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 solve the problems step by step.

### Part 1: Size of \( G \) and Order of the Element \( (1, 4) \)

**Size of \( G \):**

Given \( G = \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/6\mathbb{Z} \), the size of \( G \) is the product of the sizes of the two groups:

\[
|G| = |\mathbb{Z}/2\mathbb{Z}| \times |\mathbb{Z}/6\mathbb{Z}| = 2 \times 6 = 12
\]

**Order of the Element \( (1, 4) \):**

To find the order of \( (1, 4) \in G \), we need the smallest positive integer \( k \) such that \( k(1, 4) = (0, 0) \). This means \( k \) must satisfy:

\[
k \cdot 1 \equiv 0 \pmod{2} \quad \text{and} \quad k \cdot 4 \equiv 0 \pmod{6}
\]

For the first condition, \( k \) must be even because \( k \equiv 0 \pmod{2} \).

For the second condition, \( k \cdot 4 \equiv 0 \pmod{6} \). Simplifying this, we get:

\[
k \cdot 4 \equiv 0 \pmod{6} \implies k \cdot 4 = 6m \text{ for some integer } m
\]

Thus, \( k \) must be a multiple of \( \frac{6}{\gcd(4, 6)} = \frac{6}{2} = 3 \).

Combining the two conditions, \( k \) must be the least common multiple (LCM) of 2 and 3, which is:

\[
\text{lcm}(2, 3) = 6
\]

Therefore, the order of the element \( (1, 4) \) is 6.

### Part 2: Maximal Possible Order of \( g \) and Non-cyclicity of \( G \) When \( \gcd(2, n) > 1 \)

If \( \gcd(2, n) > 1 \), it means \( n \) is even (since 2 is prime). Let \( n = 2m \) for some integer \( m \).

**Maximal Order of \( g \):**

Given \( g = (a, b) \in \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z} \), the order of \( g \) is the least common multiple (LCM) of the orders of \( a \) and \( b \):

\[
\text{ord}(g) = \text{lcm}(\text{ord}(a), \text{ord}(b))
\]

- The order of \( a \in \mathbb{Z}/2\mathbb{Z} \) is either 1 (if \( a = 0 \)) or 2 (if \( a = 1 \)).
- The order of \( b \in \mathbb{Z}/n\mathbb{Z} \) is the smallest positive integer \( k \) such that \( k \cdot b \equiv 0 \pmod{n} \).

Since \( n = 2m \), if \( b \) is odd, its order is \( n \). If \( b \) is even, let \( b = 2b' \), then the order of \( b \) is the smallest positive integer \( k \) such that \( k \cdot 2b' \equiv 0 \pmod{2m} \), which is \( m \).

The maximal possible value of \( \text{ord}(g) \) is when \( \text{ord}(a) = 2 \) and \( \text{ord}(b) = n \), which gives:

\[
\text{lcm}(2, n) = \frac{2n}{\gcd(2, n)} = \frac{2 \cdot 2m}{2} = 2m = n
\]

Therefore, the maximal possible order of \( g \) is \( n \).

**Proof that \( G \) Cannot be Cyclic When \( \gcd(2, n) > 1 \):**

To be cyclic, \( G \) must have an element whose order is \( |G| = 2n \).

However, as calculated, the maximal order of any element in \( G \) when \( \gcd(2, n) > 1 \) is \( n \). Since \( n < 2n \), \( G \) cannot be cyclic.

### Part 3: Cyclicity of \( G \) When \( 2 \nmid n \)

If \( 2 \nmid n \), \( n \) is odd. Let \( n = 2k + 1 \) for some integer \( k \).

Since \( n \) is odd, any element \( b \in \mathbb{Z}/n\mathbb{Z} \) can have order \( n \).

Consider \( G = \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z} \). 

Let \( g = (1, 1) \). The order of \( g \) is:

\[
\text{ord}(g) = \text{lcm}(\text{ord}(1 \in \mathbb{Z}/2\mathbb{Z}), \text{ord}(1 \in \mathbb{Z}/n\mathbb{Z})) = \text{lcm}(2, n)
\]

Since \( 2 \nmid n \), \( \gcd(2, n) = 1 \), thus:

\[
\text{lcm}(2, n) = 2n
\]

Therefore, \( g \) has order \( 2n \), which is the size of \( G \). Hence, \( G \) is cyclic.

### Summary

1. The size of \( G \) is 12, and the order of \( (1, 4) \) is 6.
2. If \( \gcd(2, n) > 1 \), the maximal possible order of \( g \in G \) is \( n \), so \( G \) cannot be cyclic.
3. If \( 2 \nmid n \), \( G \) is cyclic.

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

1. What are the generators of a cyclic group?
2. How do you determine the order of an element in a direct product of groups?
3. What is the structure theorem for finitely generated abelian groups?
4. How does the Chinese Remainder Theorem apply to group theory?
5. What are the properties of cyclic groups?
6. How can you determine if a given group is isomorphic to another group?
7. What are the applications of cyclic groups in number theory?
8. Can non-abelian groups be cyclic?

**Tip:** To check if a group is cyclic, find an element whose order equals the size of the group.

Explore a group theory problem involving Z/2Z × Z/nZ structure and cyclic properties. Understand the size of the group, the order of specific elements, and conditions under which the group can be cyclic or not. Suitable for advanced undergraduate and graduate students.

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