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 $G$.

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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: $\mathbb{Z}_{26}$

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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.