = b a^k b^{-1}\) is true, it helps in breaking down the problem into smaller, manageable parts for further verification.

Understanding the role of the inductive hypothesis allows the induction method to proceed logically, showing the progression from \( n = k \) to \( n = k+1 \). This is essentially what allows the entire domain of positive integers to be encompassed with your proof.

Base Case

When using mathematical induction to prove a statement, the base case verifies the claim for the initial step. This is usually the simplest part and represents the smallest value of \( n \) that you need to consider. For most problems, including our current exercise, the base case occurs when \( n = 1 \).

In our equation \( (b a b^{-1})^1 = b a^1 b^{-1} \), you want to show this statement holds true before you apply induction. This step acts as the foundation for the entire proof. Without a correct base case, the later steps would not logically follow.

• The base case is often quite trivial, as it merely checks that the original expression is valid for \( n = 1 \).
• It sets the stage for the inductive step by providing the first true example needed to confirm that the pattern works continuously across all positive integers.

Successfully proving the base case is critical because it signifies the starting point from which you begin your mathematical reach through subsequent numbers using induction.

Identity Element

The identity element is a fundamental concept found in various areas of mathematics, including group theory. In our specific problem, the identity element is crucial when simplifying expressions.

When you encounter terms like \( (b a b^{-1}) \), understanding the identity element helps you simplify expressions. The identity element, represented as \( e \) in most contexts, is a special element that, when combined with any element of the set, leaves it unchanged.

• In our exercise, \( b^{-1} b = e \) utilizes the identity element.
• Because \( b^{-1} b = e \), which is essentially 1, it simplifies the expression \( b a^k (b^{-1} b) a b^{-1} \) to \( b a^{k+1} b^{-1} \).

By recognizing and applying the identity element, you can make complex expressions more manageable, which is an essential step in using mathematical induction effectively. This helps you ensure that the expressions remain valid across the different stages of the proof.