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

Step 1: Factorize the order of the group

The order of the group $G$ is given by $|G| = 42$. First, we factorize 42:

$42 = 2 \times 3 \times 7$

Step 2: Use Sylow's theorems

According to Sylow's theorems:

1. The number of Sylow-$p$ subgroups (where $p$ is a prime divisor of the order of the group) divides the order of the group and is congruent to 1 modulo $p$.
2. Let $n_p$ denote the number of Sylow-$p$ subgroups.

For Sylow-7 subgroups:

• $p = 7$
• $|G| = 42$, and the number of Sylow-7 subgroups $n_7$ divides $\frac{42}{7} = 6$.
• Additionally, $n_7 \equiv 1 \pmod{7}$.

Step 3: Possible values of $n_7$

Since $n_7$ must divide 6 and satisfy $n_7 \equiv 1 \pmod{7}$, we consider the divisors of 6:

• Divisors of 6: 1, 2, 3, 6

We now check each:

• $n_7 = 1$: $1 \equiv 1 \pmod{7}$ (satisfies the condition)
• $n_7 = 2$: $2 \not\equiv 1 \pmod{7}$ (does not satisfy the condition)
• $n_7 = 3$: $3 \not\equiv 1 \pmod{7}$ (does not satisfy the condition)
• $n_7 = 6$: $6 \not\equiv 1 \pmod{7}$ (does not satisfy the condition)

Conclusion

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

Therefore, there is exactly 1 Sylow-7 subgroup in $G$.

Would you like further details or have any questions?

Here are 5 related questions to expand on this topic:

1. How would you determine the number of Sylow-3 subgroups for a group of order 42?
2. What can be inferred about the structure of $G$ if it has exactly one Sylow-7 subgroup?
3. How does the existence of a unique Sylow subgroup influence the structure of a group?
4. Can a group of order 42 be simple? Why or why not?
5. How can Sylow's theorems help determine whether a group of order 42 is abelian?

Tip: Remember, if there is exactly one Sylow-$p$ subgroup in a group, it must be normal in the group.