Problem 23 - A Computational Introduction to Number Theory and Algebra

Question

Let \(\rho_{i}: G_{i} \rightarrow G_{i}^{\prime},\) for \(i=1, \ldots, k,\) be group homomorphisms. Show that the map $$ \begin{aligned} \rho: \quad G_{1} \times \cdots \times G_{k} & \rightarrow G_{1}^{\prime} \times \cdots \times G_{k}^{\prime} \\ \left(a_{1}, \ldots, a_{k}\right) & \mapsto\left(\rho_{1}\left(a_{1}\right), \ldots, \rho_{k}\left(a_{k}\right)\right) \end{aligned} $$ is a group homomorphism. Also show that if each \(\rho_{i}\) is an isomorphism, then so is \(\rho .\)

Step-by-Step Solution

Answer

Question: Show that the map 𝜌:𝐺1×⋯×𝐺𝑘→𝐺1′×⋯×𝐺𝑘′ defined by 𝜌(𝑎1,...,𝑎𝑘)=(𝜌1(𝑎1),...,𝜌𝑘(𝑎𝑘)) is a group homomorphism. Additionally, prove that if each 𝜌𝑖 is an isomorphism, then the map 𝜌 is also an isomorphism. Solution: To show that 𝜌 is a group homomorphism, we checked that 𝜌 preserves the identity element and the operation. We found that 𝜌(𝑒1,𝑒2,...,𝑒𝑘)=(𝑒′1,𝑒′2,...,𝑒′𝑘) and 𝜌(𝑎1𝑏1,𝑎2𝑏2,...,𝑎𝑘𝑏𝑘)=𝜌(𝑎1,𝑎2,...,𝑎𝑘)𝜌(𝑏1,𝑏2,...,𝑏𝑘). Therefore, 𝜌 is a group homomorphism. Additionally, we proved that if each 𝜌𝑖 is an isomorphism, then 𝜌 is both injective and surjective, making it an isomorphism as well.

Step 1: Show that the identity element is preserved

Let \((e_1, e_2, \ldots, e_k)\) be the identity element in the group \(G_1 \times \cdots \times G_k\), where \(e_i\) is the identity element in \(G_i\). Then, we have: $$ \begin{aligned} \rho(e_1, e_2, \ldots, e_k) &= \left(\rho_1(e_1), \rho_2(e_2), \ldots, \rho_k(e_k)\right) \end{aligned} $$ Since each \(\rho_i\) is a group homomorphism, \(\rho_i(e_i)\) is the identity element in \(G_i'\), say \(e'_i\). Therefore, we get: $$ \begin{aligned} \rho(e_1, e_2, \ldots, e_k) &= \left(e'_1, e'_2, \ldots, e'_k\right) \end{aligned} $$ This shows that \(\rho\) preserves the identity element.

Step 2: Show that the operation is preserved

Let \((a_1, a_2, \ldots, a_k), (b_1, b_2, \ldots, b_k) \in G_1 \times \cdots \times G_k\). We need to show that \(\rho((a_1, a_2, \ldots, a_k)(b_1, b_2, \ldots, b_k)) = \rho(a_1, a_2, \ldots, a_k) \rho(b_1, b_2, \ldots, b_k)\). First, we have: $$ \begin{aligned} (a_1, a_2, \ldots, a_k)(b_1, b_2, \ldots, b_k) &= (a_1b_1, a_2b_2, \ldots, a_kb_k) \end{aligned} $$ Now, let's compare applying \(\rho\) to both sides: $$ \begin{aligned} \rho(a_1b_1, a_2b_2, \ldots, a_kb_k) &= \left(\rho_1(a_1b_1), \rho_2(a_2b_2), \ldots, \rho_k(a_kb_k)\right) \\ \rho(a_1, a_2, \ldots, a_k) \rho(b_1, b_2, \ldots, b_k) &= \left(\rho_1(a_1), \rho_2(a_2), \ldots, \rho_k(a_k)\right) \left(\rho_1(b_1), \rho_2(b_2), \ldots, \rho_k(b_k)\right) = \left(\rho_1(a_1)\rho_1(b_1), \rho_2(a_2)\rho_2(b_2), \ldots, \rho_k(a_k)\rho_k(b_k)\right) \end{aligned} $$ Since each \(\rho_i\) is a group homomorphism, we know that \(\rho_i(a_i b_i) = \rho_i(a_i) \rho_i(b_i)\). This gives us: $$ \begin{aligned} \rho(a_1b_1, a_2b_2, \ldots, a_kb_k) &= \rho(a_1, a_2, \ldots, a_k) \rho(b_1, b_2, \ldots, b_k) \end{aligned} $$ This shows that \(\rho\) preserves the operation, hence \(\rho\) is a group homomorphism.

Step 3: Show that if each \(\rho_i\) is an isomorphism, then \(\rho\) is an isomorphism

Let's assume that each \(\rho_i\) is an isomorphism. We need to show that \(\rho\) is injective and surjective. Injectivity: Let \((a_1, a_2, \ldots, a_k), (b_1, b_2, \ldots, b_k) \in G_1 \times \cdots \times G_k\) such that \(\rho(a_1, a_2, \ldots, a_k) = \rho(b_1, b_2, \ldots, b_k)\). This implies that: $$ \left(\rho_1(a_1), \rho_2(a_2), \ldots, \rho_k(a_k)\right) = \left(\rho_1(b_1), \rho_2(b_2), \ldots, \rho_k(b_k)\right) $$ Since each \(\rho_i\) is an isomorphism and therefore injective, this means that \(a_i = b_i\) for all \(i\). Thus, \((a_1, a_2, \ldots, a_k) = (b_1, b_2, \ldots, b_k)\) and \(\rho\) is injective. Surjectivity: Let \((c_1, c_2, \ldots, c_k) \in G_1' \times \cdots \times G_k'\). Since each \(\rho_i\) is an isomorphism and therefore surjective, for each \(i\), there exists an element \(a_i \in G_i\) such that \(\rho_i(a_i) = c_i\). Then, we have: $$ \rho(a_1, a_2, \ldots, a_k) = \left(\rho_1(a_1), \rho_2(a_2), \ldots, \rho_k(a_k)\right) = \left(c_1, c_2, \ldots, c_k\right) $$ This means that \(\rho\) is surjective. Since \(\rho\) is injective and surjective, it is an isomorphism.

Key Concepts

Group Theory

Group theory is an area of abstract algebra that studies algebraic structures known as groups. These structures consist of a set of elements along with an operation that combines any two elements to form a third element. The operation must satisfy four key properties: closure, associativity, the existence of an identity element, and the existence of inverse elements.

Groups are omnipresent in mathematics and are applied in various fields such as physics, cryptography, and even Rubik's Cube solving! Understanding the basic properties of these mathematical entities allows us to analyze symmetry, mathematical operations, and transformations more deeply.

Group Isomorphism

In the realm of group theory, the concept of group isomorphism provides a way to say when two groups are essentially the same, even if their elements and operations look different. A group isomorphism is a bijective group homomorphism—a relationship between two groups that preserves the group operation and is both injective (one-to-one) and surjective (onto).

When two groups are isomorphic, they have the same group structure, which means they will have the same number of elements of each order, the same group tables (up to relabeling of elements), and their elements will follow similar algebraic rules. Therefore, any property that is preserved under group homomorphisms is also preserved for isomorphic groups.

Identity Element

The identity element in group theory is a unique element within a group that, when combined with any element of the group, leaves that element unchanged. For a group \(G\) with an operation \(\ast\), the identity element \(e\) satisfies the condition that for every element \(a \in G\), \(e \ast a = a \ast e = a\).

In the context of group homomorphisms, it's essential that this specific element is preserved. That is, a group homomorphism \(\rho : G \rightarrow G'\) must satisfy \(\rho(e) = e'\), where \(e'\) is the identity element of \(G'\). If this condition is not met, then \(\rho\) fails to be a group homomorphism, which emphasizes just how crucial the identity element is in the structure of a group.

Injective and Surjective Mappings

Mappings, or functions between sets, can be classified in several ways depending on how they relate elements from the source set to the target set. Two important types of mappings in the context of group theory are injective (one-to-one) and surjective (onto) mappings.

An injective mapping ensures that different elements in the domain map to different elements in the codomain. In other words, if \(a \eq b\), then \(f(a) \eq f(b)\). This property is crucial when discussing isomorphisms since it guarantees that the mapping is a precise matching between elements, without collapsing different elements to the same image.

A surjective mapping makes sure that every element in the codomain is an image of at least one element in the domain. Surjectivity ensures that the mapping covers the entire codomain. When a mapping is both injective and surjective, it is termed bijective, and it is this bijective property that is one component of a group isomorphism, indicating a perfect correspondence between the two groups involved.