To determine the number of subgroups of order 4 in the group $G = \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/26\mathbb{Z}$, we need to explore the structure of $G$ and identify which subsets form subgroups of the desired order.

1. Structure of $G$:

   • $\mathbb{Z}/2\mathbb{Z}$ has elements $\{0, 1\}$.
   • $\mathbb{Z}/26\mathbb{Z}$ has elements $\{0, 1, 2, \ldots, 25\}$.

   Hence, $G$ consists of pairs $(a, b)$ where $a \in \mathbb{Z}/2\mathbb{Z}$ and $b \in \mathbb{Z}/26\mathbb{Z}$.

2. Order of elements:

   • The order of an element $(a, b)$ in $G$ is the least common multiple (LCM) of the orders of $a$ and $b$.
   • The order of $a \in \mathbb{Z}/2\mathbb{Z}$ is 1 if $a = 0$ and 2 if $a = 1$.
   • The order of $b \in \mathbb{Z}/26\mathbb{Z}$ depends on the value of $b$. Specifically, the order of $b$ is the smallest positive integer $k$ such that $kb \equiv 0 \pmod{26}$.

3. Finding elements of order 4:

   • To have a subgroup of order 4, we need an element whose order is 4. This requires that $\text{LCM}(\text{order of } a, \text{order of } b) = 4$.

   • Case 1: $a = 0$.

     • Here, the order of $(0, b)$ is just the order of $b$. We need $b$ to have order 4.
     • An element $b$ has order 4 in $\mathbb{Z}/26\mathbb{Z}$ if $b$ is such that $26/\gcd(26, b) = 4$. Thus, $\gcd(26, b) = 26/4 = 6.5$, which is not possible. So, no elements $b$ in $\mathbb{Z}/26\mathbb{Z}$ have order 4.

   • Case 2: $a = 1$.

     • Here, the order of $(1, b)$ is $\text{LCM}(2, \text{order of } b)$. To achieve order 4, the order of $b$ must be 4.
     • The order of $b$ in $\mathbb{Z}/26\mathbb{Z}$ must be 4. As shown above, there are no elements in $\mathbb{Z}/26\mathbb{Z}$ with order 4.

Since there are no elements of order 4 in $\mathbb{Z}/26\mathbb{Z}$, there can be no elements in $G$ with order 4, and thus no subgroups of order 4 in $G$.

4. Double-check:
   • The elements in $G$ form