Given that $G$ is a group of order 42, we need to determine the number of Sylow-7 subgroups.

First, factorize the order of the group: $|G| = 42 = 2 \times 3 \times 7$

According to the Sylow theorems, the number of Sylow-$p$ subgroups, denoted $n_p$, must satisfy two conditions:

1. $n_p \equiv 1 \pmod{p}$
2. $n_p$ divides $|G| / p$

For the Sylow-7 subgroups:

• $p = 7$
• $|G| / 7 = 6$

So, $n_7$ must divide 6 and $n_7 \equiv 1 \pmod{7}$.

The divisors of 6 are 1, 2, 3, and 6. We need to check which of these satisfy $n_7 \equiv 1 \pmod{7}$:

• $n_7 = 1$: $1 \equiv 1 \pmod{7}$ ✔️
• $n_7 = 2$: $2 \equiv 2 \pmod{7}$ ❌
• $n_7 = 3$: $3 \equiv 3 \pmod{7}$ ❌
• $n_7 = 6$: $6 \equiv 6 \pmod{7}$ ❌

The only value that satisfies both conditions is $n_7 = 1$.

Therefore, the number of Sylow-7 subgroups in $G$ is: $n_7 = 1$

This means there is exactly one Sylow-7 subgroup in $G$.

Do you have any questions or want more details?

Here are some related questions to further your understanding:

1. What is the significance of the Sylow theorems in group theory?
2. How do we determine the number of Sylow-$p$ subgroups for other prime factors?
3. What does it mean for a subgroup to be normal in a group?
4. Can there be more than one Sylow-2 subgroup in $G$? Why or why not?
5. How does the existence of a unique Sylow-7 subgroup affect the structure of $G$?
6. What is the order of the Sylow-3 subgroups in $G$?
7. How are the Sylow subgroups used to analyze the structure of finite groups?
8. Can the number of Sylow-7 subgroups ever be zero? Why or why not?

Tip: Understanding the interplay between Sylow subgroups and group order helps in analyzing the internal structure and classification of finite groups.