Math Problem Statement
if each element, except the identity of a group be of order 2, then prove that the group is abelian.
Solution
We are asked to prove that if each element of a group , except the identity element, has order 2, then the group is abelian.
Step-by-Step Proof:
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Given Conditions:
- Let be a group, and represent the identity element of .
- For every where , we have . This means that each non-identity element of is its own inverse.
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Show that is Abelian: To prove that is abelian, we need to show that for any two elements , we have .
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Consider the product : Since and , let’s compute the product : Since by the condition that all non-identity elements have order 2, we now have:
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Manipulating the equation: Now, multiply both sides of the equation by on the left and on the right: Simplifying this, since and , we get: Thus, , which shows that is abelian.
Conclusion:
Therefore, the group is abelian, as we have shown that for any two elements , we have .
Would you like further clarification or a more detailed explanation of any part of the proof?
Related Questions:
- What is the definition of an abelian group?
- Can a group have more than one element of order 2?
- What does it mean for a group element to have order 2?
- How do we prove that a group is cyclic?
- What are the properties of cyclic groups?
Tip: The key to proving that a group is abelian in such cases is to use the properties of group elements, such as their orders and inverses, to manipulate the group's structure algebraically.
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Math Problem Analysis
Mathematical Concepts
Group Theory
Abelian Groups
Order of Elements
Formulas
Order of an element: a^n = e implies the order of a is n
(ab)^2 = e implies ab = ba for groups where all elements have order 2
Theorems
Basic properties of groups
Definition of an abelian group
Suitable Grade Level
Undergraduate Mathematics or Advanced High School (Grades 11-12)