Let n ≥ 2 be an integer, and let G = Z/2Z × Z/nZ, where we use component-wise
addition as the group operation.
1. Consider the case n = 6, i.e., G = Z/2Z × Z/6Z. Determine the size of |G| and the
order of the element (1, 4) ∈ G.
2. Assume now that g ∈ G and gcd(2, n) > 1. Determine the maximal possible value
of ord(g). Use this to prove that G cannot be cyclic in this case.
3. Prove that G is cyclic if 2 ∤ n.

Let's tackle each part of the problem step by step.

### Part 1

**Determine the size of \( |G| \) and the order of the element \( (1, 4) \in G \).**

Given \( G = \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/6\mathbb{Z} \):

1. **Size of \( |G| \):**

The size of the group \( G \) is the product of the sizes of \( \mathbb{Z}/2\mathbb{Z} \) and \( \mathbb{Z}/6\mathbb{Z} \):
\[
|G| = |\mathbb{Z}/2\mathbb{Z}| \times |\mathbb{Z}/6\mathbb{Z}| = 2 \times 6 = 12
\]

2. **Order of the element \( (1, 4) \):**

The order of an element \( (a, b) \in \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/6\mathbb{Z} \) is the least common multiple (LCM) of the orders of \( a \) and \( b \).

- The order of \( 1 \in \mathbb{Z}/2\mathbb{Z} \) is 2.
- The order of \( 4 \in \mathbb{Z}/6\mathbb{Z} \) is the smallest positive integer \( k \) such that \( 4k \equiv 0 \pmod{6} \).

We find \( k \) by solving:
\[
4k \equiv 0 \pmod{6}
\]
The smallest \( k \) is 3 because \( 4 \times 3 = 12 \equiv 0 \pmod{6} \).

Therefore, the order of \( (1, 4) \) is:
\[
\text{lcm}(2, 3) = 6
\]

### Part 2

**Assume now that \( g \in G \) and \( \gcd(2, n) > 1 \). Determine the maximal possible value of \( \text{ord}(g) \). Use this to prove that \( G \) cannot be cyclic in this case.**

If \( \gcd(2, n) > 1 \), let \( d = \gcd(2, n) \). Then \( d \) must be either 2 or greater.

1. **Maximal possible order of an element \( g \in G \):**

Since \( G = \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z} \), the order of an element \( (a, b) \) is \( \text{lcm}(\text{ord}(a), \text{ord}(b)) \).

- \( \text{ord}(a) \) is at most 2 (the order of elements in \( \mathbb{Z}/2\mathbb{Z} \)).
- \( \text{ord}(b) \) is at most \( n/d \), where \( d = \gcd(2, n) \).

Since \( \gcd(2, n) > 1 \), \( \text{ord}(b) \) must be a divisor of \( n/d \), and the maximal possible order is \( n/2 \).

2. **Prove that \( G \) cannot be cyclic if \( \gcd(2, n) > 1 \):**

For \( G \) to be cyclic, there must be an element whose order is equal to the size of the group \( |G| \).

- The size of the group \( |G| \) is \( 2n \).
- If \( \gcd(2, n) > 1 \), the maximal possible order of any element \( g \in G \) is \( n/2 \).

Since \( n/2 < 2n \), no element in \( G \) can have order \( 2n \). Therefore, \( G \) cannot be cyclic.

### Part 3

**Prove that \( G \) is cyclic if \( 2 \nmid n \).**

If \( 2 \nmid n \), then \( n \) is odd. In this case, \( \mathbb{Z}/2\mathbb{Z} \) and \( \mathbb{Z}/n\mathbb{Z} \) are both cyclic groups with coprime orders (2 and \( n \), respectively).

Since the orders are coprime, the direct product of these two cyclic groups is also cyclic. Specifically, \( G = \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z} \) is cyclic with an element of order \( 2n \).

Therefore, \( G \) is cyclic when \( 2 \nmid n \).

---

Would you like details on any part of this solution or have any questions? Here are some related questions to consider:

1. What are some examples of elements in \( G \) when \( n = 6 \)?
2. How would the order of an element change if \( n \) were a different even number?
3. What is the structure of \( G \) when \( n \) is a prime number?
4. How does the LCM of two numbers relate to the concept of group order?
5. Can \( G \) be cyclic if \( n \) is a power of 2?
6. What happens to the order of \( G \) if \( n \) is a product of distinct primes?
7. How do we determine if a group is cyclic in general?
8. What are some real-world applications of cyclic groups?

**Tip:** When working with groups, it is useful to understand the orders of elements and the structure of the group to determine properties like whether it is cyclic or not.

Explore a group theory problem involving G = Z/2Z × Z/nZ. Determine the size of |G|, the order of the element (1, 4) ∈ G, and conditions under which G is cyclic. Learn the applications of group theory in algebraic structures.

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