et G be a group of order 42.

Determine the number of Sylow-7 subgroups of G.


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.

Learn how to determine the number of Sylow-7 subgroups in a group of order 42 using Sylow's theorems. This problem involves understanding group theory and prime factorization. Suitable for advanced undergraduate students studying algebra and group theory.

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