Give the order of the element in the factor group 20+<12> in Z60/ <12>

To find the order of the element \( 20 + \langle 12 \rangle \) in the factor group \( \mathbb{Z}_{60} / \langle 12 \rangle \), we need to understand what this notation means and how to compute it.

### Step-by-step Solution

1. **Understanding the Groups and Notation:**
   - \( \mathbb{Z}_{60} \) is the set of integers modulo 60.
   - \( \langle 12 \rangle \) represents the subgroup generated by 12 in \( \mathbb{Z}_{60} \). This subgroup consists of all multiples of 12 modulo 60, i.e., \( \{0, 12, 24, 36, 48\} \).
   - The factor group \( \mathbb{Z}_{60} / \langle 12 \rangle \) consists of all cosets of \( \langle 12 \rangle \) in \( \mathbb{Z}_{60} \).

2. **Cosets in the Factor Group:**
   - A coset of \( \langle 12 \rangle \) in \( \mathbb{Z}_{60} \) is of the form \( a + \langle 12 \rangle \), where \( a \) is an integer in \( \mathbb{Z}_{60} \).
   - Specifically, \( 20 + \langle 12 \rangle \) is the coset containing all elements of the form \( 20 + 12k \) for integers \( k \).

3. **Finding the Order of an Element:**
   - The order of an element \( g + \langle 12 \rangle \) in the factor group \( \mathbb{Z}_{60} / \langle 12 \rangle \) is the smallest positive integer \( n \) such that \( n(g + \langle 12 \rangle) = \langle 12 \rangle \) (the identity coset).
   - Equivalently, this means finding the smallest \( n \) such that \( ng \in \langle 12 \rangle \).

4. **Applying this to \( 20 + \langle 12 \rangle \):**
   - We need to find the smallest \( n \) such that \( 20n \) is a multiple of 12 modulo 60.

5. **Calculation:**
   - Start by finding the multiples of 12 modulo 60: \( 0, 12, 24, 36, 48 \).
   - We want to find the smallest \( n \) such that \( 20n \equiv 0, 12, 24, 36, \) or \( 48 \pmod{60} \).

   Let's check each case:
   - \( n = 1: 20 \times 1 = 20 \notin \langle 12 \rangle \)
   - \( n = 2: 20 \times 2 = 40 \notin \langle 12 \rangle \)
   - \( n = 3: 20 \times 3 = 60 \equiv 0 \pmod{60} \)

   Here, \( 20 \times 3 = 60 \) is a multiple of 12 (since 0 is a multiple of any number). Thus, \( n = 3 \) is the smallest integer that satisfies the condition.

### Conclusion
The order of the element \( 20 + \langle 12 \rangle \) in the factor group \( \mathbb{Z}_{60} / \langle 12 \rangle \) is **3**.

Would you like further details on this calculation or have any questions?

Here are some related questions for further exploration:
1. What is the factor group \( \mathbb{Z}_{60} / \langle 12 \rangle \) isomorphic to?
2. How would you find the order of the element \( 30 + \langle 12 \rangle \) in \( \mathbb{Z}_{60} / \langle 12 \rangle \)?
3. What is the significance of the subgroup \( \langle 12 \rangle \) in \( \mathbb{Z}_{60} \)?
4. How would the order change if we considered \( \mathbb{Z}_{60} / \langle 6 \rangle \) instead?
5. What is a coset, and how do you determine its elements in a group?

**Tip:** When working with factor groups, always check if the subgroup is normal in the larger group, as it ensures that the factor group is well-defined.

Explore how to find the order of the element 20 + <12> in the factor group Z60/<12>. Detailed step-by-step solution using group theory concepts. Learn about cosets, factor groups, and Lagrange's Theorem. Suitable for undergraduate students.

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