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.