To find the number of Sylow-7 subgroups of a group $G$ of order 42, we use Sylow's theorems.

First, note that the order of the group $|G| = 42$ can be factored as:

$42 = 2 \times 3 \times 7$

According to Sylow's theorems:

1. The number of Sylow-$p$ subgroups, denoted by $n_p$, for a prime $p$ dividing the order of the group, divides $|G|$ and is congruent to 1 modulo $p$.

For the Sylow-7 subgroups, $p = 7$ and $|G| = 42$. The number of Sylow-7 subgroups $n_7$ must satisfy:

$n_7 \equiv 1 \pmod{7}$ $n_7 \text{ divides } \frac{42}{7} = 6$

The divisors of 6 are 1, 2, 3, and 6. Among these, the only number congruent to 1 modulo 7 is $n_7 = 1$.

Thus, the number of Sylow-7 subgroups of $G$ is $n_7 = 1$.

This implies that the Sylow-7 subgroup is unique and is therefore normal in $G$.

Would you like more details or have any questions?

Here are 8 related questions you might consider:

1. How do Sylow's theorems help in determining the structure of a group?
2. What can be said about the uniqueness of a Sylow subgroup if the number is 1?
3. How do we determine the number of Sylow-2 subgroups for a group of order 42?
4. What is the significance of a normal Sylow subgroup in the structure of a group?
5. Can a group of order 42 be simple? Why or why not?
6. What are the possible orders of elements in a group of order 42?
7. What can we infer about the group structure if all Sylow subgroups are normal?
8. How can we construct a group of order 42?

Tip: Always verify that the number of Sylow subgroups satisfies both divisibility and congruence conditions according to Sylow's theorems.