To determine the number of Sylow-7 subgroups of a group $G$ of order 42, we can apply Sylow's theorems. Let's go through the steps:

Step 1: Determine the Order of the Sylow-7 Subgroups

The order of a Sylow-7 subgroup is determined by the largest power of 7 that divides the order of the group $|G|$. The order of $G$ is 42, and we can factorize it as follows:

$|G| = 42 = 2 \times 3 \times 7$

So, the order of a Sylow-7 subgroup $P_7$ must be 7, as it is the largest power of 7 that divides 42.

Step 2: Use Sylow's Theorem to Determine the Number of Sylow-7 Subgroups

Sylow's theorem tells us that the number of Sylow-7 subgroups, denoted by $n_7$, must satisfy two conditions:

1. $n_7$ divides the index $|G| / |P_7| = 42 / 7 = 6$.
2. $n_7 \equiv 1 \mod 7$.

Step 3: List the Possible Divisors and Find $n_7$

The divisors of 6 are 1, 2, 3, and 6. We need to find which of these values satisfy $n_7 \equiv 1 \mod 7$.

• $n_7 = 1$: $1 \equiv 1 \mod 7$
• $n_7 = 2$: $2 \not\equiv 1 \mod 7$
• $n_7 = 3$: $3 \not\equiv 1 \mod 7$
• $n_7 = 6$: $6 \not\equiv 1 \mod 7$

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

Conclusion

There is exactly one Sylow-7 subgroup in $G$.

Would you like any additional details or have any questions?

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Here are 5 related questions to deepen your understanding:

1. How many Sylow-2 subgroups can exist in a group of order 42?
2. If $G$ is a group of order 42, what can you conclude about the structure of $G$ if there is only one Sylow-7 subgroup?
3. How does the existence of a unique Sylow-7 subgroup affect the normality of this subgroup?
4. What can Sylow's theorems tell us about the number of Sylow-3 subgroups in $G$?
5. What is the significance of the condition $n_p \equiv 1 \mod p$ in Sylow's theorems?

Tip: The uniqueness of a Sylow $p$-subgroup often implies that the subgroup is normal in the group, which can provide insights into the group's structure.