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Chapter 6: Problem 25

Let \(\rho: G \rightarrow G^{\prime}\) be a group homomorphism. Let \(H\) be a subgroup of \(G,\) and let \(\tau: H \rightarrow G^{\prime}\) be the restriction of \(\rho\) to \(H\). Show that \(\tau\) is a group homomorphism and that \(\operatorname{Ker} \tau=\operatorname{Ker} \rho \cap H\).

Short Answer

Expert verified
In this exercise, we have proven two things: 1. 蟿, the restriction of 蟻 to H, is a group homomorphism. We showed that 蟿 preserves the group operation by using the properties of 蟻 as a group homomorphism. 2. The kernel of 蟿 (Ker蟿) is equal to the intersection of the kernel of 蟻 (Ker蟻) and H. We showed this by proving that Ker蟿 is a subset of the intersection of Ker蟻 and H, and that the intersection of Ker蟻 and H is a subset of Ker蟿.

Step by step solution

01

Prove that 蟿 is a group homomorphism

To prove that 蟿 is a group homomorphism, we need to show that it preserves the group operation. That is, for any elements h1 and h2 in H, 蟿(h1 * h2) = 蟿(h1) * 蟿(h2). Since 蟿 is the restriction of 蟻 to H, we have 蟿(h) = 蟻(h) for all h in H. So we can write: 蟿(h1 * h2) = 蟻(h1 * h2) Since 蟻 is a group homomorphism, we know that 蟻(h1 * h2) = 蟻(h1) * 蟻(h2). Thus, 蟿(h1 * h2) = 蟻(h1) * 蟻(h2) = 蟿(h1) * 蟿(h2) This shows that 蟿 preserves the group operation and is therefore a group homomorphism.
02

Show that Ker蟿 is a subset of the intersection of Ker蟻 and H

Recall that Ker蟿 = {h 鈭� H : 蟿(h) = e鈥瞹 and Ker蟻 = {g 鈭� G : 蟻(g) = e鈥瞹, where e鈥� is the identity in G鈥�. Let h be an element in Ker蟿. This means that 蟿(h) = e鈥�, or equivalently, 蟻(h) = e鈥�, since 蟿 is the restriction of 蟻 to H. This implies that h is in Ker蟻. Also, h is an element of H by definition. Thus, h is in the intersection of Ker蟻 and H. Therefore, Ker蟿 鈯� (Ker蟻 鈭� H).
03

Show that the intersection of Ker蟻 and H is a subset of Ker蟿

Now let h be an element in the intersection of Ker蟻 and H. This means that h is in both Ker蟻 and H. Thus, 蟻(h) = e鈥� and h is an element of H. Since 蟿 is the restriction of 蟻 to H, we have 蟿(h) = 蟻(h) = e鈥�. This implies that h is an element of Ker蟿. Therefore, (Ker蟻 鈭� H) 鈯� Ker蟿.
04

Conclude that Ker蟿 = Ker蟻 鈭� H

From steps 2 and 3, we have shown that Ker蟿 鈯� (Ker蟻 鈭� H) and (Ker蟻 鈭� H) 鈯� Ker蟿. Therefore, we can conclude that Ker蟿 = Ker蟻 鈭� H. This completes the proof.

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Key Concepts

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

Kernel of a Homomorphism
In group theory, a homomorphism is a function between two groups that respects the group operation. The kernel of a homomorphism is a set of all elements in the original group that are mapped to the identity element of the target group.

Mathematically, if we have a homomorphism \( \rho: G \rightarrow G' \), the kernel, denoted as \( \operatorname{Ker}(\rho) \), is defined as \( \{ g \in G \ : \ \rho(g) = e' \} \), where \( e' \) is the identity element in \( G' \).

Key properties of the kernel include:
  • It is always a subgroup of the original group \( G \).
  • It helps determine whether the homomorphism is injective. Specifically, a homomorphism is injective if and only if its kernel contains only the identity element of \( G \).
Understanding the kernel is crucial in comprehending the structure-preserving features of group homomorphisms.
Subgroup
A subgroup is a subset of a group that itself forms a group under the same operation. If \( H \) is a subgroup of a group \( G \), denoted \( H \leq G \), then \( H \) must satisfy the following properties:
  • Closure: For any elements \( a, b \in H \), the result of the operation \( a * b \) must also be in \( H \).
  • Identity: The identity element of \( G \) is also in \( H \).
  • Inverses: For every element \( a \in H \), its inverse \( a^{-1} \) is also in \( H \).
Subgroups are integral to group theory as they help explore the intrinsic properties of the group \( G \).

One crucial aspect involves the concept of subgroup restrictions. When a homomorphism \( \rho: G \rightarrow G' \) is restricted to a subgroup \( H \), this results in a new homomorphism from \( H \) to \( G' \), denoted \( \tau: H \rightarrow G' \). This is possible because \( H \) inherits the group structure from \( G \).
Intersection of Kernels
In group theory, intersections are used to find commonalities between sets. The intersection of kernels specifically concerns the elements that are common in the kernels of two or more homomorphisms.

If \( \tau: H \rightarrow G' \) is a homomorphism obtained by restricting \( \rho: G \rightarrow G' \) to a subgroup \( H \), the kernel of \( \tau \), \( \operatorname{Ker}(\tau) \), is the intersection of the kernel of \( \rho \) with \( H \).

This can be represented as \( \operatorname{Ker}(\tau) = \operatorname{Ker}(\rho) \cap H \). This identity reveals important structural information about how \( H \) and \( \rho \) interact.
  • \( \operatorname{Ker}(\tau) \subseteq H \): Because \( \tau \) is defined only for elements in \( H \).
  • \( \operatorname{Ker}(\tau) \subseteq \operatorname{Ker}(\rho) \): Because any element in \( \operatorname{Ker}(\tau) \) must also map to the identity in \( G' \) under \( \rho \).
Understanding this intersection gives a deeper insight into the arrangement and properties of subgroups and homomorphisms.

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Most popular questions from this chapter

}(I, G)\( be the set of f鈥� # Let \)I\( be a set and \)G\( be an abelian group, and consider the group \)\operatorname{Map}(I, G)\( of functions \)f: I \rightarrow G .\( Let Map \)^{\\#}(I, G)\( be the set of functions \)f \in \operatorname{Map}(I, G)\( such that \)f(i) \neq 0_{G}\( for at most finitely many \)i \in I\(. Show that \)\operatorname{Map}^{\\#}(I, G)\( is a subgroup of \)\operatorname{Map}(I, G)$

Let \(G\) be an abelian group of order \(n,\) and let \(m\) be an integer. Show that \(m G=G\) if and only if \(\operatorname{gcd}(m, n)=1\)

Let \(\rho: G \rightarrow G^{\prime}\) be a surjective group homomorphism. Let \(S\) be the set of all subgroups of \(G\) that contain Ker \(\rho,\) and let \(S^{\prime}\) be the set of all subgroups of \(G^{\prime} .\) Show that the sets \(\mathcal{S}\) and \(\mathcal{S}^{\prime}\) are in one-to-one correspondence, via the map that sends \(H \in \mathcal{S}\) to \(\rho(H) \in \mathcal{S}^{\prime} .\) Also show that this correspondence preserves inclusions; that is, for all \(H_{1}, H_{2} \in \mathcal{S},\) we have \(H_{1} \subseteq H_{2} \Longleftrightarrow \rho\left(H_{1}\right) \subseteq \rho\left(H_{2}\right) .\)

Let \(\rho: G \rightarrow G^{\prime}\) be a group homomorphism. Let \(H, K\) be subgroups of \(G\) and let \(m\) be a positive integer. Show that \(\rho(H+K)=\rho(H)+\rho(K)\) and \(\rho(m H)=m \rho(H) .\)

Write down the cosets of \(\left(\mathbb{Z}_{35}^{*}\right)^{2}\) in \(\mathbb{Z}_{35}^{*}\), along with the multiplication table for the quotient group \(\mathbb{Z}_{35}^{*} /\left(\mathbb{Z}_{35}^{*}\right)^{2}\).
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