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Chapter 16: Problem 2

Let \(\alpha: \mathscr{F}(\mathbb{R}) \rightarrow \mathbb{R}\) be defined by \(\alpha(f)=f(1)\) and let \(\beta: \mathscr{F}(\mathbb{R}) \rightarrow \mathbb{R}\) be defined by \(\beta(f)=f(2)\) Let \(J\) be the set of all the functions from \(\mathbb{R}\) to \(\mathbb{R}\) whose graph passes through the point \((1,0)\) and let \(K\) be the set of all the functions whose graph passes through \((2,0)\). Use the \(\mathrm{FHT}\) to prove that \(\mathbb{R} \cong \mathscr{F}(\mathbb{R}) / J\) and \(\mathbb{R} \cong \mathscr{F}(\mathbb{R}) / K\).

Short Answer

Expert verified
\( \mathbb{R} \cong \mathscr{F}(\mathbb{R}) / J \) and \( \mathbb{R} \cong \mathscr{F}(\mathbb{R}) / K \) by FHT.

Step by step solution

01

Understanding the Fundamental Homomorphism Theorem (FHT)

The Fundamental Homomorphism Theorem states that if \( \phi: G \rightarrow H \) is a group homomorphism with kernel \( \ker(\phi) \), then \( G/\ker(\phi) \) is isomorphic to the image of \( \phi \). In symbols, \( G/\ker(\phi) \cong \phi(G) \). Here, we will apply this theorem to function spaces and the given homomorphisms.
02

Define the Homomorphism for \( \alpha \)

Consider the function \( \alpha: \mathscr{F}(\mathbb{R}) \rightarrow \mathbb{R} \) defined as \( \alpha(f) = f(1) \). This is a surjective homomorphism since every real number can be obtained by evaluating some function at 1. The kernel of \( \alpha \) is the set of functions \( f \) such that \( f(1) = 0 \), which precisely defines the set \( J \).
03

Application of the FHT to \( \alpha \)

The image of \( \alpha \) is all of \( \mathbb{R} \) because any real number can be obtained as \( \alpha(f) = f(1) \). By the Fundamental Homomorphism Theorem, we have \( \mathscr{F}(\mathbb{R}) / J \cong \alpha(\mathscr{F}(\mathbb{R})) = \mathbb{R} \).
04

Define the Homomorphism for \( \beta \)

Consider the function \( \beta: \mathscr{F}(\mathbb{R}) \rightarrow \mathbb{R} \) defined as \( \beta(f) = f(2) \). Similarly to \( \alpha \), this is a surjective homomorphism. The kernel of \( \beta \) is the set of functions \( f \) such that \( f(2) = 0 \), which precisely defines the set \( K \).
05

Application of the FHT to \( \beta \)

The image of \( \beta \) is all of \( \mathbb{R} \) because any real number can be obtained as \( \beta(f) = f(2) \). By the Fundamental Homomorphism Theorem, we have \( \mathscr{F}(\mathbb{R}) / K \cong \beta(\mathscr{F}(\mathbb{R})) = \mathbb{R} \).

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Key Concepts

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

Group Homomorphism
A group homomorphism is a function between two groups that preserves the group structure. In simple terms, if you perform a group operation on elements in the first group, a homomorphism ensures the same operation on the transformed elements in the second group. For example, if \( \phi : G \rightarrow H \) is a homomorphism, then for any two elements \( a, b \in G \), the property \( \phi(a * b) = \phi(a) \cdot \phi(b) \) holds, where \( * \) and \( \cdot \) represent group operations in \( G \) and \( H \) respectively.
- Homomorphisms maintain the group operation.- They are functions that "map" groups succinctly.- They help identify similarities between different algebraic structures.
In our problem, \( \alpha(f) = f(1) \) and \( \beta(f) = f(2) \) are homomorphisms from the space of functions \( \mathscr{F}(\mathbb{R}) \) to \( \mathbb{R} \), illustrating how evaluations at specific points can act as group homomorphisms.
Kernel
The kernel of a group homomorphism is the set of elements in the domain that map to the identity element in the codomain. For a homomorphism \( \phi : G \rightarrow H \), the kernel is defined as \( \ker(\phi) = \{ g \in G : \phi(g) = e_H \} \), where \( e_H \) is the identity element in \( H \).
- The kernel helps determine injectivity: \( \phi \) is injective if \( \ker(\phi) \) is the trivial group.- It plays a crucial role in the Fundamental Homomorphism Theorem.
In our example:
  • The kernel of \( \alpha \) is all functions with \( f(1) = 0 \), forming set \( J \).
  • Similarly, the kernel of \( \beta \) is all functions with \( f(2) = 0 \), forming set \( K \).
Here, the kernels represent constraints where functions pass through specific points on the graph.
Isomorphism
An isomorphism is a bijective homomorphism. That is, it is both injective (no element in the codomain is mapped to by more than one element in the domain) and surjective (every element in the codomain is mapped to by some element in the domain). Isomorphisms establish an equivalence between groups, implying they have the same structure.
- Isomorphic groups are essentially 'the same' in terms of their group structure.- The notation \( G \cong H \) denotes that group \( G \) is isomorphic to group \( H \).
Through the Fundamental Homomorphism Theorem, after recognizing elements in the kernel, we express the original group quotiented by the kernel as isomorphic to the image of the homomorphism.
In the given exercise, \( \mathscr{F}(\mathbb{R}) / J \cong \mathbb{R} \) and \( \mathscr{F}(\mathbb{R}) / K \cong \mathbb{R} \), confirming that these quotients are structurally identical to the real numbers.
Function Spaces
Function spaces are collections of functions that act as 'elements' in a larger set. They allow for the study of functions in an algebraic structure akin to how numbers are treated in algebra.
- One common function space is \( \mathscr{F}(\mathbb{R}) \), representing all real-valued functions of a real variable.- Function spaces can support operations like addition of functions and scalar multiplication.
In this problem, we examine \( \mathscr{F}(\mathbb{R}) \), which denotes functions mapping real numbers to real numbers. Function spaces can be quite vast and diverse, containing all kinds of functions, from simple polynomials to more complex transcendental functions. Evaluating function spaces under homomorphisms like \( \alpha \) and \( \beta \) gives insightful structures captured by the functions' behaviors at specific points.

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Most popular questions from this chapter

Let \(G\) be a finite group, and \(K\) a \(p\)-Sylow subgroup of \(G\). Let \(X\) be the set of all the conjugates of \(K\). See Exercise M2. If \(C_{1}, C_{2} \in X\), let \(C_{1} \sim C_{2}\) iff \(C_{1}=a C_{2} a^{-1}\) for some \(a \in K\). Prove that \(\sim\) is an equivalence relation on \(X\).

If \(G\) is a group and \(p\) is any prime divisor of \(|G|\), it will be shown here that \(G\) has at least one element of order \(p\). This has already been shown for abelian groups in Chapter 15 , Exercise \(\mathrm{H} 4\). Thus, assume here that \(G\) is not abelian. The argument will proceed by induction; thus, let \(|G|=k\), and assume our claim is true for any group of order less than \(k\). Let \(\mathbf{C}\) be the center of \(G\), let \(C_{a}\) be the centralizer of \(a\) for each \(a \in G\), and let \(k=c+k_{s}+\cdots+k_{t}\) be the class equation of \(G\), as in Chapter 15, Exercise G2. Prove : If \(p\) is a factor of \(\left|C_{a}\right|\) for any \(a \in G\), where \(a \notin \mathbf{C}\), we are done. (Explain why.)

Let \(f\) be a homomorphism from \(G\) onto \(H\) with kernel \(K\) : $$ G \underset{K}{R} H $$ If \(S\) is any subgroup of \(H\), let \(S^{*}=\\{x \in G: f(x) \in S\\}\). Prove: $$ S \cong S^{*} / K $$

Let \(G\) be a group; let \(H\) and \(K\) be subgroups of \(G\), with \(H\) a normal subgroup of \(G\). Prove the following : By the FHT, \(H /(H \cap K) \cong H K / H\). (This is referred to as the first isomorphism theorem.)

As a provisional definition, let us call a finite abelian group "decomposable" if there are elements \(a_{1}, \ldots, a_{n} \in G\) such that: (D1) For every \(x \in G\), there are integers \(k_{1}, \ldots, k_{n}\) such that \(x=a_{1}^{k_{1}} a_{2}^{k_{2}} \cdots a_{n}^{k_{n}}\). (D2) If there are integers \(l_{1}, \ldots, l_{n}\) such that \(a_{1}^{l_{1}} a_{2}^{l_{2}} \cdots a_{n}^{l_{n}}=e\) then \(a_{1}^{l_{1}}=a_{2}^{t_{2}}=\cdots\) \(=a_{n}^{l_{n}}=e\) If (D1) and (D2) hold, we will write \(G=\left[a_{1}, a_{2}, \ldots, a_{n}\right]\). Prove: \(G \cong\left\langle a_{1}\right\rangle \times G^{\prime}\). Conclude that \(G \cong\left\langle a_{1}\right\rangle \times\left\langle a_{2}\right\rangle \times \cdots \times\left\langle a_{n}\right\rangle .\)
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