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We are asked to prove that \( G_n = \{ a \in G \mid a = x^n \text{ for some } x \in G \} \) is a subgroup of the group \( G \), which is abelian.

### Steps to prove that \( G_n \) is a subgroup of \( G \):

To prove that \( G_n \) is a subgroup of \( G \), we need to show that:
1. The identity element of \( G \) belongs to \( G_n \).
2. \( G_n \) is closed under the group operation.
3. \( G_n \) is closed under taking inverses.

#### Step 1: Identity element in \( G_n \)
- Let \( e \) be the identity element of \( G \). 
- Since \( G \) is abelian, for any \( x \in G \), we have \( e = x^0 \) for \( n = 0 \).
- Therefore, \( e \in G_n \), so the identity element is in \( G_n \).

#### Step 2: Closure under the group operation
- Let \( a_1, a_2 \in G_n \), so by definition, there exist \( x_1, x_2 \in G \) such that \( a_1 = x_1^n \) and \( a_2 = x_2^n \).
- We need to show that \( a_1 a_2 \in G_n \). Since \( G \) is abelian, we have:
  \[
  a_1 a_2 = x_1^n x_2^n = (x_1 x_2)^n.
  \]
- Because \( x_1 x_2 \in G \) (since \( G \) is a group and closed under the group operation), we see that \( a_1 a_2 \in G_n \).

#### Step 3: Closure under inverses
- Let \( a \in G_n \). By definition, there exists \( x \in G \) such that \( a = x^n \).
- We need to show that \( a^{-1} \in G_n \). Since \( G \) is abelian and the inverse of an element in a group is also in the group, we have:
  \[
  a^{-1} = (x^n)^{-1} = (x^{-1})^n.
  \]
- Since \( x^{-1} \in G \), it follows that \( a^{-1} \in G_n \).

### Conclusion:
Since \( G_n \) 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, \( G_n \) is a subgroup of \( G \).

Would you like more details on any of these steps?

Here are 5 related questions:
1. What is the definition of a subgroup in group theory?
2. Why is the closure under inverses important in proving a subgroup?
3. How does the property of \( G \) being abelian simplify this proof?
4. Can we generalize this proof to non-abelian groups?
5. What would change in this proof if \( G_n \) was not defined with powers of \( n \)?

**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.

Let G be an abelian group. For a fixed positive integer n, let G_n = {a ∈ G | a = x^n for some x ∈ G}. Prove that G_n is a subgroup of G.

Learn how to prove that the set G_n = {a ∈ G | a = x^n for some x ∈ G} forms a subgroup of an abelian group G. This proof involves verifying the identity element, closure under the group operation, and closure under inverses. Suitable for undergraduate students studying group theory.

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