Problem 4
Question
We have seen in the text that \(\mathbb{R}\) is isomorphic to \(\mathbb{R}^{+} .\)Prove that \(\mathbb{R}\) is not isomorphic to \(\mathbb{R}^{*}\) (the multiplicative group of the nonzero real numbers). (HINT: Consider the properties of the number \(-1\) in \(\mathbb{R}^{*}\). Does \(\mathbb{R}\) have any element with those properties?)
Step-by-Step Solution
Verified Answer
\( \mathbb{R} \) cannot be isomorphic to \( \mathbb{R}^{*} \) because \( \mathbb{R}^{*} \) has an element of order 2 (\(-1\)), but \( \mathbb{R} \) does not.
1Step 1: Understanding Isomorphisms
An isomorphism between two groups is a bijective homomorphism. This means each element of the first group can be paired with one and only one element of the second group, and the group operations are preserved.
2Step 2: Considering the Group Properties of \( \mathbb{R} \)
The group \( \mathbb{R} \) (under addition) consists of all real numbers, not including zero. In \( \mathbb{R} \), every element can be added to zero to get itself, and each element has an inverse (its negative). Importantly, integer multiples are well-defined.
3Step 3: Considering the Group Properties of \( \mathbb{R}^{*} \)
The group \( \mathbb{R}^{*} \) (under multiplication) consists of all non-zero real numbers. In this group, every element \( x \) has a multiplicative inverse \( x^{-1} \), and the group operation (multiplication) must respect these inverses.
4Step 4: Investigate \(-1\) in \( \mathbb{R}^{*} \)
In the multiplicative group \( \mathbb{R}^{*} \), the element \(-1\) is unique because its square is 1, i.e., \((-1)\cdot(-1) = 1\). This means that it has order 2, which is a distinct property.
5Step 5: Analyze \( \mathbb{R} \) for Similar Elements
In the additive group \( \mathbb{R} \), no element \( x \) other than 0 has the property that \( 2x = 0 \). In other terms, \( x+x = 0 \) implies \( x = 0 \). Therefore, there is no element of order 2 in \( \mathbb{R} \).
6Step 6: Conclusion on Isomorphism
Since \( \mathbb{R}^{*} \) has an element of order 2 (\(-1\)), but \( \mathbb{R} \) does not have any non-zero element with order 2, no isomorphism can exist between \( \mathbb{R} \) and \( \mathbb{R}^{*} \). This is due to a mismatch in their group structures regarding elements of specific orders.
Key Concepts
Real numbersAdditive groupsMultiplicative groupsElement order in groups
Real numbers
Real numbers, denoted by \( \mathbb{R} \), represent a continuous set of numbers that include all the rational and irrational numbers. This encompasses well-known numbers like 0, positive and negative integers, fractions, and numbers like \( \pi \) and \( e \). Real numbers can be visually represented on a number line, which extends infinitely in both the positive and negative directions.
- Rational Numbers: These can be expressed as the ratio of two integers, for instance, \( \frac{1}{2} \) or 2.
- Irrational Numbers: These cannot be expressed as such a ratio, including numbers like \( \sqrt{2} \) and \( \pi \).
Additive groups
Additive groups are algebraic structures where the group operation is addition. The group \( \mathbb{R} \) under addition includes all real numbers. In this setting, the main properties are:
- **Closure**: The sum of any two real numbers is a real number.
- **Associativity**: For any real numbers \( a, b, \) and \( c,\) the equation \( (a + b) + c = a + (b + c) \) holds.
- **Identity Element**: The number 0 is the identity element because adding 0 to any number leaves it unchanged.
- **Inverse Element**: Every real number \( a \) has a corresponding inverse \( -a \) such that \( a + (-a) = 0 \).
Multiplicative groups
Multiplicative groups focus on multiplication as the group operation. The group \( \mathbb{R}^{*} \) includes all non-zero real numbers. Here, the properties are slightly different from additive groups:
- **Closure**: The product of any two non-zero real numbers is a non-zero real number.
- **Associativity**: The multiplication of real numbers is associative.
- **Identity Element**: The number 1 acts as the identity element because any number multiplied by 1 remains unchanged.
- **Inverse Element**: Each non-zero number \( x \) has a multiplicative inverse \( 1/x \) such that \( x \cdot 1/x = 1 \).
Element order in groups
The order of an element in a group is defined as the smallest positive integer \( n \) such that the \( n \)th power (or sum, in additive groups) of the element equals the identity element of the group. For instance:
- In \( \mathbb{R} \) (under addition), only the zero element has a finite order, specifically order one.
- In \( \mathbb{R}^{*} \) (under multiplication), \(-1\) has an order of 2 because \((-1) \cdot (-1) = 1\), the multiplicative identity.
Other exercises in this chapter
Problem 4
Prove that the following groups are isomorphic. Show that \(f(x, y)=(-1)^{y} x\) is an isomorphism from \(\mathbb{R}^{+} \times \mathbb{Z}_{2}\) to \(\mathbb{R}
View solution Problem 4
If a group \(G\) is generated, say, by \(a, b\), and \(c\), then a set of equations involving \(a, b\), and \(c\) is called a set of defining equations for \(G\
View solution Problem 5
Let \(c\) be a fixed element of \(G .\) Let \(H\) be a group with the same set as \(G\), and with the operation \(x * y=x c y .\) Prove that the function \(f(x)
View solution Problem 5
Prove that \(\mathbb{Z}\) is not isomorphic to \(\mathbb{Q}\).
View solution