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Chapter 9: Problem 4

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?)

Short Answer

Expert verified
\( \mathbb{R} \) cannot be isomorphic to \( \mathbb{R}^{*} \) because \( \mathbb{R}^{*} \) has an element of order 2 (\(-1\)), but \( \mathbb{R} \) does not.

Step by step solution

01

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.
02

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.
03

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.
04

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.
05

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 \( \mathbb{R} \).
06

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 \( \pi \).
The set of real numbers is central to many mathematical concepts, as it allows for the introduction of calculus, analysis, and many other fields. Understanding real numbers is foundational for comprehending more complex structures like groups.
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 \( a + (-a) = 0 \).
In the context of the problem, understanding these properties will help see why certain real numbers in \( \mathbb{R} \) do not behave the same under multiplication, making it impossible for \( \mathbb{R} \) to be isomorphic to \( \mathbb{R}^{*} \).
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 \( x \cdot 1/x = 1 \).
Within \( \mathbb{R}^{*} \), elements have distinct behaviors, such as the number \(-1\), whose square returns the identity (1), a property not shared by elements in typical additive groups. This underlines the difference in group structure and why \( \mathbb{R} \) and \( \mathbb{R}^{*} \) cannot be isomorphic.
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.
This concept of element order is crucial for understanding group isomorphisms. Two groups are isomorphic if their elements can be paired in a way that preserves group structure. Since \( \mathbb{R}^{*} \) has elements of order 2 but \( \mathbb{R} \) does not, this prevents them from being isomorphic. Elements of differing orders between these groups highlight the fundamental differences in their structures.

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Most popular questions from this chapter

If \(G\) is a group, an automorphism of \(G\) is an isomorphism from \(G\) to \(G\). We have seen (Exercise A1) that the identity function \(\varepsilon(x)=x\), is an automorphism of \(G\). However, many groups have other automorphisms besides this obvious one. If \(G\) is any group, and \(a\) is any element of \(G\), prove that \(f(x)=a x a^{-1}\) is an automorphism of \(G\).

Recall that an isomorphism \(f\) from \(G_{1}\) to \(G_{2}\) is a one-to-one correspondence between \(G_{1}\) and \(G_{2}\) satisfying \(f(a b)=f(a) f(b) \cdot f\) matches every element of \(G_{1}\) with a corresponding element of \(G_{2}\). It is important to note that: (i) \(f\) matches the neutral element of \(G_{1}\) with the neutral element of \(G_{2}\). (ii) If \(f\) matches an element \(x\) in \(G_{1}\) with \(y\) in \(G_{2}\), then, necessarily, \(f\) matches \(x^{-1}\) with \(y^{-1}\). That is, if \(x \leftrightarrow y\), then \(x^{-1} \leftrightarrow y^{-1}\). (iii) \(f\) matches a generator of \(G_{1}\) with a generator of \(G_{2}\). The details of these statements are now left as an exercise. Let \(G_{1}\) and \(G_{2}\) be groups, and let \(f: G_{1} \rightarrow G_{2}\) be an isomorphism. If \(G_{1}\) is a cyclic group with generator \(a\), prove that \(G_{2}\) is also a cyclic group, with generator \(f(a)\).

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 \(f(x)=c^{-1} x\) is an isomorphism from \(G\) to \(H\).

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 \(\mathbb{R}^{*}\). (REMARK: To combine elements of \(\mathbb{R}^{+} \times \mathbb{Z}_{2}\), one multiplies first components, adds second components.) Conclude that \(\mathbb{R}^{*} \cong \mathbb{R}^{+} \times \mathbb{Z}_{2}\).

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 \(G\) if these equations completely determine the table of \(G\). (See end of Chapter 5.) If \(G^{\prime}\) is another group, generated by elements \(a^{\prime}, b^{\prime}\), and \(c^{\prime}\) satisfying the same defining equations as \(a, b\), and \(c\), then \(G^{\prime}\) has the same table as \(G\) (because the tables of \(G\) and \(G^{\prime}\) are completely determined by the defining equations, which are the same for \(G\) as for \(G^{\prime}\) ). Consequently, if we know generators and defining equations for two groups \(G\) and \(G^{\prime}\), and if we are able to match the generators of \(G\) with those of \(G^{\prime}\) so that the defining equations are the same, we may conclude that \(G \cong G^{\prime}\). Prove that the following pairs of groups \(G, G^{\prime}\) are isomorphic. \(G=S_{3} ; G^{\prime}=\\{e, a, b, a b, a b a, a b a b\\}\) where \(a^{2}=e, b^{2}=e\), and \(b a b=a b a\).
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