Problem 22
Question
Verify that the "is isomorphic to" relation on abelian groups is an equivalence relation; that is, for all abelian groups \(G_{1}, G_{2}, G_{3},\) we have: (a) \(G_{1} \cong G_{1}\) (b) \(G_{1} \cong G_{2}\) implies \(G_{2} \cong G_{1}\); (c) \(G_{1} \cong G_{2}\) and \(G_{2} \cong G_{3}\) implies \(G_{1} \cong G_{3}\).
Step-by-Step Solution
Verified Answer
To show that the "is isomorphic to" relation on abelian groups is an equivalence relation, we need to demonstrate reflexivity, symmetry, and transitivity. We find that every abelian group is isomorphic to itself using the identity function, proving reflexivity. Additionally, we show that if an isomorphism exists between two groups, the inverse bijection also serves as an isomorphism between the inverse groups, proving symmetry. Lastly, we prove transitivity by using the composition of isomorphisms, which also preserves the group operation. In conclusion, the "is isomorphic to" relation on abelian groups is an equivalence relation because it satisfies reflexivity, symmetry, and transitivity.
1Step 1: Reflexivity - \(G_1 \cong G_1\)
We want to show that every abelian group \(G_1\) is isomorphic to itself. To prove this, we need to find an isomorphism between the group and itself. In this case, the identity function \(f: G_1 \to G_1\) is such a function, where \(f(g) = g\) for all \(g \in G_1\). This function is both injective and surjective, and it preserves the group operation, as \(f(g_1 * g_2) = g_1 * g_2 = f(g_1) * f(g_2)\). Therefore, \(G_1 \cong G_1\).
2Step 2: Symmetry - \(G_1 \cong G_2\) implies \(G_2 \cong G_1\)
We want to show that if \(G_1 \cong G_2\), then \(G_2 \cong G_1\). Suppose \(f: G_1 \to G_2\) is an isomorphism between \(G_1\) and \(G_2\). To show that \(G_2 \cong G_1\), we need to find an isomorphism between \(G_2\) and \(G_1\). The inverse of the bijection \(f\), denoted as \(f^{-1}: G_2 \to G_1\), is such a function. Since \(f\) is an isomorphism, its inverse function \(f^{-1}\) is also a bijection, and it preserves the group operation: \(f^{-1}(g_1 *' g_2) = f^{-1}(f(f^{-1}(g_1)) * f(f^{-1}(g_2))) = f^{-1}(f(f^{-1}(g_1) * f^{-1}(g_2))) = f^{-1}(g_1) * f^{-1}(g_2)\). Therefore, \(G_2 \cong G_1\).
3Step 3: Transitivity - \(G_1 \cong G_2\) and \(G_2 \cong G_3\) implies \(G_1 \cong G_3\)
We want to show that if \(G_1 \cong G_2\) and \(G_2 \cong G_3\), then \(G_1 \cong G_3\). Suppose there are isomorphisms \(f: G_1 \to G_2\) and \(g: G_2 \to G_3\). To obtain an isomorphism between \(G_1\) and \(G_3\), we can use the composition of the two isomorphisms, denoted as \(g \circ f: G_1 \to G_3\). The composition of two bijections is a bijection, and we only need to check the preservation of the group operation: \((g \circ f)(g_1 * g_2) = g(f(g_1 * g_2)) = g(f(g_1) *' f(g_2)) = g(f(g_1)) *'' g(f(g_2)) = (g \circ f)(g_1) * (g \circ f)(g_2)\). Therefore, \(G_1 \cong G_3\).
In conclusion, the "is isomorphic to" relation on abelian groups is an equivalence relation, as it satisfies reflexivity, symmetry, and transitivity.
Key Concepts
IsomorphismAbelian GroupsGroup Theory
Isomorphism
An isomorphism in mathematics, particularly in group theory, signifies a structural similarity between two groups. An isomorphism is essentially a function that maps elements of one group to elements of another in a way that perfectly maintains the group's operations and structure. This concept is significant because it allows mathematicians to consider two groups as identical in terms of their group-theoretic properties.
An important property of isomorphisms is that they are bijections, meaning they are both injective (one-to-one) and surjective (onto). This ensures that each element in one group corresponds to exactly one element in the other group and vice versa. Beyond maintaining structure, the preservation of operations means that if two elements are combined in the group (using the group's operation), the function respects this combination when applied to those elements. Mathematically, a function \( f: G_1 \to G_2 \) is an isomorphism if \( f \) meets the criteria of bijection and if for any elements \( a, b \in G_1 \), the equation \( f(a \cdot b) = f(a) \cdot' f(b) \) holds true, where \( \cdot \) and \( \cdot' \) are the operations in groups \( G_1 \) and \( G_2 \) respectively.
Understanding isomorphisms is crucial because they help identify when two groups can essentially be considered the same in terms of their structural and operational properties. This concept underpins many fundamental ideas in algebra.
An important property of isomorphisms is that they are bijections, meaning they are both injective (one-to-one) and surjective (onto). This ensures that each element in one group corresponds to exactly one element in the other group and vice versa. Beyond maintaining structure, the preservation of operations means that if two elements are combined in the group (using the group's operation), the function respects this combination when applied to those elements. Mathematically, a function \( f: G_1 \to G_2 \) is an isomorphism if \( f \) meets the criteria of bijection and if for any elements \( a, b \in G_1 \), the equation \( f(a \cdot b) = f(a) \cdot' f(b) \) holds true, where \( \cdot \) and \( \cdot' \) are the operations in groups \( G_1 \) and \( G_2 \) respectively.
Understanding isomorphisms is crucial because they help identify when two groups can essentially be considered the same in terms of their structural and operational properties. This concept underpins many fundamental ideas in algebra.
Abelian Groups
Abelian groups, named after the mathematician Niels Henrik Abel, are a special class of groups in mathematics. What sets them apart from other groups is that their group operation is commutative. This means that for any two elements \( a \) and \( b \) in the group, the equation \( a \cdot b = b \cdot a \) holds. This property is crucial because it simplifies the complexity involved in various mathematical problems and proofs.
Various common mathematical structures are inherently abelian. For example:
Various common mathematical structures are inherently abelian. For example:
- The set of integers with addition is an abelian group.
- The set of real numbers with addition forms an abelian group.
- The group of integers modulo \( n \), under addition, is an abelian group.
Group Theory
Group theory is a branch of mathematics that studies the algebraic structures known as groups. At its heart, group theory focuses on understanding symmetrical relationships, like rotations and reflections, that can be mathematically formalized.
A group consists of a set of elements together with a binary operation that combines any two elements in the set to form another element in the set. This combination process must satisfy four key properties:
A group consists of a set of elements together with a binary operation that combines any two elements in the set to form another element in the set. This combination process must satisfy four key properties:
- Closure: The result of the operation on any two elements of the group is still an element within the group.
- Associativity: When performing the group operation on multiple elements, the result does not depend on how they are grouped, i.e., \((a \cdot b) \cdot c = a \cdot (b \cdot c)\).
- Identity Element: There exists an element in the group that, when used in the group operation with any element, results in the original element.
- Inverse Element: For every element, there exists another element which combines with the original element to yield the identity element.
Other exercises in this chapter
Problem 20
Let \(G\) be an abelian group, and let \(\sim\) be an equivalence relation on \(G\). Further, suppose that for all \(a, a^{\prime}, b \in G,\) if \(a \sim a^{\p
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Let \(H\) be a subgroup of an abelian group \(G,\) and let \(a, b \in G\) Show that \([a+b]_{H}=\left\\{x+y: x \in[a]_{H}, y \in[b]_{H}\right\\}\)
View solution Problem 23
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} \time
View solution Problem 24
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)=\r
View solution