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.