et G be a group of order 42.

Determine the number of Sylow-7 subgroups of G.

Let H1 ⊆ G be a Sylow-3 subgroup of G. Show that G has a subgroup H2 satisfying
|H2| = 2 and H1 ∩ H2 = {e}.

Let's analyze and solve the problem step by step.

### Part 1: Determining the number of Sylow-7 subgroups of \( G \)

Given that \( G \) is a group of order 42, we can use the Sylow theorems to determine the number of Sylow \( p \)-subgroups.

First, we factorize the order of the group:
\[ |G| = 42 = 2 \times 3 \times 7 \]

By the Sylow theorems, the number of Sylow-7 subgroups, denoted \( n_7 \), must satisfy:
1. \( n_7 \equiv 1 \pmod{7} \)
2. \( n_7 \) divides 6 (the cofactor \( 42 / 7 \))

The divisors of 6 are 1, 2, 3, and 6. Let's check which of these satisfy \( n_7 \equiv 1 \pmod{7} \):

- \( n_1 \equiv 1 \pmod{7} \) ✔️
- \( n_2 \equiv 2 \pmod{7} \) ❌
- \( n_3 \equiv 3 \pmod{7} \) ❌
- \( n_6 \equiv 6 \pmod{7} \) ❌

Thus, the only possible number is:
\[ n_7 = 1 \]

Therefore, there is exactly one Sylow-7 subgroup in \( G \).

### Part 2: Finding a subgroup \( H_2 \) of order 2 with \( H_1 \cap H_2 = \{e\} \)

Let \( H_1 \subseteq G \) be a Sylow-3 subgroup of \( G \). We know:
\[ |H_1| = 3 \]

To find a subgroup \( H_2 \) of order 2, we note that by Sylow theorems, there must be a Sylow-2 subgroup of \( G \). Let's denote this Sylow-2 subgroup by \( H_2 \). The order of \( H_2 \) is:
\[ |H_2| = 2 \]

We need to show that \( H_1 \cap H_2 = \{e\} \).

By the properties of Sylow subgroups, a Sylow \( p \)-subgroup is unique if and only if it is normal in \( G \). Here, we already have:
- The unique Sylow-7 subgroup, which is normal in \( G \).
- Sylow-3 subgroup \( H_1 \) (may not be unique or normal unless stated otherwise).

Now, consider the Sylow-2 subgroup \( H_2 \). Since \( H_2 \) is of order 2 and the order of \( H_1 \) is 3, their intersection must be the identity element because 2 and 3 are coprime. 

Hence:
\[ H_1 \cap H_2 = \{e\} \]

This implies that:
- The identity element \( e \) is the only common element between \( H_1 \) and \( H_2 \).

Thus, we have found the required subgroup \( H_2 \) of order 2 that satisfies \( H_1 \cap H_2 = \{e\} \).

### Summary

- The number of Sylow-7 subgroups in \( G \) is exactly 1.
- There exists a Sylow-2 subgroup \( H_2 \) in \( G \) such that \( H_1 \cap H_2 = \{e\} \), where \( H_1 \) is a Sylow-3 subgroup.

Do you want more details or have any questions?

Here are some related questions to further your understanding:

1. What are the Sylow theorems and how are they used to determine the number of Sylow subgroups?
2. Why must the number of Sylow \( p \)-subgroups be congruent to 1 modulo \( p \)?
3. What does it mean for a subgroup to be normal in a group?
4. How do you determine the orders of Sylow subgroups for a given group?
5. Why is the intersection of Sylow-3 and Sylow-2 subgroups trivial?
6. Can there be more than one Sylow-3 subgroup in \( G \)? Why or why not?
7. How does the coprimality of 2 and 3 affect the structure of \( G \)?
8. What role do normalizers play in the context of Sylow subgroups?

**Tip:** Understanding the fundamental properties of Sylow subgroups is crucial in group theory, especially in analyzing the structure of finite groups.

Explore the determination of Sylow-7 subgroups and the existence of a Sylow-2 subgroup satisfying specific conditions in a group of order 42. Understand how group theory principles and Sylow theorems apply in this context.

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