Problem 2
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 \(G\).) \(G=\langle\mathbf{R},+\rangle, H=\\{\log n: n \in \mathbb{Z}, n>0\\} . \quad H\) is \(\square\) is not \(\square\) a subgroup of \(G\)
Step-by-Step Solution
Verified Answer
The set \(H\) is not a subgroup of \(G\) because it does not contain inverse elements.
1Step 1: Verify Closure Property
A subgroup must be closed under the group operation, which in this case is addition. Take two arbitrary elements of \(H\), say \(\log n\) and \(\log m\) where \(n, m \in \mathbb{Z}\) and \(n, m > 0\). Check if \(\log n + \log m\) is in \(H\). Since \(\log n + \log m = \log(nm)\), if \(nm\) is a positive integer, \(\log(nm)\) is in \(H\). The product of two positive integers is a positive integer, so \(\log n + \log m \in H\). Hence, \(H\) is closed under addition.
2Step 2: Check for Identity Element
The identity element of \(G\), the group of real numbers under addition, is \(0\). Check if the identity element \(0\) belongs to \(H\). We need \(\log n = 0\). This happens when \(n = 1\) as \(\log 1 = 0\). Since \(1\) is a positive integer, \(0 \in H\). Thus, \(H\) has the identity element of \(G\).
3Step 3: Verify Inverse Elements
For every element \(\log n \in H\), there must be an element \(-\log n\) in \(H\) such that their sum is \(0\). The inverse of \(\log n\) is \(-\log n = \log(n^{-1})\). For \(-\log n\) to be in \(H\), \(n^{-1}\) must be a positive integer. However, \(n^{-1}\) is not an integer for \(n > 1\). Thus, inverses do not always exist in \(H\), failing the subgroup test.
Key Concepts
Closure PropertyIdentity ElementInverse ElementsReal Numbers Under Addition
Closure Property
The closure property is one of the key conditions for a set to be a subgroup under a particular operation. For a set like \(H\) to be closed under an operation such as addition, combining any two elements of \(H\) using the operation must result in another element of \(H\). This ensures the operation, when applied to any elements of the set, doesn’t produce elements outside of the original set.
In this exercise, we have \(H = \{\log n : n \in \mathbb{Z}, n > 0\}\), and our operation is addition. To check for closure, consider two arbitrary elements: \(\log n\) and \(\log m\). Adding these gives us \(\log n + \log m = \log(nm)\).
This result must also be a part of \(H\). Since both \(n\) and \(m\) are positive integers, their product \(nm\) is also a positive integer. Therefore, \(\log(nm)\) belongs to \(H\), confirming that \(H\) is closed under addition.
In this exercise, we have \(H = \{\log n : n \in \mathbb{Z}, n > 0\}\), and our operation is addition. To check for closure, consider two arbitrary elements: \(\log n\) and \(\log m\). Adding these gives us \(\log n + \log m = \log(nm)\).
This result must also be a part of \(H\). Since both \(n\) and \(m\) are positive integers, their product \(nm\) is also a positive integer. Therefore, \(\log(nm)\) belongs to \(H\), confirming that \(H\) is closed under addition.
Identity Element
The identity element is an integral part of group theory. It is a special element of a group that leaves any element of the group unchanged when combined with it under the group operation. For the group of real numbers under addition, that element is \(0\).
To verify if \(H\) contains this identity element, we check whether \(0\) can be expressed using elements of \(H\). Simply, we need an expression like \(\log n = 0\), and the solution to this is \(n = 1\) because \(\log 1 = 0\). Since \(1\) is a positive integer, \(0\) is indeed in \(H\).
This confirms that \(H\) satisfies the identity element condition of being a subgroup, as it contains the identity element of \(G\).
To verify if \(H\) contains this identity element, we check whether \(0\) can be expressed using elements of \(H\). Simply, we need an expression like \(\log n = 0\), and the solution to this is \(n = 1\) because \(\log 1 = 0\). Since \(1\) is a positive integer, \(0\) is indeed in \(H\).
This confirms that \(H\) satisfies the identity element condition of being a subgroup, as it contains the identity element of \(G\).
Inverse Elements
An important aspect of subgroup qualification is the existence of inverse elements. For every element in the subgroup, there must be another element that, when combined with the original, results in the identity element. In terms of addition, for each element \(\log n\) in \(H\), there needs to be a \(-\log n\) such that \(\log n + (-\log n) = 0\).
The mathematically derived inverse of \(\log n\) would be \(-\log n = \log(n^{-1})\). This would mean that \(n^{-1}\) must also be a positive integer, which is a requirement \(H\) must fulfill to provide inverse elements. However, here lies the problem: for any integer \(n > 1\), \(n^{-1}\) is not an integer, because the reciprocal of integers greater than 1 are fractions.
This finding reveals that \(-\log n\) is not generally an element of \(H\), meaning \(H\) does not satisfy this subgroup condition.
The mathematically derived inverse of \(\log n\) would be \(-\log n = \log(n^{-1})\). This would mean that \(n^{-1}\) must also be a positive integer, which is a requirement \(H\) must fulfill to provide inverse elements. However, here lies the problem: for any integer \(n > 1\), \(n^{-1}\) is not an integer, because the reciprocal of integers greater than 1 are fractions.
This finding reveals that \(-\log n\) is not generally an element of \(H\), meaning \(H\) does not satisfy this subgroup condition.
Real Numbers Under Addition
The real numbers under addition, denoted \(\langle \mathbb{R}, + \rangle \), form a well-known mathematical group. Here, we use addition as the operation across the set of all real numbers. This makes each real number and its negative exist within the system, always producing another real number when added.
The group is characterized by several properties:
The group is characterized by several properties:
- Closure: Adding any two real numbers always results in another real number.
- Identity Element: The real number \(0\) acts as the identity, leaving any real number unchanged when added to it.
- Inverse Elements: Every real number \(a\) has an inverse \(-a\), such that \(a + (-a) = 0\).
Other exercises in this chapter
Problem 2
Let \(G\) be an abelian group. Let \(n\) be a fixed integer, and let \(H=\left\\{x \in G: x^{n}=e\right\\}\). Prove that \(H\) is a subgroup of \(G\).
View solution Problem 2
In each of the following, show that \(H\) is a subgroup of \(G\). \(G=\langle\mathscr{F}(\mathbb{R}),+\rangle, H=\\{f \in \mathscr{F}(\mathbb{R}): f(-x)=-f(x)\\
View solution Problem 3
Describe the subgroup of \(\mathbb{Z}_{12}\) generated by 6 and \(9 .\)
View solution Problem 3
By the center of a group \(G\) we mean the set of all the elements of \(G\) which commute with every element of \(G\), that is, $$ C=\\{a \in G: a x=x a \text {
View solution