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(5) Prove that every group of order 3 is cyclic.

(5) Prove that every group of order 3 is cyclic.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Abstract Algebra

### Problem Statement **(5) Prove that every group of order 3 is cyclic.** ### Explanation To prove that every group of order 3 is cyclic, consider the following: 1. **Definition of a Group:** - A group is a set equipped with an operation that satisfies four properties: closure, associativity, identity, and inverses. 2. **Definition of Group Order:** - The order of a group is the number of elements in the group. 3. **Definition of a Cyclic Group:** - A group is cyclic if there exists an element (called a generator) in the group such that every other element can be written as a power of this generator. ### Proof Outline 1. **Let G be a Group of Order 3:** - By definition, this means that G has exactly 3 elements. 2. **Lagrange's Theorem:** - The order of any subgroup of a finite group divides the order of the group. Hence, subgroups of order 1 and 3 are possible for G. Since the identity element always forms a trivial subgroup of order 1, we consider other non-identity elements. 3. **Existence of a Nontrivial Element:** - Let \( g \) be a non-identity element in G. The order of \( g \) is the smallest positive integer n such that \( g^n = e \) (the identity element). 4. **Apply Lagrange's Theorem:** - The order of \( g \) must divide 3. So, the possible orders of \( g \) are 1 or 3. 5. **Generator of the Group:** - The element \( g \) cannot have order 1, as that would make it the identity element. Hence, \( g \) must have order 3. - Therefore, \( g, g^2, \) and \( e \) (where \( g^3 = e \)) are distinct elements of the group G. 6. **Conclusion:** - This list \( \{e, g, g^2\} \) includes all three elements of G, showing that every element of G can be expressed as a power of \( g \). - Hence, \( G \) is a cyclic group with \( g \) as a generator. Thus, every group
Transcribed Image Text:### Problem Statement **(5) Prove that every group of order 3 is cyclic.** ### Explanation To prove that every group of order 3 is cyclic, consider the following: 1. **Definition of a Group:** - A group is a set equipped with an operation that satisfies four properties: closure, associativity, identity, and inverses. 2. **Definition of Group Order:** - The order of a group is the number of elements in the group. 3. **Definition of a Cyclic Group:** - A group is cyclic if there exists an element (called a generator) in the group such that every other element can be written as a power of this generator. ### Proof Outline 1. **Let G be a Group of Order 3:** - By definition, this means that G has exactly 3 elements. 2. **Lagrange's Theorem:** - The order of any subgroup of a finite group divides the order of the group. Hence, subgroups of order 1 and 3 are possible for G. Since the identity element always forms a trivial subgroup of order 1, we consider other non-identity elements. 3. **Existence of a Nontrivial Element:** - Let \( g \) be a non-identity element in G. The order of \( g \) is the smallest positive integer n such that \( g^n = e \) (the identity element). 4. **Apply Lagrange's Theorem:** - The order of \( g \) must divide 3. So, the possible orders of \( g \) are 1 or 3. 5. **Generator of the Group:** - The element \( g \) cannot have order 1, as that would make it the identity element. Hence, \( g \) must have order 3. - Therefore, \( g, g^2, \) and \( e \) (where \( g^3 = e \)) are distinct elements of the group G. 6. **Conclusion:** - This list \( \{e, g, g^2\} \) includes all three elements of G, showing that every element of G can be expressed as a power of \( g \). - Hence, \( G \) is a cyclic group with \( g \) as a generator. Thus, every group
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