Math Problem Statement

solve this

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.

Solution

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

Steps to prove that GnG_n is a subgroup of GG:

To prove that GnG_n is a subgroup of GG, we need to show that:

  1. The identity element of GG belongs to GnG_n.
  2. GnG_n is closed under the group operation.
  3. GnG_n is closed under taking inverses.

Step 1: Identity element in GnG_n

  • Let ee be the identity element of GG.
  • Since GG is abelian, for any xGx \in G, we have e=x0e = x^0 for n=0n = 0.
  • Therefore, eGne \in G_n, so the identity element is in GnG_n.

Step 2: Closure under the group operation

  • Let a1,a2Gna_1, a_2 \in G_n, so by definition, there exist x1,x2Gx_1, x_2 \in G such that a1=x1na_1 = x_1^n and a2=x2na_2 = x_2^n.
  • We need to show that a1a2Gna_1 a_2 \in G_n. Since GG is abelian, we have: a1a2=x1nx2n=(x1x2)n.a_1 a_2 = x_1^n x_2^n = (x_1 x_2)^n.
  • Because x1x2Gx_1 x_2 \in G (since GG is a group and closed under the group operation), we see that a1a2Gna_1 a_2 \in G_n.

Step 3: Closure under inverses

  • Let aGna \in G_n. By definition, there exists xGx \in G such that a=xna = x^n.
  • We need to show that a1Gna^{-1} \in G_n. Since GG is abelian and the inverse of an element in a group is also in the group, we have: a1=(xn)1=(x1)n.a^{-1} = (x^n)^{-1} = (x^{-1})^n.
  • Since x1Gx^{-1} \in G, it follows that a1Gna^{-1} \in G_n.

Conclusion:

Since GnG_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, GnG_n is a subgroup of GG.

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 GG being abelian simplify this proof?
  4. Can we generalize this proof to non-abelian groups?
  5. What would change in this proof if GnG_n was not defined with powers of nn?

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