Given a group $G$ of order 42, we want to show that $G$ has a subgroup $H_2$ of order 2 such that $H_1 \cap H_2 = \{e\}$, where $H_1 \subseteq G$ is a Sylow-3 subgroup of $G$.

To prove this, we need to consider the properties of Sylow subgroups and use Sylow's theorems. Let's break it down step-by-step:

Step 1: Sylow Subgroups of $G$

The order of $G$ is 42. We can factorize 42 as: $42 = 2 \cdot 3 \cdot 7$

By Sylow's theorems:

• The number of Sylow-2 subgroups ($n_2$) divides 21 (since $\frac{42}{2} = 21$) and is congruent to 1 modulo 2.
• The number of Sylow-3 subgroups ($n_3$) divides 14 (since $\frac{42}{3} = 14$) and is congruent to 1 modulo 3.
• The number of Sylow-7 subgroups ($n_7$) divides 6 (since $\frac{42}{7} = 6$) and is congruent to 1 modulo 7.

Step 2: Properties of Sylow-3 Subgroups

Let $H_1$ be a Sylow-3 subgroup of $G$. The order of $H_1$ is 3. By Sylow's theorems, the number of Sylow-3 subgroups $n_3$ must divide 14 and be congruent to 1 modulo 3. The possible values for $n_3$ are 1 or 7.

Step 3: Existence of a Sylow-2 Subgroup

Let $H_2$ be a Sylow-2 subgroup of $G$. The order of $H_2$ is 2. By Sylow's theorems, the number of Sylow-2 subgroups $n_2$ must divide 21 and be congruent to 1 modulo 2. The possible values for $n_2$ are 1 or 3 or 7 or 21.

Step 4: Intersection of $H_1$ and $H_2$

To show that H_1 \cap \(H_2 = {e} ), consider the following:

• Since $H_1$ is of order 3, every element of $H_1$ has an order dividing 3.
• Since $H_2$ is of order 2, every element of $H_2$ has an order dividing 2.
• The only element common to both orders 3 and 2 is the identity element $e$.

Therefore, $H_1 \cap H_2 = \{e\}$.

Conclusion

We have shown that $G$ contains a Sylow-2 subgroup $H_2$ of order 2 and a Sylow-3 subgroup $H_1$ of order 3, such that the intersection $H_1 \cap H_2 = \{e\}$. This satisfies the condition given in the problem statement.

Would you like more details on any part of this proof or do you have any questions?

Here are some related questions you might find interesting:

1. How does Sylow's first theorem guarantee the existence of a Sylow $p$-subgroup?
2. What are the possible values for the number of Sylow-7 subgroups in a group of order 42?
3. Can the number of Sylow subgroups give any information about the structure of the group?
4. How does Lagrange's theorem help in understanding the orders of subgroups?
5. What is the significance of the condition $n_p \equiv 1 \ (\text{mod} \ p)$ in Sylow's theorems?
6. How can we use Sylow's theorems to prove that a group of order 30 is not simple?
7. What is a normal Sylow subgroup, and how is it identified?
8. In what cases can the intersection of two Sylow subgroups be non-trivial?

Tip: When working with Sylow subgroups, always check how the number of these subgroups (as constrained by Sylow's theorems) affects the overall structure and properties of the group.