Problem 1 - A Book of Abstract Algebra

Question

Let \(G\) be the subset of \(S_{4}\) consisting of the permutations $$ \begin{array}{ll} \varepsilon=\left(\begin{array}{llll} 1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4 \end{array}\right) & f=\left(\begin{array}{llll} 1 & 2 & 3 & 4 \\ 2 & 1 & 4 & 3 \end{array}\right) \\ g=\left(\begin{array}{llll} 1 & 2 & 3 & 4 \\ 3 & 4 & 1 & 2 \end{array}\right) & h=\left(\begin{array}{llll} 1 & 2 & 3 & 4 \\ 4 & 3 & 2 & 1 \end{array}\right) \end{array} $$ Show that \(G\) is a group of permutations, and write its table. \begin{tabular}{l|llll} & \(\varepsilon\) & \(f\) & \(g\) & \(h\) \\ \hline\(\varepsilon\) & & & \\ \(f\) & & & \\ \(g\) & & & \\ \(h\) & & & \end{tabular}

Step-by-Step Solution

Answer

The set \(G\) is a group as it satisfies closure, identity, inverses, and associativity. The group table documents all permutation products.

Step 1: Understanding the Group Requirements

To show that a set is a group, we must verify four requirements: closure, identity, inverses, and associativity. Closure means the product of any two elements in the set is also in the set. Identity means there is an element, denoted by \(\varepsilon\), such that multiplying it with any other element leaves that element unchanged. Inverses means for each element, there exists another element, which when multiplied together, gives \(\varepsilon\). Finally, associativity means any grouping of three elements in a product leads to the same result.

Step 2: Identity Element

The permutation \(\varepsilon\) is the identity element since it does not change any element in a set upon multiplication. Verifying the identity property, we see that multiplying \(\varepsilon\) by any element in \(G\) yields the element itself: \(\varepsilon \cdot x = x \cdot \varepsilon = x\) for all \(x \in G\).

Step 3: Inverses for Each Element

For each permutation in \(G\), find an element that acts as its inverse. \(\varepsilon\) is self-inversed because \(\varepsilon^2 = \varepsilon\). The inverse of \(f\) is itself, as \(f \cdot f = \varepsilon\). The inverse of \(g\) is also \(g\), since \(g \cdot g = \varepsilon\). Similarly, \(h\) is self-inversed, as \(h \cdot h = \varepsilon\). Thus, each element has an inverse in \(G\).

Step 4: Closure Verification

Multiply each combination of two elements in \(G\) and check if the result remains in \(G\). For example:\(f \cdot g = (1\ 3)(2\ 4) = h\) (still in \(G\)) and this holds for all element pairs. After checking each product, we confirm closure.

Step 5: Associativity Check

To demonstrate associativity, notice that within permutations, the operation is inherently associative. This is due to the nature of function composition as it's always associative; hence this condition is satisfied automatically.

Step 6: Constructing the Group Table

Construct a table showing all possible products of group elements:\[\begin{array}{c|cccc} & \varepsilon & f & g & h \\hline\varepsilon & \varepsilon & f & g & h \f & f & \varepsilon & h & g \g & g & h & \varepsilon & f \h & h & g & f & \varepsilon \\end{array}\]

Step 7: Verifying the Group Table

Check each entry against the conditions: closure and correct inverses, with respect to our initial definition of the group elements. Ensure every element's product is in \(G\). The completed table consistently reflects the conditions of a group given the operations defined.

Key Concepts

Permutations in S4

In group theory, permutations are arrangements or rearrangements of a set. Specifically, when we are dealing with permutations in the context of the symmetric group, such as those in \( S_4 \), we are looking at all possible arrangements of four elements. In our exercise, we use permutations \( \varepsilon, f, g, \) and \( h \). These permutations are transformations, essentially swelling the positions of elements 1, 2, 3, and 4 among themselves:

• \( \varepsilon \) is the identity permutation, meaning no change is done: \( (1 o 1, 2 o 2, 3 o 3, 4 o 4) \).
• \( f \) swaps \( 1 o2 \) and \( 2 o1 \), as well as \( 3 o4 \) and \( 4 o3 \).
• \( g \) cyclically moves \( 1 o3 \), \( 2 o4 \), \( 3 o1 \), and \( 4 o2 \).
• \( h \) reverses everything: \( 1 o4 \), \( 2 o3 \), \( 3 o2 \), and \( 4 o1 \).

You can imagine these permutations as shuffling cards, where each card is a number, and the shuffle represents a particular permutation.

Group Properties

For a set with an operation to be considered a group, it needs to adhere to specific properties:

• Closure: For any two elements in the group, their operation still results in an element within the group.
• Identity Element: There is one element, the identity, that, when combined with any other element, leaves the other element unchanged.
• Inverses: Every element has an inverse, such that when they are combined, the result is the identity.
• Associativity: The way in which elements are grouped does not affect the outcome of the operation.

In our problem, we show these properties using the permutations \(\varepsilon, f, g, h\). Each element is closed under the operation of the set, they all have an inverse, \(\varepsilon\) serves as the identity, and the operation is associative as is the nature of permutation composition.

Cayley Table

The Cayley table is like a multiplication table but for group elements. It helps visualize how elements combine and ensures group properties like closure and inverses are satisfied. In our problem, we constructed a table for permutations \(\varepsilon, f, g, h\). Each row and column corresponds to an element of the group, and the intersection shows the result of multiplying the row and column element:

• The first row and column correspond to multiplying by \(\varepsilon\), functioning like the mathematical identity, leaving other elements unchanged.
• Other cells in the table are filled by the operation results, such as \(f \cdot g = h\), ensuring that results remain within the group.

This table is crucial for confirming all group properties, as it visually represents closure and inversion.

Closure in Group

The concept of closure is pivotal in group theory, especially when showcasing a set as a group. Closure means that if you take any two elements from the group and perform the group operation, the result will still be inside the group. For permutation operations, like those provided by \(\varepsilon, f, g, h\), closure was verified by checking products:

• \(f \cdot g\) resulted in \(h\), which is inside the set \(G\).
• This pattern holds for every possible element pair combination, affirming closure.

Without closure, even with other group properties satisfied, a set cannot be a group because some operations would 'leave' the group.

Associative Property in Groups

Associativity in groups ensures that the grouping of elements during the operation does not influence the outcome. This is crucial because it allows for flexibility in operations, which is inherently satisfied in function composition, and by extension, permutation operations.

In simpler terms, for any three elements \(a, b,\) and \(c\) in the group:

• \((a \cdot b) \cdot c = a \cdot (b \cdot c)\).

In our exercise, this associative nature carries through all operations involving \(\varepsilon, f, g, h\). Every calculation respects this property automatically by the nature of permutations, saving us from having to prove it for every specific case.