Math Problem Statement
solve this
Solution
We are asked to prove that is a subgroup of the group , which is abelian.
Steps to prove that is a subgroup of :
To prove that is a subgroup of , we need to show that:
- The identity element of belongs to .
- is closed under the group operation.
- is closed under taking inverses.
Step 1: Identity element in
- Let be the identity element of .
- Since is abelian, for any , we have for .
- Therefore, , so the identity element is in .
Step 2: Closure under the group operation
- Let , so by definition, there exist such that and .
- We need to show that . Since is abelian, we have:
- Because (since is a group and closed under the group operation), we see that .
Step 3: Closure under inverses
- Let . By definition, there exists such that .
- We need to show that . Since is abelian and the inverse of an element in a group is also in the group, we have:
- Since , it follows that .
Conclusion:
Since contains the identity element, is closed under the group operation, and is closed under inverses, it satisfies the three conditions of being a subgroup. Therefore, is a subgroup of .
Would you like more details on any of these steps?
Here are 5 related questions:
- What is the definition of a subgroup in group theory?
- Why is the closure under inverses important in proving a subgroup?
- How does the property of being abelian simplify this proof?
- Can we generalize this proof to non-abelian groups?
- What would change in this proof if was not defined with powers of ?
Tip: When proving a subgroup, it's useful to always verify identity, closure under the operation, and closure under inverses—these are the core criteria.
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Math Problem Analysis
Mathematical Concepts
Group Theory
Subgroups
Abelian Groups
Formulas
a = x^n for some x ∈ G
Theorems
Subgroup Criterion (identity element, closure under operation, closure under inverses)
Suitable Grade Level
Undergraduate - Group Theory