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Chapter 10: Problem 3

Let \(a, b\), and \(c\) be elements of a group \(G\). Prove the following: If \(a^{k}=e\) where \(k\) is odd, then the order of \(a\) is odd.

Short Answer

Expert verified
If \( a^k = e \) with odd \( k \), then \( a \) has an odd order.

Step by step solution

01

Understand Group Order and Element Order

The order of an element in a group is the smallest positive integer \( n \) such that \( a^n = e \), where \( e \) is the identity element of the group. To find the order of an element \( a \) in group \( G \), we need to examine the powers of \( a \) and identify the smallest integer where \( a^n \) returns to the identity.
02

Examine the Assumption

We are given that \( a^k = e \) where \( k \) is an odd integer. This suggests that \( k \) is a cycle length that repeats without returning to \( e \) any earlier since \( a^k \) is the first repetition to \( e \). We need to show that the smallest such \( n \) that satisfies \( a^n=e \) is also odd.
03

Assume to the Contrary

Assume, for contradiction, that the order \( n \) of \( a \) is even. This would mean that \( a^n = e \) for some even \( n \). Since \( a^k = e \) where \( k \) is odd and \( n \) is the smallest such number, it must mean \( n \) divides \( k \).
04

Demonstrate the Contradiction

Given that \( n \) divides \( k \), and \( k \) is odd, it follows that \( n \) cannot be even because an even \( n \) cannot be a divisor of an odd \( k \). Therefore, our assumption that \( n \) could be even is false, leaving the only option that \( n \) is odd.
05

Conclude the Proof

The contradiction shows that our assumption of an even \( n \) leads to logical inconsistency. Consequently, the order of \( a \) must be the odd integer as assumed \( k \). Thus, the proof is complete, confirming that if \( a^k = e \) and \( k \) is odd, then the order of \( a \) is necessarily odd as well.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Order of an Element
In group theory, the order of an element is a fundamental concept that helps us understand the behavior of the element in relation to the identity element of a group. Specifically, it is the smallest positive integer, called the order, denoted by \( n \), such that raising the element \( a \) to the \( n \)-th power equals the identity element \( e \) of the group, i.e., \( a^n = e \).
  • The order gives insight into how many times you need to apply the element to get back to the starting point represented by the identity.
  • Knowing the order can reveal important structures about the group, like its cyclic nature.
  • It also helps in simplifying calculations within the group, contributing to understanding group symmetry and operations.
In the problem given, the assertion is that if an element \( a \) satisfies \( a^k = e \) and \( k \) is an odd integer, then the order of \( a \) must also be odd.
Identity Element
An identity element in group theory is a special member of the group that leaves other elements unchanged when combined with them. For any element \( a \) in a group \( G \), the identity element \( e \) satisfies the equation \( a \cdot e = e \cdot a = a \).
  • This property makes the identity element a critical part of group structure.
  • It acts as the neutral or starting point from which other operations are measured.
  • In a mathematical sense, it 'resets' operations, making it possible to isolate contributions from other group members.
In context, the identity element's role is silent yet pivotal, especially when proving the nature of other elements, such as in establishing the oddness of the order of \( a \) based on powers equating to the identity.
Contradiction Proof
A contradiction proof is an elegant technique used to show that a particular assumption cannot be true. This method involves assuming the opposite of what you want to prove, and then showing that this assumption leads to a logical inconsistency or impossibility.
  • The goal is to derive a contradiction from the assumption, thus proving the assumption is false.
  • Once a contradiction is presented, the original statement is confirmed as true.
  • This form of proof is particularly useful when dealing with odd and even integers, divisibility, and similar scenarios.
For the exercise given, a contradiction proof was used to demonstrate that assuming the order \( n \) is even leads to an inconsistency. As \( n \) being even would contradict the condition \( a^k = e \) where \( k \) is odd, it forces the conclusion that \( n \) must be odd.
Cyclic Group
A cyclic group is a type of group in which all elements can be generated by repeatedly applying an operation to a single element, known as a generator. If such an element exists, the group is said to be cyclic, and every member of the group is a power of the generator.
  • Cyclic groups are some of the simplest and most foundational structures in group theory.
  • They can be finite or infinite, depending on how many distinct elements they consist of.
  • The order of the group (the number of its elements) is the same as the order of any of its generators.
In the context of proving properties like the order of an element, considering whether the group is cyclic can provide useful insights, as the repetitive nature of the elements aligns with concepts like element order and identity.

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Most popular questions from this chapter

From elementary arithmetic we know that every integer may be written uniquely as a product of prime numbers. Two integers \(m\) and \(n\) are said to be relatively prime if they have no prime factors in common. (For example, 15 and 8 are relatively prime.) Here is a useful fact: If \(m\) and \(n\) are relatively prime, and \(m\) is a factor of \(n k\), then \(m\) is a factor of \(k\). (Indeed, all the prime factors of \(m\) are factors of \(n k\) but not of \(n\), hence are factors of \(k\).) Let \(a\) be an element of order \(n\) in a group \(G\). Prove the following: If \(a^{m}\) has order \(n\), then \(m\) and \(n\) are relatively prime. [HINT: Assume \(m\) and \(n\) have a common factor \(q>1\), hence we can write \(m=m^{\prime} q\) and \(n=n^{\prime} q\). Explain why \(\left(a^{m}\right)^{n^{\prime}}=e\), and proceed from there.]

Let \(a\) and \(b\) be elements of a group \(G .\) Let ord \((a)=m\) and \(\operatorname{ord}(b)=n ; \operatorname{lcm}(m, n)\) denotes the least common multiple of \(m\) and \(n .\) Prove the following: Let \(a\) and \(b\) commute. There is an element \(c\) in \(G\) whose order is \(\operatorname{lcm}(m, n)\).

Let \(a, b\), and \(c\) be elements of a group \(G\). Prove the following: The order of \(a^{-1}\) is the same as the order of \(a\).

Let \(a\) and \(b\) be elements of a group \(G .\) Let ord \((a)=m\) and \(\operatorname{ord}(b)=n ; \operatorname{lcm}(m, n)\) denotes the least common multiple of \(m\) and \(n .\) Prove the following: 4 Let \(a\) and \(b\) commute. If \(m\) and \(n\) are relatively prime, then ord \((a b)=m n\).

What is the order of 6 in \(\mathbb{Z}_{16}\) ?
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