How many subgroups of order 13 does G = 52 have? And also find the subgroups when H * H and Z2 and Z26

To solve this, let's break it down:

### Step 1: **Finding the number of subgroups of order 13 in \( G \):**

We are given that the group \( G \) has order 52 (i.e., \( |G| = 52 \)). According to **Lagrange's Theorem**, the order of any subgroup of a finite group divides the order of the group. Therefore, the possible orders of subgroups in \( G \) must divide 52. The divisors of 52 are:

\[
1, 2, 4, 13, 26, 52
\]

We are asked to find how many subgroups of order 13 exist in \( G \).

Now, consider the **Sylow Theorems**, which tell us how to count subgroups of a particular order. A **Sylow \( p \)-subgroup** is a subgroup whose order is the largest power of a prime that divides the order of the group. Here, 13 is a prime number and divides 52. So, we are dealing with Sylow 13-subgroups.

According to the third Sylow Theorem, the number of Sylow 13-subgroups, denoted by \( n_{13} \), must satisfy two conditions:
1. \( n_{13} \equiv 1 \mod 13 \)
2. \( n_{13} \) divides \( |G| \) (which is 52).

The divisors of 52 are \( 1, 2, 4, 13, 26, 52 \), so the possible values of \( n_{13} \) are one of these. However, it must satisfy \( n_{13} \equiv 1 \mod 13 \).

- \( n_{13} = 1 \) satisfies this condition.

Hence, there is **exactly 1 subgroup of order 13** in \( G \).

---

### Step 2: **Finding the subgroups for \( H * H \), \( \mathbb{Z}_2 \), and \( \mathbb{Z}_{26} \):**

1. **\( H * H \):**
   - If we interpret \( H * H \) as the direct product of two groups \( H \) and \( H \), we would need to know the structure of \( H \). Assuming \( H \) is a cyclic group, say \( \mathbb{Z}_n \), then \( H * H \) would be \( \mathbb{Z}_n \times \mathbb{Z}_n \), which is the product of two cyclic groups.
   
   To compute subgroups, it depends on the order and structure of \( H \). If you provide more details on \( H \), I can give you more information.

2. **\( \mathbb{Z}_2 \):**
   - \( \mathbb{Z}_2 \) is the cyclic group of order 2, consisting of two elements: \( \{0, 1\} \). It has exactly one non-trivial proper subgroup, which is the trivial group \( \{0\} \).

3. **\( \mathbb{Z}_{26} \):**
   - \( \mathbb{Z}_{26} \) is the cyclic group of order 26. The subgroups of a cyclic group are determined by the divisors of its order. The divisors of 26 are \( 1, 2, 13, 26 \), so \( \mathbb{Z}_{26} \) has subgroups of orders 1, 2, 13, and 26.

   The specific subgroups of \( \mathbb{Z}_{26} \) would be:
   - The trivial subgroup: \( \{0\} \)
   - Subgroup of order 2: \( \{0, 13\} \)
   - Subgroup of order 13: Generated by \( 2 \)
   - The full group itself: \( \mathbb{Z}_{26} \)

---

Would you like further details on any of these, or do you have other specific questions?

### Follow-Up Questions:
1. What is the significance of the Sylow Theorems in group theory?
2. Can you explain how to find all the subgroups of a cyclic group?
3. How do direct products of groups affect the subgroup structure?
4. How do the orders of elements in a group relate to its subgroups?
5. How are Sylow subgroups related to normal subgroups in a group?

### Tip:
In any finite group, the number of subgroups of a particular order is always constrained by the Sylow Theorems, which ensure the existence and uniqueness in many cases.

Learn how to find subgroups of order 13 in a group of order 52 using Sylow Theorems and explore subgroup structures in cyclic groups and direct products of groups. This problem delves into group theory concepts and is suitable for advanced high school students.

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