Problem 8 - Algebra: Pure and Applied

Question

Find the order of the indicated element in the indicated quotient group. $$ 3+\langle 6\rangle \text { in } \mathbb{Z}_{15} /\langle 6\rangle $$

Step-by-Step Solution

Answer

The order of 3 in \( \mathbb{Z}_{15}/\langle 6\rangle \) is 2.

Step 1: Identify the quotient group

The quotient group \( \mathbb{Z}_{15}/\langle 6\rangle \) consists of the cosets of \( \mathbb{Z}_{15} \) by the subgroup \( \langle 6 \rangle \). The subgroup \( \langle 6 \rangle \) contains all multiples of 6 in \( \mathbb{Z}_{15} \), specifically \( \langle 6 \rangle = \{0, 6, 12\} \).

Step 2: Identify the cosets

The element \( 3 + \langle 6 \rangle \) represents the coset \( \{3, 9, 15\} \). Note that 15 is equivalent to 0 in \( \mathbb{Z}_{15} \), so the coset simplifies to \( \{3, 9, 0\} \).

Step 3: Determine the group order

Compute the order of the group \( \mathbb{Z}_{15}/\langle 6\rangle \). Since \(|\langle 6 \rangle| = 3\), the index of \( \langle 6 \rangle \) in \( \mathbb{Z}_{15} \) is \(15/3 = 5\), so \( \mathbb{Z}_{15}/\langle 6\rangle \) has 5 elements.

Step 4: Calculate the order of the coset

The order of the element \( 3 + \langle 6 \rangle \) is the smallest positive integer \( n \) such that \( n(3 + \langle 6 \rangle) = \langle 6 \rangle \). Using iterative addition, we have:- \( 1(3 + \langle 6 \rangle) = 3 + \langle 6 \rangle \)- \( 2(3 + \langle 6 \rangle) = 6 + \langle 6 \rangle = \langle 6 \rangle \) (since 6 is a member of the subgroup \( \langle 6 \rangle \))Thus, the order of \( 3 + \langle 6 \rangle \) is 2.

Key Concepts

Quotient Group

A quotient group is an important concept in group theory, allowing us to form a new group from an existing group by using a subgroup. It's like creating a simplified version where certain elements are considered equivalent. In the context of our exercise, we have the group \( \mathbb{Z}_{15} \) and a subgroup \( \langle 6 \rangle \).

• To create a quotient group, we take the whole group \( \mathbb{Z}_{15} \)
• We then identify the subgroup \( \langle 6 \rangle \), which includes all multiples of the number 6.

The elements of the quotient group \( \mathbb{Z}_{15} / \langle 6 \rangle \) are not just single numbers; they are collections of numbers called cosets. Each coset is essentially an entire set of elements that we treat as one entity in this new group formation.

Cosets

Cosets form the building blocks of quotient groups. They represent collections of elements within the larger group that we iterate over. Each coset in a quotient group is an aggregate of shifts over the subgroup. Here, cosets basically bundle together elements that are considered equivalent under subgroup operations.In our example, find the cosets formed by adding the elements of \( \mathbb{Z}_{15} \) to \( \langle 6 \rangle \).

• The coset represented by \( 3 + \langle 6 \rangle \) is the collection \( \{3, 9, 0\} \).
• This is because, when you take the element 3 and keep adding any element from the subgroup \( \langle 6 \rangle \), you get either 3, 9, or 0, effectively visualizing the cycles of influence of the subgroup.

Understanding cosets is crucial because operations in the quotient group operate on these cosets rather than on individual elements. It's like simplifying the entire group operations by considering these collective behaviors.

Order of an Element

The order of an element in a group describes how the group's operation affects it, specifically the smallest number of operations required to bring the element back to an equivalent of the identity element, which is the subgroup here. In quotient groups, we're usually finding how many times the coset can be iterated within the operations of the group before reaching the identity.In our specific case, the order of the coset \( 3 + \langle 6 \rangle \) in \( \mathbb{Z}_{15}/\langle 6 \rangle \) is identified as:

• Calculate multiple additions: Starting with \( 3 + \langle 6 \rangle \), iterate through until reaching a member of \( \langle 6 \rangle \),
• Here, \( 2(3 + \langle 6 \rangle) = 6 + \langle 6 \rangle = \langle 6 \rangle \), indicating that two rounds bring us to the neutral set, which acts as the identity.

Thus, the order of \( 3 + \langle 6 \rangle \) is 2. This means it takes two operations to land back at the equivalent of the identity element within this structure.