Given a group $G$ of order 42, let's analyze the problems one by one.

1. Determining the number of Sylow-7 subgroups of $G$

By Sylow theorems, the number of Sylow-$p$ subgroups $n_p$ for a prime $p$ dividing the order of the group $|G|$ satisfies:

1. $n_p \equiv 1 \mod p$
2. $n_p$ divides $|G|$

Here, $|G| = 42$ and $42 = 2 \cdot 3 \cdot 7$. We focus on the Sylow-7 subgroups.

For the Sylow-7 subgroups:

• $p = 7$
• $|G| = 42$

The number $n_7$ of Sylow-7 subgroups must divide $\frac{42}{7} = 6$ and $n_7 \equiv 1 \mod 7$.

The divisors of 6 are 1, 2, 3, and 6. Checking these mod 7:

• $1 \equiv 1 \mod 7$
• $2 \not\equiv 1 \mod 7$
• $3 \not\equiv 1 \mod 7$
• $6 \not\equiv 1 \mod 7$

Thus, $n_7 = 1$.

So, the number of Sylow-7 subgroups of $G$ is $1$.

2. Existence of a subgroup $H_2$ with $|H_2| = 2$ and $H_1 \cap H_2 = \{ e \}$

Let $H_1 \subseteq G$ be a Sylow-3 subgroup. The order of $H_1$ is 3 (since $3$ is the largest power of 3 dividing 42).

We need to show there exists a subgroup $H_2$ of order 2 such that $H_1 \cap H_2 = \{ e \}$.

Step 1: Existence of Sylow-2 Subgroup

By the Sylow theorems:

• $p = 2$
• $|G| = 42$

The number $n_2$ of Sylow-2 subgroups must divide $\frac{42}{2^1} = 21$ and $n_2 \equiv 1 \mod 2$.

The divisors of 21 are 1, 3, 7, and 21. Checking these mod 2:

• $1 \equiv 1 \mod 2$
• $3 \equiv 1 \mod 2$
• $7 \equiv 1 \mod 2$
• $21 \equiv 1 \mod 2$

So, $n_2$ can be 1, 3, 7, or 21.

Step 2: Intersection of Sylow Subgroups

If $n_2 = 1$, there is a unique Sylow-2 subgroup, $H_2$, of order 2, and it intersects trivially with any other Sylow subgroup, including $H_1$.

If $n_2 > 1$, there are multiple Sylow-2 subgroups. Still, since Sylow subgroups of different orders intersect trivially, we can find at least one Sylow-2 subgroup $H_2$ such that $H_2 \cap H_1 = \{ e \}$.

Thus, in any case, we can find a Sylow-2 subgroup $H_2$ with $|H_2| = 2$ and $H_1 \cap H_2 = \{ e \}$.

Summary

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

Would you like more details or have any questions about these solutions?

Here are some related questions for further exploration:

1. What are Sylow theorems and how are they applied in group theory?
2. How do the Sylow theorems help in determining the structure of a finite group?
3. Can you determine the number of Sylow-3 subgroups of a group of order 42?
4. How can you prove that a group of order 42 is not simple?
5. What is the significance of a subgroup's order in group theory?
6. How do intersections of Sylow subgroups help in understanding the group's structure?
7. What are some other applications of Sylow theorems in algebra?
8. How do the properties of Sylow subgroups influence the composition series of a group?

Tip: When working with Sylow theorems, always check the divisors of the group order and their congruence properties to determine the possible numbers of Sylow subgroups.