Let 𝐺 be a group of order 52 = 4 ⋅ 13.
1. How many subgroups of order 13 does 𝐺 have?
Let now 𝐺 = ℤ/2ℤ × ℤ/26ℤ. (The operation being addition.)
2. How many subgroups of order 4 does 𝐺 have?
3. List the elements of each subgroup of order 4.

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.

Explore the analysis of subgroups in a group of order 52. Learn how to determine the number of subgroups of order 13 and order 4 in the group, and list elements of each subgroup of order 4. This problem involves key concepts from group theory and the application of Sylow theorems.

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