Proving H is a Subgroup of G According to G's Abelian Property

In the study of group theory, one often encounters the question: given an abelian group (G) and a subset (H) of (G) defined by (H {a in G | a^3 e}), how do we show that (H) is a subgroup of (G)? This article provides a detailed explanation of the conditions required for (H) to be a subgroup and how these conditions can be satisfied.

Understanding the Structure of G

The structure of (G) plays a crucial role in showing that (H) is a subgroup. By the elementary divisor theorem, finite abelian groups can be expressed as a direct sum of cyclic subgroups of the form (mathbb{Z}/p^t), where (p) is a prime and there can be multiple repetitions of the same prime. For instance, if (p 3) appears in the decomposition of (G), the group (H) can be viewed as consisting of the sums (mathbb{Z}/3) one copy for each time (p 3) appears. This is enough to conclude that (H) is a subgroup of (G).

Definition and Structure of H

For clarity, we define (H {a in G | a^3 e}). This set (H) consists of all elements in (G) whose order divides 3. This definition implies that (H) is the set generated by all elements in (G) that satisfy the property (a^3 e).

Verifying the Subgroup Criteria

Closure

To prove (H) is a subgroup, we need to verify the subgroup criteria. First, consider the closure property. We must show that the product of any two elements in (H) is also in (H). Let (a, b in H). Then, we have (a^3 e) and (b^3 e). We can write the product (ab) and check if ((ab)^3 e). [(ab)^3 a^3b^3 ee e]

Thus, (ab in H). This demonstrates closure.

Associativity

Since (H) is a subset of (G) and (G) is a group, the associative law is inherited by (H). This means that for any three elements (a, b, c in H), [(ab)c a(bc)]

This is straightforward and does not require further proof.

Identity Element

The identity element (e) of (G) is also in (H) because (e^3 e). Therefore, the identity element of (G) acts as the identity element in (H) as well.

Inverse Elements

The final criterion to check is the existence of inverses. For any (a in H), there exists an element (a^{-1} in G) such that (a a^{-1} e). Since (a^3 e), we have: [(a^{-1})^3 e]

This shows that (a^{-1} in H). Thus, every element in (H) has an inverse in (H).

Conclusion

By verifying that (H) satisfies the subgroup criteria, we have shown that (H {a in G | a^3 e}) is indeed a subgroup of (G). The fact that (G) is abelian is not crucial here, as (H) will remain a subgroup regardless of whether (G) itself is abelian. These observations are compactly captured by the elementary divisor theorem, providing a deeper understanding of the structure of finite abelian groups.