Let n ≥ 2 be an integer, and let G = Z/2Z × Z/nZ, where we use component-wise addition as the group operation

Assume now that g ∈ G and gcd(2, n) > 1. Determine the maximal possible value of ord(g)

To determine the maximal possible value of the order of an element \( g \in G \), where \( G = \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z} \) and \(\gcd(2, n) > 1\), we need to carefully analyze the structure of \( G \) and the properties of its elements.

### Step 1: Structure of \( G \)
The group \( G \) consists of ordered pairs \((a, b)\), where \( a \in \mathbb{Z}/2\mathbb{Z} \) and \( b \in \mathbb{Z}/n\mathbb{Z} \). Each component is added modulo its respective modulus. Hence, the elements of \( G \) are \((0, b)\) and \((1, b)\) for \( b \in \mathbb{Z}/n\mathbb{Z} \).

### Step 2: Order of an element in \( G \)
The order of an element \( g = (a, b) \in G \) is the smallest positive integer \( k \) such that \( k \cdot (a, b) = (0, 0) \). This is equivalent to finding the least common multiple (LCM) of the orders of the components \( a \) and \( b \).

#### Order of \( a \)
- If \( a = 0 \), then the order of \( a \) is 1.
- If \( a = 1 \), then the order of \( a \) is 2 (since \( 1 + 1 = 0 \) in \(\mathbb{Z}/2\mathbb{Z}\)).

#### Order of \( b \)
The order of \( b \) in \( \mathbb{Z}/n\mathbb{Z} \) is the smallest positive integer \( m \) such that \( m \cdot b = 0 \mod n \). This is given by \( \frac{n}{\gcd(b, n)} \).

### Step 3: Finding the maximal possible order
To find the maximal possible order of an element \( (a, b) \), we need to maximize the LCM of the orders of \( a \) and \( b \).

- For \( a = 0 \), the order is simply the order of \( b \), which is \( \frac{n}{\gcd(b, n)} \). This achieves its maximum when \(\gcd(b, n)\) is minimized, ideally \( \gcd(b, n) = 1 \), giving an order of \( n \).
- For \( a = 1 \), the order is \( \text{LCM}(2, \frac{n}{\gcd(b, n)}) \).

Given that \(\gcd(2, n) > 1\), \( n \) must be even. Hence, the possible values of \(\gcd(b, n)\) can vary, but we are interested in the case where \( \frac{n}{\gcd(b, n)} \) contributes maximally.

### Evaluating LCM for maximal order
To maximize \(\text{LCM}(2, \frac{n}{\gcd(b, n)})\):
- If \(\gcd(b, n) = 2\), then \(\frac{n}{\gcd(b, n)} = \frac{n}{2}\).
- The LCM is \( \text{LCM}(2, \frac{n}{2}) \).

For \( n = 2k \) (an even number),
- If \(\gcd(b, n) = 2\), then \(\frac{n}{\gcd(b, n)} = k\),
- Thus, \( \text{LCM}(2, k) \).

This \( \text{LCM} \) achieves its maximal value when \( k \) is a multiple of 2, hence the largest possible value of \( k \) is \( \frac{n}{2} \).

Thus, the maximal possible order of any element in \( G \) occurs when \( k \) is maximized, which gives \( n \). This achieves \( n \) in some specific cases, but we need \( \text{LCM}(2, k) \) to be maximal, so:

### Final Result
The maximal possible value of the order of \( g \in G \) where \( \gcd(2, n) > 1 \) is:

\[ \boxed{n} \]

since the LCM of the orders of components when maximized in the scenarios above yields \( n \) when conditions permit.

Explore the maximal possible order of an element g in the group G = Z/2Z × Z/nZ, where gcd(2, n) > 1. Understand how to determine the order based on the structure of the group and properties of its elements. Suitable for advanced undergraduate students and beyond.

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