Is Z12 isomorphic to the external direct product of Z6 and Z2?

Yes, \( \mathbb{Z}_{12} \) is isomorphic to the external direct product of \( \mathbb{Z}_6 \times \mathbb{Z}_2 \). To understand why, we can explore the properties of these two groups:

### 1. **Group Structure of \( \mathbb{Z}_{12} \):**
   - \( \mathbb{Z}_{12} \) is the cyclic group of order 12, generated by the element 1 (mod 12). Every element of \( \mathbb{Z}_{12} \) can be written as \( \{0, 1, 2, \dots, 11\} \), and the group operation is addition modulo 12.

### 2. **Group Structure of \( \mathbb{Z}_6 \times \mathbb{Z}_2 \):**
   - \( \mathbb{Z}_6 \) is the cyclic group of order 6, with elements \( \{0, 1, 2, 3, 4, 5\} \), and the group operation is addition modulo 6.
   - \( \mathbb{Z}_2 \) is the cyclic group of order 2, with elements \( \{0, 1\} \), and the group operation is addition modulo 2.
   - The external direct product \( \mathbb{Z}_6 \times \mathbb{Z}_2 \) consists of pairs \( (a, b) \), where \( a \in \mathbb{Z}_6 \) and \( b \in \mathbb{Z}_2 \). The group operation is component-wise addition modulo 6 and modulo 2, respectively.

### 3. **Checking the Isomorphism:**
   - To check if two groups are isomorphic, they must have the same order (number of elements) and an isomorphism must exist between their group structures.
   - The order of \( \mathbb{Z}_{12} \) is 12.
   - The order of \( \mathbb{Z}_6 \times \mathbb{Z}_2 \) is \( 6 \times 2 = 12 \).
   
   Since both groups have the same number of elements, we need to check if their group structures are compatible. One way to check this is to verify if they are both cyclic and if they share the same group type.
   
   - \( \mathbb{Z}_{12} \) is cyclic and generated by 1 mod 12.
   - \( \mathbb{Z}_6 \times \mathbb{Z}_2 \) is also cyclic because the element \( (1, 1) \) generates all 12 elements of the group.

Thus, the two groups are isomorphic. This is an example of the structure theorem for finite abelian groups, which implies that the cyclic group \( \mathbb{Z}_{12} \) can be decomposed as a direct product of \( \mathbb{Z}_6 \) and \( \mathbb{Z}_2 \).

### Conclusion:
Yes, \( \mathbb{Z}_{12} \cong \mathbb{Z}_6 \times \mathbb{Z}_2 \), meaning these two groups are isomorphic.

Would you like more details or have any questions? Here are some related questions you might explore:

1. How do you construct an explicit isomorphism between \( \mathbb{Z}_{12} \) and \( \mathbb{Z}_6 \times \mathbb{Z}_2 \)?
2. What is the structure theorem for finite abelian groups?
3. Can you provide another example of two isomorphic groups?
4. Is \( \mathbb{Z}_{12} \) isomorphic to \( \mathbb{Z}_4 \times \mathbb{Z}_3 \)?
5. How does the decomposition of a group as a direct product affect its subgroups?

**Tip:** When trying to establish an isomorphism between groups, always check both the group orders and the generators to ensure compatibility in structure.

Discover why Z12 is isomorphic to the external direct product of Z6 and Z2 with a detailed explanation rooted in group theory. This solution explores cyclic groups, the structure theorem for finite abelian groups, and isomorphism properties. Suitable for undergraduate students studying algebra.

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