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

Let's solve this problem step by step.

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

To find the number of Sylow 3-subgroups of \( G \), we use the Sylow theorems.

- **Step 1**: The order of the group \( G \) is 45. Since 45 can be factored as \( 45 = 3^2 \times 5 \), the number of Sylow 3-subgroups, denoted by \( n_3 \), must satisfy the following conditions:
  1. \( n_3 \equiv 1 \pmod{3} \) (This is a condition from the Sylow theorem).
  2. \( n_3 \) divides \( 15 \) (This is \( \frac{|G|}{3^2} = \frac{45}{9} = 5 \)).

The divisors of 5 are 1 and 5. So, \( n_3 \) can be either 1 or 5. But it must also satisfy \( n_3 \equiv 1 \pmod{3} \). Both 1 and 5 satisfy this condition. Therefore, \( n_3 \) can be either 1 or 5.

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

Given that \( H \) is a Sylow 3-subgroup, \( |H| = 9 \) (since \( |H| = 3^2 \)).

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

By Cauchy's theorem, which states that if a prime \( p \) divides the order of a finite group, then the group has an element of order \( p \), we can conclude that \( H \) has an element of order 3. Since \( H \) has order 9, which is \( 3^2 \), there must exist at least one element in \( H \) with order 3.

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

Let's assume for contradiction that \( H \) is simple. 

- **Step 1**: A simple group has no nontrivial normal subgroups other than the group itself and the trivial group.

- **Step 2**: The group \( H \) is of order 9. We know that \( H \) has elements of order 3. The subgroup generated by any element of order 3 has order 3.

- **Step 3**: Let \( x \in H \) be an element of order 3. Then \( \langle x \rangle \) is a subgroup of \( H \) of order 3.

- **Step 4**: The subgroup \( \langle x \rangle \) is normal in \( H \) because the index \( [H : \langle x \rangle] = 9/3 = 3 \), and in a group of prime index, every subgroup is normal.

Since \( H \) has a nontrivial normal subgroup \( \langle x \rangle \) of order 3, \( H \) cannot be simple (because simple groups have no nontrivial normal subgroups).

### Summary:

1. The number of Sylow 3-subgroups of \( G \) is either 1 or 5.
2. 
   a. \( H \), a Sylow 3-subgroup of \( G \), has an element of order 3.
   b. \( H \) cannot be simple because it has a nontrivial normal subgroup of order 3.

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Would you like further details or have any questions?

Here are some related questions to deepen your understanding:

1. What is the significance of Sylow theorems in group theory?
2. How does Cauchy's theorem apply to groups of different orders?
3. Can you find an example of a simple group of small order?
4. How would the problem change if the group had order 60 instead of 45?
5. Can \( H \) have any other normal subgroups besides those of order 3?

**Tip:** Always check the orders of subgroups when applying Sylow theorems; they often provide key insights into the structure of the group.

Explore a group theory problem involving the determination of Sylow 3-subgroups and the existence of elements of order 3 in a group of order 45. Learn about Sylow theorems, Cauchy's theorem, and the structure of simple groups.

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