Let G be a group of order 45.

1. Determine the number of Sylow 3-subgroups of G.

2. Let H ⊆ G be a Sylow 3-subgroup of G.

a. Show that H has an element of order 3.

b. Use the above element to show that H cannot be simple

Given that \( G \) is a group of order \( 45 \), we can analyze the number and structure of Sylow 3-subgroups of \( G \). Let’s break down the problem step by step.

### Problem Setup
The order of \( G \) is \( |G| = 45 \). The prime factorization of 45 is:
\[
45 = 3^2 \times 5.
\]
By Sylow’s theorems, the number of Sylow \( p \)-subgroups (where \( p \) is a prime dividing the order of \( G \)) must satisfy specific properties.

### Part 1: Determine the Number of Sylow 3-Subgroups

Let \( n_3 \) be the number of Sylow 3-subgroups of \( G \). According to Sylow’s theorems:

1. \( n_3 \equiv 1 \pmod{3} \).
2. \( n_3 \) divides \( \frac{|G|}{3^2} = \frac{45}{9} = 5 \).

So \( n_3 \) is a divisor of 5, meaning \( n_3 \in \{1, 5\} \). Additionally, \( n_3 \equiv 1 \pmod{3} \), which implies that \( n_3 \) must be 1 (since 5 does not satisfy this congruence).

Thus, \( n_3 = 1 \). Therefore, there is exactly **one Sylow 3-subgroup** of \( G \). 

### Part 2: Let \( H \subseteq G \) be a Sylow 3-subgroup of \( G \).

Since \( G \) has only one Sylow 3-subgroup \( H \), it follows that \( H \) is normal in \( G \).

#### Part 2a: Show that \( H \) Has an Element of Order 3

The order of \( H \) is \( 9 \) (since \( H \) is a Sylow 3-subgroup and \( 9 = 3^2 \)). By Lagrange’s theorem, the possible orders of elements in \( H \) are divisors of 9, i.e., \( 1, 3, \) and \( 9 \).

By Cauchy’s theorem, since \( H \) is a group of order \( 9 \) and \( 3 \) divides \( 9 \), \( H \) must contain an element of order 3.

#### Part 2b: Use the Above Element to Show that \( H \) Cannot Be Simple

To show that \( H \) cannot be simple, recall that a group is simple if it has no proper, nontrivial normal subgroups.

Notice that \( H \) is a group of order 9, which is a \( p \)-group (where \( p = 3 \)). Every \( p \)-group has a nontrivial center, so the center \( Z(H) \) is nontrivial.

In fact, for a group of order \( 9 = 3^2 \), the center \( Z(H) \) must have order 3 or 9. If \( |Z(H)| = 9 \), then \( Z(H) = H \), implying that \( H \) is abelian. If \( |Z(H)| = 3 \), then \( Z(H) \) is a nontrivial proper subgroup of \( H \).

In either case, \( H \) cannot be simple because either:

- \( H \) is abelian, and every abelian group of order 9 has nontrivial proper subgroups (e.g., cyclic subgroups generated by elements of order 3).
- \( Z(H) \) is a proper nontrivial normal subgroup of \( H \), contradicting simplicity.

Therefore, \( H \) cannot be simple.

### Conclusion

1. The number of Sylow 3-subgroups of \( G \) is \( 1 \).
2. \( H \) contains an element of order 3.
3. \( H \) cannot be simple because it has a nontrivial proper normal subgroup.

### Follow-Up Questions

1. How does the structure of \( G \) change if it has 5 Sylow 3-subgroups?
2. Can you describe the structure of a group of order 45 if it is cyclic?
3. How many Sylow 5-subgroups does \( G \) have?
4. What can you say about the center of \( G \) based on its Sylow subgroups?
5. What are the possible isomorphism classes for a group of order 45?

### Tip

When analyzing \( p \)-groups, always consider their center, as it plays a crucial role in determining whether the group is simple and in understanding its subgroup structure.

Explore the number of Sylow 3-subgroups in a group of order 45 and analyze the structure of Sylow 3-subgroups. Understand why a Sylow 3-subgroup cannot be simple using group theory concepts such as Sylow theorems, Lagrange's theorem, and Cauchy's theorem.

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