Problem 20 - Abstract Algebra: Theory and Applications

Question

Let \(H\) and \(K\) be subgroups of a group \(G\). Define a relation \(\sim\) on \(G\) by \(a \sim b\) if there exists an \(h \in H\) and a \(k \in K\) such that \(h a k=b\). Show that this relation is an equivalence relation. The corresponding equivalence classes are called double cosets. Compute the double cosets of \(H=\\{(1),(123),(132)\\}\) in \(A_{4}\).

Step-by-Step Solution

Answer

Question: Verify if the given relation ~ is an equivalence relation on group G and, if so, find the double cosets of H in A_4. Answer: The relation ~ is indeed an equivalence relation, as it is reflexive, symmetric, and transitive. The double cosets of H in A_4 are [(1),(123),(132)], [(12)(34)], [(13)(24)], [(14)(23)], [(124),(142),(234),(243)], and [(134),(143)].

Step 1: Verify if the relation is reflexive

To show that the relation is reflexive, we need to check whether \(a \sim a\) for all \(a \in G\). If there exists \(h \in H\) and \(k \in K\) such that \(hak=a\), then the relation is reflexive. Since the identity element \(e \in H\) and \(e \in K\), we have \(a = eae\), and therefore \(a \sim a\) for all \(a \in G\).

Step 2: Verify if the relation is symmetric

To show that the relation is symmetric, we need to check whether \(a \sim b\) implies \(b \sim a\). If \(hak=b\), we need to find \(h' \in H\) and \(k' \in K\) such that \(h'bk'=a\). We can multiply both sides of \(hak=b\) by \(h^{-1}\) and \(k^{-1}\) on the left and right, respectively. We get \(a = h^{-1}bk^{-1}\). Since \(h^{-1} \in H\) and \(k^{-1} \in K\), the relation is indeed symmetric.

Step 3: Verify if the relation is transitive

To show that the relation is transitive, we need to check whether \(a \sim b\) and \(b \sim c\) imply \(a \sim c\). If \(hak=b\) and \(h'bk'=c\), we need to find \(h'' \in H\) and \(k'' \in K\) such that \(h''ak''=c\). We can multiply the second equation by \(a^{-1}\) on the right and the first equation by \(k\) on the right. Then we get \(h'a^{-1}=h''k''\), and so \(h''=hh'a^{-1}\) and \(k''=ak'\). Since \(h'' \in H\) and \(k'' \in K\), the relation is transitive. Since the relation \(\sim\) is reflexive, symmetric, and transitive, it is indeed an equivalence relation.

Step 4: Calculate the double cosets of H in A_4

First, we list out all elements of \(A_4\), the group of even permutations of 4 elements: \(A_4 = \{(1),(12)(34),(13)(24),(14)(23),(123),(132),(124),(142),(134),(143),(234),(243)\}\). Now we must compute the double cosets by combining elements from \(H\) and \(A_4\) according to the relation \(\sim\). Since \(H \subseteq A_4\), we'll iterate through \(a \in A_4\) and for each \(a\), find \(H \sim a \sim K\). 1. For \(a = (1)\), we have double coset \(H \sim (1) \sim K = \{(1), (123), (132)\}\). 2. For \(a = (12)(34)\), we have double coset \(H \sim (12)(34) \sim K = \{(12)(34)\}\). 3. For \(a = (13)(24)\), we have double coset \(H \sim (13)(24) \sim K = \{(13)(24)\}\). 4. For \(a = (14)(23)\), we have double coset \(H \sim (14)(23) \sim K = \{(14)(23)\}\). 5. For \(a = (123)\), we already computed their double coset with \(a=(1)\) as H. 6. For \(a = (132)\), we already computed their double coset with \(a=(1)\) as H. 7. For \(a = (124)\), we have double coset \(H \sim (124) \sim K = \{(124), (142), (234), (243)\}\). 8. For \(a = (142)\), we already computed their double coset with \(a=(124)\). 9. For \(a = (134)\), we have double coset \(H \sim (134) \sim K = \{(134), (143)\}\). 10. For \(a=(143)\), we already computed their double coset with \(a=(134)\). 11. For \(a=(234)\), we already computed their double coset with \(a=(124)\). 12. For \(a=(243)\), we already computed their double coset with \(a=(124)\). So, the double cosets of \(H\) in \(A_4\) are @{@"[(1),(123),(132)]","[(12)(34)]","[(13)(24)]","[(14)(23)]","[(124),(142),(234),(243)]" and "[(134),(143)]}@".

Key Concepts

Equivalence Relation

An equivalence relation is a fundamental concept in mathematics that helps to understand how elements within a set can be considered 'equivalent' in some way. This relation is characterized by three essential properties: it must be reflexive, symmetric, and transitive.

• Reflexive: Every element is equivalent to itself. For any element a in the set, a is related to a (denoted a \(\sim\) a).
• Symmetric: If one element is equivalent to another, then that second element is equivalent to the first. In formal terms, for any two elements a and b, if a \(\sim\) b, then b \(\sim\) a.
• Transitive: If one element is equivalent to a second, and that second element is equivalent to a third, then the first and third elements are equivalent. That is, if a \(\sim\) b and b \(\sim\) c, it follows that a \(\sim\) c.

These properties ensure that we can group together elements that are equivalent to each other into distinct sets, known as equivalence classes. In the context of the problem, double cosets formed under these conditions group elements of a group G into equivalence classes defined by subgroups H and K. By verifying the properties of reflexive, symmetric, and transitive for the relation defined by H and K, we establish it as an equivalence relation, organizing the group G into distinct double cosets.

Subgroups in Group Theory

Subgroups play a central role in the study of group theory, which is a major branch of algebra focused on the algebraic structures known as groups. These structures consist of a set equipped with an operation that combines any two of its elements to form a third element, satisfying four fundamental properties: closure, associativity, identity, and invertibility.

A subgroup is defined as a subset of a parent group that forms a group under the same operation as the parent group. In short, it must meet the following criteria:

• Closure: If two elements belong to the subgroup, their product (as defined by the group operation) must also belong to the subgroup.
• Associativity: The operation within the subgroup must satisfy associativity, which is inherited from the parent group.
• Identity: The subgroup must contain the identity element of the parent group, which leaves any element it is combined with unchanged.
• Invertibility: For every element in the subgroup, there must be an inverse element that also belongs to the subgroup.

Finding subgroups is an essential part of understanding the structure of a group, as they provide insight into the group's composition and behavior. In the double coset problem, H and K are both subgroups of a larger group G, and recognizing how they interact with each other leads to a deeper understanding of the structure of G itself.

Permutation Groups

Permutation groups are a class of groups where the elements are permutations of a set, and the group operation is the composition of permutations. A permutation is essentially a reordering of the elements of a set. The study of permutation groups is pivotal in many areas of mathematics, including algebra, combinatorics, and even in solving Rubik's Cube puzzles.

The symmetric group on n symbols, denoted by S_n, consists of all possible permutations of n objects and is a key example of a permutation group. Another important permutation group is the alternating group on n symbols, A_n, which contains only the even permutations (those that can be achieved by an even number of swaps of two elements) from the symmetric group.

In the context of our problem, A₄ is the group of all even permutations of four elements. It is relevant because, given a subgroup H, we aim to find the double cosets of H in A₄. As demonstrated in the exercise, by applying properties from the concept of equivalence relations to permutation groups through the subgroups, students can determine distinct equivalence classes, which in this case are the double cosets.