Verifying G_n as a Subgroup Using the Subgroup Test in Abelian Groups

In the realm of group theory, understanding the properties of subgroups is fundamental. Particularly, verifying that G_n {a in G mid a X^n text{ for some } X in G } is a subgroup of an abelian group G is a crucial step in this theory.

Understanding the Group and Subgroup Concepts

First, let's clarify the concepts involved. An abelian group, also known as a commutative group, is a set G with a binary operation (multiplication) that is closed, associative, has an identity element, and for which each element has an inverse. The requirement that G_n is a subgroup of G involves verifying the following criteria:

Closure Under the Group Operation

For any two elements in G_n, their product should also be in G_n. Given a, b in G_n, we can express them as a X^n and b Y^n for some X, Y in G.

Step 1: Verification of Closure

We need to check if a cdot b in G_n:

Let a, b in G_n. Then, we have:a  X^n and b  Y^nTherefore,a cdot b  X^n cdot Y^nSince G is abelian, we can rewrite this as:a cdot b  (X cdot Y)^nLet Z  X cdot Y. Then Z in G and a cdot b  Z^n. Therefore, a cdot b in G_n.

Existence of the Identity Element

The identity element e of the group G must also be in G_n. Given that e is the identity, we can express it as:

e  0^n

where 0 in G is the identity element in the abelian group. Thus, e in G_n.

Existence of Inverses

For every element in G_n, its inverse should also be in G_n. Given a in G_n, we can express it as a X^n for some X in G. The inverse of a is:

a^{-1}  (X^n)^{-1}  (X^{-1})^n

Let Y X^{-1}. Since Y in G, we have:

a^{-1}  Y^n

Thus, a^{-1} in G_n.

The Subgroup Test Simplified

To further simplify the verification process, we can use the subgroup test. According to this test, G_n is a subgroup of G if and only if for any a, b in G_n, the product a cdot b^{-1} in G_n.

Verification Using the Subgroup Test

Let a, b in G_n, and express them as a X^n and b Y^n. We need to check if a cdot b^{-1} in G_n:

a cdot b^{-1}  X^n cdot (Y^n)^{-1}  X^n cdot Y^{-n}  X cdot Y^{-1}^n in G_n

This implies that G_n leq G if the abelian nature of G allows us to rearrange the elements as needed.

Conclusion

By verifying the closure under the group operation, the presence of the identity element, and the existence of inverses, we have demonstrated that G_n is a subgroup of G. Moreover, the subgroup test provides a concise and elegant way to confirm this relationship.