Problem 4 - A Book of Abstract Algebra

Question

In each of the following, determine whether or not \(H\) is a subgroup of \(G\). (Assume that the operation of \(H\) is the same as that of \(G\).) \(G=\left\langle\mathbb{R}^{*}, \cdot\right\rangle, H=\left\\{2^{n} 3^{m}: m, n \in \mathbb{Z}\right\\} . \quad H\) is \(\square \quad\) is not \(\square \quad\) a subgroup of \(G\)

Step-by-Step Solution

Answer

H is a subgroup of G.

Step 1: Verify the existence of identity element in H

First, identify the identity element of the group \( G = \langle \mathbb{R}^*, \cdot \rangle \). In this case, the identity element is 1 because multiplying any real number by 1 leaves it unchanged. Check if the identity element is in \( H \). Set \( n = 0 \) and \( m = 0 \) in the expression \( 2^n3^m \). This gives \( 2^0 3^0 = 1 \), which is indeed in \( H \). Therefore, \( H \) contains the identity element.

Step 2: Verify the closure property of H

For closure, suppose \( a, b \in H \). We need to show that the product \( ab \in H \). Assume \( a = 2^{n_1}3^{m_1} \) and \( b = 2^{n_2}3^{m_2} \), where \( n_1, n_2, m_1, m_2 \in \mathbb{Z} \). Their product is \( ab = 2^{n_1}3^{m_1} \cdot 2^{n_2}3^{m_2} = 2^{n_1+n_2} 3^{m_1+m_2} \). Since \( n_1+n_2 \) and \( m_1+m_2 \) are integers, \( ab \in H \). Thus, \( H \) is closed under the group operation.

Step 3: Verify the presence of inverses in H

We must show that for each \( a \in H \), there is an inverse \( a^{-1} \in H \). If \( a = 2^n 3^m \), then \( a^{-1} = 2^{-n} 3^{-m} \). This expression is still in the form \( 2^x 3^y \) where \( x = -n \) and \( y = -m \), which are integers. Therefore, \( a^{-1} \in H \). Hence, \( H \) has inverses for all its elements.

Step 4: Conclusion about subgroup criteria

Since \( H \) contains the identity, is closed under multiplication, and contains inverses for all its elements, \( H \) satisfies all the subgroup criteria. Therefore, \( H \) is a subgroup of \( G \).

Key Concepts

Subgroup Criterion

To determine whether a set is a subgroup of a given group, it must satisfy certain criteria. These are known as the subgroup criteria. Specifically, for a subset \( H \) of a group \( G \), \( H \) must have:

• An identity element that is the same as in \( G \).
• Closure under the group operation.
• An inverse for every element in \( H \).

These conditions ensure that \( H \) behaves like a group on its own under the operation defined by \( G \). If all these conditions are fulfilled, \( H \) can be considered a subgroup of \( G \). It essentially means that within the larger group \( G \), the subset \( H \) has all the necessary properties to form a smaller group by itself. This forms the basis of subgroup detection in group theory.

Closure Property

In group theory, the closure property is one of the essential conditions that a subset must satisfy to be considered a subgroup. The closure property ensures that performing the group operation (like multiplication or addition) on any two elements within the subset results in another element of the same subset.For example, if we have a group \( G = \langle\mathbb{R}^*, \cdot\rangle\) and a subset \( H = \{2^n 3^m : n, m \in \mathbb{Z}\} \), checking closure involves confirming that for any \( a = 2^{n_1} 3^{m_1} \) and \( b = 2^{n_2} 3^{m_2} \) in \( H \), their product \( ab = 2^{n_1+n_2} 3^{m_1+m_2} \) is also in \( H \).Since the exponents \( n_1+n_2 \) and \( m_1+m_2 \) are integers (closed under addition), the closure property is satisfied. Thus, the operation never takes us out of the set \( H \), keeping the subset consistent with group behavior.

Identity Element

An identity element in the context of group theory is an element that, when used in a group operation with any element of the group, leaves that element unchanged. This identity element is crucial for a subset to qualify as a subgroup.In the group \( G = \langle \mathbb{R}^*, \cdot \rangle \), the identity element is 1, because multiplying any number by 1 does not change the number. To check if a subset \( H \) contains this identity, we need to verify that 1 can be expressed in terms of the subset's notation.For \( H = \{ 2^n 3^m : n, m \in \mathbb{Z} \} \), setting \( n = 0 \) and \( m = 0 \) gives \( 2^0 3^0 = 1 \). Since 1 is present in \( H \), it meets the identity element criterion, allowing us to further validate \( H \) as potentially being a subgroup of \( G \).

Inverse Element

Having an inverse element for each element in a group is a vital property that allows the group to maintain its structure. In simple terms, for every element \( a \) in the group, there should be an element \( a^{-1} \) such that the product \( a \cdot a^{-1} = 1 \).To check for inverses in a subset \( H \) of a group \( G \), every element of \( H \) should be able to produce an inverse that is also in \( H \). For instance, consider an element \( a = 2^n 3^m \) within the subset \( H = \{ 2^n 3^m : n, m \in \mathbb{Z} \} \). The inverse would be \( a^{-1} = 2^{-n} 3^{-m} \).Since \( a^{-1} \) also follows the structure of \( 2^x 3^y \), where \( x \) and \( y \) are integers (namely \( x = -n \) and \( y = -m \)), it is evident that this inverse is within \( H \). This satisfies the inverse condition essential for \( H \) to be a subgroup of \( G \), completing the subgroup criteria.