Problem 3 - A Book of Abstract Algebra

Question

Let \(G\) be an abelian group. Let \(H=\left\\{x \in G: x=y^{2}\right.\) for some \(\left.y \in G\right\\}\), that is, let \(H\) be the set of all the elements of \(G\) which have a square root. Prove that \(H\) is a subgroup of \(G\).

Step-by-Step Solution

Answer

H is a subgroup of G because it includes the identity, is closed under operation, and contains inverses.

Step 1: Verify Identity Element

We need to show that the identity element of the group \(G\), denoted as \(e\), belongs to \(H\). Since \(e = e^2\), where \(e\) itself is an element of \(G\), the identity element \(e\) is a perfect square in \(G\). Thus, \(e \in H\).

Step 2: Show Closure Under Group Operation

Let's consider two elements \(a, b \in H\). By the definition of \(H\), there exist \(x, y \in G\) such that \(a = x^2\) and \(b = y^2\). We need to show that the product \(ab\) is also in \(H\). Since \(G\) is abelian, \(ab = x^2y^2 = (xy)^2\). Therefore, \(ab\) is a square of the element \(xy\) in \(G\), and hence \(ab \in H\).

Step 3: Prove Inverses

Take any \(a \in H\). By definition, there exists an element \(x \in G\) such that \(a = x^2\). The inverse of \(a\) in the group \(G\) is \(a^{-1}\). In an abelian group, \( (x^{-1})^2 = (x^2)^{-1} = a^{-1} \). Thus, \(a^{-1}\) is also a square of the element \(x^{-1}\), and hence \(a^{-1} \in H\).

Step 4: Conclusion: H is a Subgroup

Since we have verified that \(H\) contains the identity element, is closed under group operation, and contains inverses for every element, it follows that \(H\) satisfies all subgroup criteria. Therefore, \(H\) is a subgroup of \(G\).

Key Concepts

Subgroup

In the context of group theory, a subgroup is a smaller group contained within a larger group, sharing the same operation. For a subset to qualify as a subgroup, it must itself satisfy the group properties. This means:

• Containment of the identity element.
• Closure under the group operation.
• Inclusion of inverses for each of its elements.

Identifying whether a set forms a subgroup is crucial, as it allows us to understand the structure and behavior of the original group. In our exercise, the set \(H\) comprises all elements of \(G\) with a square root. To determine if \(H\) is a subgroup, we ensure it follows these core group properties. If it does, then \(H\) inherits the group nature of \(G\), maintaining its algebraic properties.

Group Identity Element

The identity element is fundamental in group theory. It is the unique element in a group that, when combined with any other element of the group, returns that element. For a group \(G\), the identity is typically denoted as \(e\).
In simple terms:

• For any element \(a\), the equation \(a \cdot e = a\) holds true.
• In an abelian group, \(e\) satisfies \(e \cdot a = a\).

In the exercise, the identity element \(e\) of \(G\) must also be in \(H\) because \(e = e^2\). Thus, \(e\) is a perfect square in \(G\), which ensures that \(H\) covers the identity requirement necessary to be a subgroup.

Closure Property

The closure property means that when you apply the group operation to any two elements within the subgroup, the result is also within that subgroup. For a set to be a subgroup:

• This property must always hold true.
• Without closure, the set cannot maintain internal consistency under the group operation.

In our current example, since \(G\) is abelian, the multiplication of two elements \(a\) and \(b\) from \(H\) remains within \(H\). This was demonstrated as \(ab = (xy)^2\). Thanks to closure, \(ab\) forms a perfect square just like \(a\) and \(b\), establishing \(H\)'s reliability under group operations.

Inverse Element

Each element in a group must have an inverse within the group itself. The inverse of an element \(a\) is another element \(a^{-1}\) which satisfies:

• \(a \cdot a^{-1} = e\), where \(e\) is the identity element.

For \(H\) to be a subgroup, each of its elements must also have an inverse contained within \(H\). Our exercise shows that for each \(a = x^2\) in \(H\), the element \(a^{-1}\), calculated as \((x^{-1})^2\), lies in \(H\) due to the abelian nature of \(G\). This ensures every element in \(H\) is paired with an appropriate inverse, fulfilling one of the vital subgroup conditions.