Problem 1 - Algebra: Pure and Applied

Question

In Exercises 1 through 8 (a) find the stabilizer \(G_{a}\) for each \(a\) indicated, (b) find the orbit \(O_{a},(\mathrm{c})\) check the orbit-stabilizer relation \(\left|O_{a}\right|=\left[G: G_{a}\right],(\mathrm{d})\) determine whether the action is transitive, and (e) determine whether \(G\) acts faithfully on \(X\). $$ X=\\{1,2,3\\} ; G=S_{3} ; a=1,2,3 $$

Step-by-Step Solution

Answer

The stabilizer for each element is of order 2; their orbits are \(\{1, 2, 3\}\). The action is transitive and faithful, verifying the orbit-stabilizer theorem.

Step 1: Identify the Group and Action

Given the set \(X = \{1, 2, 3\}\) and the group \(G = S_3\), we recognize that \(G\) is the symmetric group on 3 elements, meaning it includes all permutations of the set \(X\). The action of \(G\) on \(X\) is defined by the application of permutations on the elements of \(X\).

Step 2: Find the Stabilizer for Each Element

The stabilizer of an element \(a\) in \(G\), denoted \(G_a\), is the set of all elements \(g \in G\) such that \(g\cdot a = a\). For each element:- For \(a = 1\): Only the identity permutation \(()\) and those permutations that keep 1 in place (e.g., \((23)\)) will stabilize 1.- For \(a = 2\): Similar reasoning applies, with stabilizers keeping 2 in place.- For \(a = 3\): Stabilizers will keep 3 in place.Since each scenario works similarly, the stabilizer for any chosen element \(a\) in \(S_3\) would be the permutations that don't change the position of \(a\). Each stabilizer is isomorphic to \(S_2\), having order 2.

Step 3: Find the Orbit of Each Element

The orbit of an element \(a\), denoted \(O_a\), is the set of elements that \(a\) can be mapped to via the action of some permutation in \(G\).- For \(a = 1\), by applying all permutations in \(S_3\), we see that any element can map to 1. Hence, \(O_1 = \{1, 2, 3\}\).- Similarly, for \(a = 2\), \(O_2 = \{1, 2, 3\}\).- For \(a = 3\), \(O_3 = \{1, 2, 3\}\).Thus, each element has the whole set \(X\) as its orbit.

Step 4: Check the Orbit-Stabilizer Relation

The Orbit-Stabilizer Theorem states \( |O_a| = [G : G_a] \), where \(|O_a|\) is the size of the orbit of \(a\), and \([G : G_a]\) is the index of the stabilizer in \(G\).- We have \(|O_1| = |O_2| = |O_3| = 3\) and \(|G| = 6\) (since \(S_3\) has 6 permutations and each stabilizer has order 2).- Thus, \([G : G_a] = \frac{6}{2} = 3\) for each \(a\).- Confirming that for every \(a\), \( |O_a| = 3 = [G : G_a] \). The relation holds for each \(a\).

Step 5: Determine Transitivity of the Action

The action is transitive if there is a single orbit for all elements of \(X\), meaning any element can reach any other via the group's action.- As all orbits are \(\{1, 2, 3\}\), the action is transitive because \(S_3\) can map any element to any other.

Step 6: Determine Faithfulness of the Action

An action is faithful if the only group element that acts as the identity on every element of \(X\) is the identity itself.- Since the only permutation in \(S_3\) that acts trivially on all of \(X\) is the identity, the action is faithful.

Key Concepts

Symmetric Group

The symmetric group, denoted as \(S_n\), is an essential object in abstract algebra. It's the group consisting of all possible permutations of \(n\) distinct objects. In our specific case, \(G = S_3\) signifies the symmetric group on three elements, namely the set \(X = \{1, 2, 3\}\).
This group encompasses all permutations of the set \(X\), which totals to 6 permutations, each representing a distinct way to arrange the three elements.
Some examples of permutations in \(S_3\) include:

• The identity permutation \(()\), which leaves all elements in their original position.
• Permutations like \((12)\), which swaps the positions of elements 1 and 2.
• More complex permutations like \((123)\), which cycles the positions through 1 to 2, 2 to 3, and finally 3 to 1.

This wealth of permutations makes \(S_3\) extraordinarily versatile when understanding symmetries and actions on sets.

Stabilizer

The stabilizer, denoted as \(G_a\), plays a critical role in our exercise. It refers to the subset of a group \(G\) that keeps an element \(a\) unchanged when the group's actions are applied. Essentially, \(G_a\) encompasses all permutations \(g\) in \(G\) for which the action of \(g\) on \(a\) leaves \(a\) unaltered.
In the scenario for \(S_3\) acting on \(X = \{1, 2, 3\}\), the stabilizer for any element \(a\) (e.g., 1, 2, or 3) will include permutations that do not affect the position of \(a\).

• For element 1, permutations like \(()\) and \((23)\) are stabilizers.
• Similarly, for element 2, permutations \(()\) and \((13)\) leave 2 in its place.
• The pattern continues for element 3 with permutations \(()\) and \((12)\).

The size of each stabilizer set in \(S_3\) is consistent, equivalent to a group \(S_2\), with 2 permutations.

Orbit-Stabilizer Theorem

The Orbit-Stabilizer Theorem is a fascinating theorem in group theory that connects orbit sizes, stabilizers, and group orders. The theorem states that the product of the size of the orbit of an element \(a\) and the order of its stabilizer equals the order of the group. Mathematically expressed as \( |O_a| \times |G_a| = |G| \).
In our exercise:

• We identified the orbit \(O_a\) for each element (1, 2, and 3) as \( \{1, 2, 3\} \). Hence, \(|O_a| = 3\).
• The stabilizer \(G_a\) for any element in \(S_3\) has a size of 2, since it's isomorphic to \(S_2\).
• The entire symmetric group \(S_3\) consists of 6 permutations making \(|G| = 6\).

Thus, confirming the Orbit-Stabilizer Theorem, \( |O_a| = \frac{|G|}{|G_a|} = \frac{6}{2} = 3 \). This relationship holds true for each element \(a\) of the set.

Transitive Action

In group theory, an action is termed transitive if every element of the set can be mapped to any other element through the group's action. Essentially, there is only one orbit encompassing all elements.
In the context of our problem, the action of \(S_3\) on \(X = \{1, 2, 3\}\) is undoubtedly transitive.

• We determined that the orbit for any element in this setup is the entire set itself, \( \{1, 2, 3\} \).
• This implies that no matter which element you start with, there's a permutation in \(S_3\) that can map it to another chosen element.

This all-encompassing reach of permutations affirms the transitive nature of the action.

Faithful Action

A group action is considered faithful if each non-identity element of the group actually acts non-trivially on some element of the set. This means there isn’t a non-identity permutation in the group that leaves the entire set unchanged.
In our exercise, the action of \(S_3\) on \(X = \{1, 2, 3\}\) is faithful. This is because:

• The only permutation in \(S_3\) that can act on all elements to leave them as they are, is the identity permutation \(()\).
• Any other permutation will change the order of elements in some way.

This definitively demonstrates the action is faithful, ensuring that the identity is only permutation that leaves each element of the set in its original position.