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.