Let's tackle each of these questions step by step.

1. Number of Subgroups of Order 13 in 𝐺

Since $G$ is a group of order 52, by the Sylow theorems, the number of subgroups of order 13 (a prime number) is denoted by $n_{13}$.

According to the Sylow theorems:

1. $n_{13}$ divides $|G|/13 = 4$.
2. $n_{13} \equiv 1 \ (\text{mod} \ 13)$.

The divisors of 4 are 1 and 4. The only number that is congruent to 1 modulo 13 is $n_{13} = 1$.

So, there is exactly one subgroup of order 13 in $G$.

2. Number of Subgroups of Order 4 in $G = \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/26\mathbb{Z}$

Let's break this down:

• The group $G = \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/26\mathbb{Z}$ has elements of the form $(a, b)$ where $a \in \mathbb{Z}/2\mathbb{Z}$ and $b \in \mathbb{Z}/26\mathbb{Z}$.
• The order of an element $(a, b)$ is the least common multiple (LCM) of the orders of $a$ and $b$.

We need subgroups of order 4, so consider the elements:

• The possible orders for $a$ are 1 and 2.
• The possible orders for $b$ are 1, 2, 13, and 26.

To have an element of order 4, $a$ must have order 2 (so $a = 1$ mod 2), and $b$ must have order 2 (so $b = 13$ mod 26). This means $b$ could be either 13 or 26 (but note that 26 is not of order 2).

The possible elements of order 2 in $\mathbb{Z}/26\mathbb{Z}$ are just $13$.

So, the subgroup is generated by $(1, 13)$, which has order 4. But since $(0, b)$ where $b$ is of order 4 doesn't exist (the element of order 4 in $\mathbb{Z}/26\mathbb{Z}$ would be 13), the subgroups of order 4 in $G$ are the subgroups generated by elements of order 4.

Now let's list the possible subgroups.

3. Listing Elements of Each Subgroup of Order 4

The element $(1, 13)$ has order 4 in $G$. The subgroup generated by this element is: $\langle (1, 13) \rangle = \{(0, 0), (1, 13), (0, 26), (1, 13)\}$

Since $\mathbb{Z}/26\mathbb{Z}$ doesn't provide any new elements that could combine with $\mathbb{Z}/2\mathbb{Z}$ to form more elements of order 4, this is the only subgroup of order 4.

Summary

1. The group $G$ has exactly one subgroup of order 13.
2. The group $G = \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/26\mathbb{Z}$ has one subgroup of order 4.
3. The elements of the subgroup of order 4 are:
   • $(0, 0)$
   • $(1, 13)$
   • $(0, 26) = (0, 0)$
   • $(1, 13)$

This results in the subgroup $\langle (1, 13) \rangle = \{(0, 0), (1, 13)\}$.

Do you want more details or have any questions?

Here are some related questions you might find interesting:

1. What is the significance of the Sylow theorems in group theory?
2. How can we determine the number of subgroups of a particular order in a finite group?
3. Can a group have more than one Sylow $p$-subgroup for some prime $p$?
4. What is the structure of the group $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/26\mathbb{Z}$?
5. How do you determine the order of an element in a direct product of groups?
6. What are the possible orders of elements in $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/26\mathbb{Z}$?
7. How do the properties of $\mathbb{Z}/p\mathbb{Z}$ affect the subgroup structure of a group?
8. What is the role of cyclic subgroups in the structure of a group?

Tip: Always check the divisibility conditions and congruences when applying the Sylow theorems to determine the number of possible subgroups of a given order in a finite group.