Abstract Algebra ### Problem Statement**(5) Prove that every group of order 3 is cyclic.**### ExplanationTo 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 Outline1. **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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Answered: (5) Prove that every group of order 3…

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

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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