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.