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

Let's proceed with the analysis step-by-step.

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

Given the order of the group \( |G| = 42 \), we can factorize it as:
\[ 42 = 2 \times 3 \times 7 \]

The Sylow-7 subgroups of \( G \) have orders equal to \( 7 \).

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

The possible divisors of 6 are 1, 2, 3, and 6. Among these, only 1 satisfies \( n_7 \equiv 1 \mod 7 \).

Therefore, there is exactly one Sylow-7 subgroup in \( G \):
\[ n_7 = 1 \]

### 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 \( G \), so \( |H_1| = 3 \).

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

The possible values for \( n_2 \) are 1, 3, 7, and 21. Among these, 1 and 21 are the only values satisfying \( n_2 \equiv 1 \mod 2 \).

To show \( H_1 \cap H_2 = \{e\} \):
- The order of \( H_1 \) is 3, and the order of any Sylow-2 subgroup \( H_2 \) is 2.
- Since 3 and 2 are coprime, their intersection \( H_1 \cap H_2 \) can only be the identity element \( e \).

This implies there is at least one Sylow-2 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 can be constructed using the Sylow-3 and Sylow-2 subgroups.

Let \( H_1 \subseteq G \) be a Sylow-3 subgroup (order 3) and \( H_2 \subseteq G \) be a Sylow-2 subgroup (order 2).

Since \( H_1 \cap H_2 = \{e\} \), the product \( H_1H_2 \) forms a subgroup. The order of \( H_1H_2 \) is:
\[ |H_1H_2| = |H_1| \times |H_2| = 3 \times 2 = 6 \]

Thus, \( G \) has a subgroup of order 6, which can be formed by combining the elements 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.

Do you want more details or have any questions?

**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 concepts in a group of order 42, including the determination of Sylow-7 subgroups, construction of Sylow-3 and Sylow-2 subgroups, and proving the existence of a subgroup of order 6. Learn the application of Sylow theorems in group classification.

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