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. (Hint: Start by concluding that H is abelian.)

Let's go through each part of the problem step by step.

### 1. Determine the number of Sylow 3-subgroups of \( G \).

Given that \( G \) is a group of order \( |G| = 45 = 3^2 \times 5 \), we apply Sylow's theorems to determine the number of Sylow 3-subgroups, denoted by \( n_3 \).

By Sylow's theorems:
- \( n_3 \) (the number of Sylow 3-subgroups) must divide \( |G| / 3^2 = 5 \).
- \( n_3 \equiv 1 \pmod{3} \).

The divisors of 5 are 1 and 5. We check which of these satisfies \( n_3 \equiv 1 \pmod{3} \):
- \( n_3 = 1 \) satisfies \( n_3 \equiv 1 \pmod{3} \).
- \( n_3 = 5 \) does not satisfy \( n_3 \equiv 1 \pmod{3} \).

Thus, \( n_3 = 1 \). This means that there is exactly one Sylow 3-subgroup in \( G \), and it is normal in \( G \).

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

Since \( H \) is a Sylow 3-subgroup of \( G \), the order of \( H \) is \( |H| = 3^2 = 9 \).

#### a. Show that \( H \) has an element of order 3.

By the Cauchy theorem, if a group \( H \) has order 9 (which is \( 3^2 \)), then there must be an element in \( H \) whose order is 3.

**Explanation**: The order of any element in \( H \) must divide 9. The possible orders of elements are 1, 3, or 9. If \( H \) is cyclic, generated by an element of order 9, then there will certainly be an element of order 3 (any element \( g \) such that \( g^3 \) generates \( H \)). If \( H \) is not cyclic, then all non-identity elements would still have orders dividing 9, and at least one of these elements would have to have order 3.

Therefore, \( H \) must contain at least one element of order 3.

#### b. Use the above element to show that \( H \) cannot be simple.

To show that \( H \) cannot be simple, we'll first conclude that \( H \) is abelian.

**Proof that \( H \) is abelian**:

\( H \) has order 9, a prime power (\( 3^2 \)). Groups of prime power order are **nilpotent**, and all groups of order \( p^2 \) (where \( p \) is a prime) are abelian. Thus, \( H \) is abelian.

Since \( H \) is abelian, consider the element \( g \in H \) of order 3 that we identified in part (a). The subgroup \( \langle g \rangle \) generated by \( g \) is a proper, non-trivial subgroup of \( H \) and is normal in \( H \) because \( H \) is abelian (every subgroup of an abelian group is normal).

Since \( H \) has a proper, non-trivial normal subgroup, \( H \) cannot be simple. A simple group has no non-trivial, proper normal subgroups, so this contradicts the simplicity of \( H \).

### Summary of Results:
1. The number of Sylow 3-subgroups of \( G \) is \( 1 \).
2. \( H \), being a Sylow 3-subgroup of \( G \), has an element of order 3, and it cannot be simple because it is abelian and contains a proper, non-trivial normal subgroup.

Would you like more details or have any questions? Here are five related questions for further exploration:

1. What is the number of Sylow 5-subgroups in \( G \)?
2. If \( n_3 = 1 \) and \( n_5 = 1 \), what can you infer about the structure of \( G \)?
3. What are the possible isomorphism classes for a group of order 45?
4. How can you use Sylow's theorems to determine whether \( G \) is simple?
5. How does the normality of the Sylow 3-subgroup influence the overall structure of \( G \)?

**Tip:** When dealing with groups of small orders, identifying the number and structure of Sylow subgroups can often lead to a full classification of the group.

Explore a group theory problem involving Sylow 3-subgroups in a group of order 45. Understand how to determine the number of Sylow 3-subgroups and demonstrate that a Sylow 3-subgroup cannot be simple. Includes detailed explanations and applications of Sylow's theorems and group structure.

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