Problem 5 - A Book of Abstract Algebra

Question

Let \(G\) be an abelian group. Let \(H\) be a subgroup of \(G\), and let \(K\) consist of all the elements \(x\) in \(G\) such that some power of \(x\) is in \(H\). That is, \(K=\left\\{x \in G:\right.\) for some integer \(\left.n>0, x^{n} \in H\right\\}\). Prove that \(K\) is a subgroup of \(G\).

Step-by-Step Solution

Answer

The set \(K\) is a subgroup of \(G\) because it is closed, contains the identity, and has inverses.

Step 1: Verify Closure

To prove that \(K\) is a subgroup of \(G\), we need to first verify closure. Assume \(a, b \in K\). Then by definition, there exist integers \(m, n > 0\) such that \(a^m \in H\) and \(b^n \in H\). Since \(H\) is a subgroup and hence closed, the product \((a^m)(b^n)\) is in \(H\). Since \(G\) is abelian, \((ab)^{mn} = a^m b^n \in H\). Thus, \(ab \in K\), proving closure.

Step 2: Verify Existence of the Identity Element

The identity element \(e\) of a group \(G\) satisfies \(e^1 = e\), which trivially belongs to \(H\) since \(H\) is a subgroup of \(G\). Thus, \(e \in K\).

Step 3: Verify Existence of Inverses

For any \(a \in K\), there exists an integer \(m > 0\) such that \(a^m \in H\). Because \(H\) is a subgroup, it contains the inverse of any element within it. Thus, \((a^m)^{-1} = a^{-m} \in H\). Given that \((a^{-1})^m = a^{-m}\), the inverse \(a^{-1}\) must be in \(K\). This shows that \(K\) contains inverses for all its elements.

Step 4: Conclude that K is a Subgroup

Having verified closure, the presence of the identity element, and the existence of inverses, we conclude that \(K\) meets the necessary conditions to be a subgroup of \(G\). Hence, \(K\) is a subgroup of \(G\).

Key Concepts

Subgroup

In abstract algebra, a subgroup is a subset of a group that itself is a group under the same operation as the original group. If you have a group \( G \) and a subset \( H \) of \( G \), then \( H \) is a subgroup if it satisfies certain conditions:

• **Closure**: For any two elements \( a, b \) in \( H \), the product \( ab \) is also in \( H \).
• **Identity**: The identity element of \( G \) is in \( H \).
• **Inverses**: For every element \( a \) in \( H \), the inverse \( a^{-1} \) is also in \( H \).

Checking these properties ensure that \( H \) can stand alone as a group, with all necessary operations behaving just like those in \( G \).To determine if a set like \( K \) is a subgroup, each of these properties must be verified.

Group Closure

Closure is a fundamental property that must be checked to determine if a subset of a group is a subgroup. In the context of groups, closure means if you perform the group operation on any two elements of the subset, the result is still within the subset.
For example, if \( a, b \in K \) and \( K \) is a candidate for a subgroup, taking two elements from \( K \) and combining them using the group operation must produce another element that is still in \( K \). In the case of abelian groups, where the operation is typically addition, closure demands that for any two elements \( a \) and \( b \), \( a + b \) should remain in \( K \).
This was shown in the solution since for any elements \( a \) and \( b \) in \( K \), their product \( ab \) is in \( K \) due to the properties of an abelian group, specifically using powers of elements that lead to results in \( H \). This essential step verifies the closure property for subgroup conditions.

Identity Element

The identity element is key in defining a group structure. Every group must have an identity element such that combining any element in the group with the identity leaves the element unchanged.
In the more straightforward terms, if \( e \) is the identity of a group \( G \), then for any element \( a \) in \( G \), the operation \( ae = ea = a \). If a subset like \( K \) is to be a subgroup, then \( e \) must also be in \( K \).
In the given solution example, since \( H \) is a subgroup of \( G \), and every subgroup must contain the identity of the parent group, \( e \) is naturally part of \( H \) and also \( K \). The presence of \( e \) ensures \( K \) has an identity element, satisfying this subgroup criterion.

Inverses in Groups

For a set to be a subgroup, each element must have an inverse that is also in the set. In the context of groups, the inverse of an element \( a \) in a group satisfies the equation \( aa^{-1} = e \), where \( e \) is the identity element.
To show \( K \) is a subgroup, it must be demonstrated that for every \( a \) in \( K \), there is an element \( a^{-1} \) such that \( a a^{-1} = e \), and importantly, \( a^{-1} \) is within \( K \) too.

• When \( a \) belongs to \( K \), it implies that some power of \( a \) belongs to \( H \), a subgroup where every element has an inverse.
• Thus, the inverse \( a^{-1} \) must match this condition, derived from how these elements are structured within \( H \). This happens because, in subgroups, the inverse of an element always exists within the subgroup.

Hence, ensuring that \( K \) holds the inverse property fortifies its stance as a subgroup of \( G \).