Let G be a group of order 42.
1. Determine the number of Sylow-7 subgroups of G.
2. Let H1 ⊆ G be a Sylow-3 subgroup of G. Show that G has a subgroup H2 satisfying
|H2| = 2 and H1 ∩ H2 = {e}.
3. Prove that G has a subgroup of order 6

To solve these problems, let's start by using Sylow theorems and other group theory principles.

### 1. Determine the number of Sylow-7 subgroups of G.

The order of the group \( |G| \) is 42. The Sylow \( p \)-subgroups of \( G \) have orders equal to the highest power of \( p \) dividing 42. 

First, factorize \( 42 \):
\[ 42 = 2 \cdot 3 \cdot 7 \]

The order of a Sylow-7 subgroup is \( 7^1 = 7 \).

According to Sylow's theorems, the number \( n_7 \) of Sylow-7 subgroups satisfies:
1. \( n_7 \equiv 1 \mod 7 \)
2. \( n_7 \) divides \( \frac{|G|}{7} = \frac{42}{7} = 6 \)

So, \( n_7 \equiv 1 \mod 7 \) and \( n_7 \) divides 6. The possible divisors of 6 are 1, 2, 3, and 6.

The only number that satisfies both conditions is \( n_7 = 1 \). Therefore, there is exactly one Sylow-7 subgroup in \( G \).

### 2. Show that \( G \) has a subgroup \( H_2 \) satisfying \( |H_2| = 2 \) and \( H_1 \cap H_2 = \{e\} \).

Let \( H_1 \subseteq G \) be a Sylow-3 subgroup of \( G \). The order of \( H_1 \) is \( 3 \).

Since \( |G| = 42 \), there is at least one Sylow-2 subgroup of \( G \) because the highest power of 2 dividing 42 is 2. A Sylow-2 subgroup \( H_2 \) of \( G \) has order 2.

From the Sylow theorems:
1. The number \( n_2 \) of Sylow-2 subgroups satisfies:
   - \( n_2 \equiv 1 \mod 2 \)
   - \( n_2 \) divides \( \frac{42}{2} = 21 \)

The possible values for \( n_2 \) are 1 or 21. Since \( n_2 \equiv 1 \mod 2 \), \( n_2 = 1 \) or \( n_2 = 21 \).

For \( H_1 \cap H_2 \):
- The order of \( H_1 \) is 3, which is relatively prime to 2 (the order of \( H_2 \)).
- Therefore, \( H_1 \cap H_2 \) can only be the identity element \( e \).

This implies that \( G \) has a subgroup \( H_2 \) of order 2 such that \( H_1 \cap H_2 = \{e\} \).

### 3. Prove that \( G \) has a subgroup of order 6.

A subgroup of order 6 must have an element of order 6. Since \( |G| = 42 \), we can use the fact that elements of \( G \) can have orders dividing 42.

Using Lagrange's theorem, possible orders of elements in \( G \) include \( 1, 2, 3, 6, 7, 14, 21, 42 \).

To form a subgroup of order 6, we can use elements from the Sylow-3 and Sylow-2 subgroups. Consider the Sylow-3 subgroup \( H_1 \) of order 3 and a Sylow-2 subgroup \( H_2 \) of order 2.

Since \( H_1 \cap H_2 = \{e\} \), the product \( H_1H_2 \) has order:
\[ |H_1||H_2| = 3 \times 2 = 6 \]

Thus, \( G \) contains a subgroup of order 6, which is the product of the Sylow-3 and Sylow-2 subgroups.

### Summary:

1. \( G \) has exactly one Sylow-7 subgroup.
2. \( G \) has a Sylow-2 subgroup \( H_2 \) such that \( H_1 \cap H_2 = \{e\} \).
3. \( G \) has a subgroup of order 6.

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**Related Questions:**

1. What are the Sylow theorems?
2. Can you give an example of a group of order 42?
3. How do Sylow subgroups help in classifying groups?
4. What is Lagrange's theorem in group theory?
5. How do you determine the order of a group element?
6. What are normal subgroups and how do they relate to Sylow subgroups?
7. Can a group have more than one Sylow subgroup of a given order?
8. How can the concept of cosets be used to understand group structure?

**Tip:** When dealing with Sylow subgroups, remember that their number has specific divisibility and congruence properties which are crucial for understanding the structure of the group.

Explore group theory problems involving Sylow-7 subgroups, Sylow-3 subgroups, and subgroups of order 6. Understand the application of Sylow theorems and Lagrange's theorem in determining subgroup structures in groups of order 42.

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