Problem 31 - A Computational Introduction to Number Theory and Algebra

Question

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 \(H_{1} \subseteq H_{2} \Longleftrightarrow \rho\left(H_{1}\right) \subseteq \rho\left(H_{2}\right) .\)

Step-by-Step Solution

Answer

The main properties proven of the given correspondence are: 1. Injectivity (One-to-One): The map is injective, meaning that if two subgroups H1 and H2 are distinct, then their images under the given map are also distinct. 2. Surjectivity (Onto): The map is surjective, meaning that for any subgroup H' in the target set, there exists a subgroup H in the source set such that the given map sends H to H'. 3. Preservation of Inclusions: The map preserves inclusions, which means that if the subgroup H1 is a subset of the subgroup H2, then the image of H1 under the map is a subset of the image of H2 under the same map.

Step 1: Injectivity (One-to-One)

To show that the map is injective, we must show that for any two distinct subgroups \(H_1\) and \(H_2\) in \(S\), their images under \(\rho\) are also distinct. In other words, if \(H_1 \neq H_2\), then \(\rho(H_1) \neq \rho(H_2)\). Suppose \(\rho(H_1)=\rho(H_2)\) for some \(H_1, H_2 \in S\) with \(H_1 \neq H_2\). Then, \(\rho^{-1}(\rho(H_1))=\rho^{-1}(\rho(H_2)) \Rightarrow H_1 = H_2\) which is a contradiction. Hence, the map is injective.

Step 2: Surjectivity (Onto)

To show that the map is surjective, we must show that for any subgroup \(H' \in S'\), there exists a subgroup \(H \in S\) such that \(\rho(H) = H'\). Let \(H' \in S'\), and consider the preimage \(\rho^{-1}(H')\). We will show that \(\rho^{-1}(H')\) is a subgroup of \(G\) that contains Ker \(\rho\). Let \(a, b \in \rho^{-1}(H')\). Then \(\rho(a), \rho(b) \in H'\). Since \(H'\) is a subgroup of \(G'\), we have \(\rho(a)\rho(b)^{-1} \in H' \Rightarrow \rho(a) \rho(b^{-1}) \in H' \Rightarrow \rho(ab^{-1}) \in H'\), so \(ab^{-1} \in \rho^{-1}(H')\). Thus, \(\rho^{-1}(H')\) is a subgroup of \(G\). Now, let \(k \in \text{Ker} \; \rho\). Then \(\rho(k) = \text{id} \in H' \Rightarrow k \in \rho^{-1}(H')\). Therefore, \(\text{Ker} \; \rho\) is a subset of \(\rho^{-1}(H')\). So, \(H:=\rho^{-1}(H') \in S\). Since \(\rho(\rho^{-1}(H')) = H'\), the map is surjective.

Step 3: Correspondence Preserves Inclusions

Now, we will show that the correspondence preserves inclusions. Let \(H_1, H_2 \in S\). We want to prove the following equivalence: \(H_1 \subseteq H_2 \Longleftrightarrow \rho(H_1) \subseteq \rho(H_2)\). (\(\Rightarrow\)) Suppose \(H_1 \subseteq H_2\). Let \(x \in \rho(H_1)\). Then, \(x = \rho(a)\) for some \(a \in H_1\). Since \(H_1 \subseteq H_2\) and \(a \in H_1\), \(a \in H_2\). Hence, \(x=\rho(a) \in \rho(H_2)\). Thus, \(\rho(H_1) \subseteq \rho(H_2)\). (\(\Leftarrow\)) Suppose \(\rho(H_1) \subseteq \rho(H_2)\). Let \(a \in H_1\). Then, \(\rho(a) \in \rho(H_1)\). Since \(\rho(H_1) \subseteq \rho(H_2)\), \(\rho(a) \in \rho(H_2)\), which means there exists \(b \in H_2\) such that \(\rho(a) = \rho(b)\). Now, since \(\rho\) is a group homomorphism, \(\rho(ab^{-1}) = \rho(a)\rho(b)^{-1} = \rho(a)\rho(a)^{-1} = e'\). Hence, \(ab^{-1} \in \text{Ker} \; \rho\). Since both \(H_1\) and \(H_2\) contain Ker \(\rho\), we have \(ab^{-1} \in H_1\). Thus, \(a \in H_1b = H_2\), and \(H_1 \subseteq H_2\). Therefore, we have proved that the given correspondence is one-to-one, onto, and preserves inclusions between the subgroups of \(G\) containing Ker \(\rho\) and the subgroups of \(G'\).

Key Concepts

Surjective Homomorphism

In group theory, a homomorphism is a map between two groups that preserves the group operation. When we call a homomorphism "surjective" or "onto," it means that every element in the target group (let's call it \( G' \)) can be mapped to by at least one element in the original group \( G \). This ensures that \( \ ho: G \rightarrow G' \) fully covers \( G' \).

To better understand surjective homomorphisms, it's important to note these features:

• Every element in \( G' \) arises from applying the homomorphism map, \( \ ho \), to some element in \( G \).
• This quality ensures that the image of \( \ ho \) is equal to the entire group \( G' \).

The significance of surjective homomorphisms in this exercise is that it allows us to establish a neat correspondence between subgroups of \( G \) and \( G' \), based on how they relate to the kernel of the homomorphism.

Subgroup Correspondence

The idea of subgroup correspondence in this context is about linking subgroups of two different groups, \( G \) and \( G' \), via a homomorphism \( \ ho \). Here, the focus is on those subgroups of \( G \) that contain the kernel of \( \ ho \) since these particular subgroups correspond directly to subgroups in \( G' \).

When we say two sets of subgroups are in "one-to-one correspondence" through \( \ ho \), it implies:

• Each subgroup of \( G \) that contains the kernel uniquely maps to a subgroup in \( G' \).
• The reverse association is also true; each subgroup in \( G' \) can be traced back to a subgroup in \( G \) containing the kernel.
• This map preserves the properties such as inclusion, meaning that if a subgroup \( H_1 \) of \( G \) includes another subgroup \( H_2 \), then \( \ ho(H_1) \) includes \( \ ho(H_2) \) in \( G' \).

This correspondence is essential for understanding how group structures in \( G \) translate into \( G' \), while also emphasizing the role of the kernel as a bridge between subgroups.

Kernel of a Homomorphism

The kernel of a homomorphism, noted as \( \text{Ker} \, \rho \), is pivotal in understanding the structure and behavior of the homomorphism \( \ ho \). It consists of all elements in \( G \) that map to the identity element in \( G' \) under \( \ ho \). It can be seen as a "measuring stick" for how similar \( G \) and \( G' \) are.

Key properties of the kernel include:

• \( \text{Ker} \, \rho \) is always a normal subgroup of \( G \).
• Because \( \ ho \) is a homomorphism, any element in the kernel maps to the identity element in \( G' \), making \( \text{Ker} \, \rho \) a crucial factor in determining what parts of \( G \) are "lost" in \( G' \).
• The kernel helps decide which subgroups of \( G \) can be associated with \( G' \) via \( \ ho \); only those subgroups of \( G \) containing \( \text{Ker} \, \rho \) are relevant to this correspondence.

Understanding the kernel not only clarifies the homomorphism's action but also sheds light on the connections between \( G \) and \( G' \) over different subgroups.