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.)
3. Prove that G has a subgroup that is isomorphic to Z/5Z.
(Note that this can be solved independently from 2.)

Let's solve each part of the problem step by step.

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

By Sylow's theorems:
- The order of \( G \) is \( |G| = 45 = 3^2 \times 5 \).
- The number of Sylow 3-subgroups, \( n_3 \), divides \( |G|/3^2 = 5 \) and \( n_3 \equiv 1 \ (\text{mod}\ 3) \).

Thus, the possible values for \( n_3 \) are the divisors of 5 that are congruent to 1 modulo 3. The divisors of 5 are 1 and 5. Among these, only 1 satisfies \( n_3 \equiv 1 \ (\text{mod}\ 3) \).

So, \( n_3 = 1 \). 

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

Since \( H \) is a Sylow 3-subgroup, \( |H| = 9 \).

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

By the Cauchy theorem, if \( H \) is a group of order 9, then it must have an element of order 3. Here's why:

The order of any element \( g \in H \) must divide the order of the group \( |H| = 9 \). So, the possible orders of elements in \( H \) are 1, 3, or 9. If \( H \) only had elements of order 1 or 3, then all non-identity elements would have order 3.

If \( H \) had an element of order 9, then it would be cyclic and generated by that element. But in either case, \( H \) has at least one element of order 3.

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

To show that \( H \) is not simple, we first need to establish that \( H \) is abelian.

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

- The order of \( H \) is 9, a prime power. For such groups, if a non-abelian group exists, it would be isomorphic to the Heisenberg group over \( \mathbb{Z}/3\mathbb{Z} \). However, such a group is not simple because it contains non-trivial, proper, normal subgroups.
- Since \( H \) has no such non-abelian structure and is a Sylow \( p \)-subgroup of a group of order 9, it must be abelian.

Given that \( H \) is abelian and has an element \( x \) of order 3, the subgroup generated by \( x \), \( \langle x \rangle \), is normal in \( H \) because \( H \) is abelian (every subgroup of an abelian group is normal).

Since \( \langle x \rangle \) is a non-trivial, proper subgroup of \( H \), \( H \) cannot be simple.

### 3. Prove that \( G \) has a subgroup that is isomorphic to \( \mathbb{Z}/5\mathbb{Z} \).

To prove that \( G \) has a subgroup isomorphic to \( \mathbb{Z}/5\mathbb{Z} \), we use Sylow's theorems again:

- The number of Sylow 5-subgroups \( n_5 \) divides \( |G|/5 = 9 \), and \( n_5 \equiv 1 \ (\text{mod}\ 5) \).

The divisors of 9 are 1, 3, and 9. The only divisor that is congruent to 1 modulo 5 is 1. Hence, \( n_5 = 1 \).

This means there is exactly one Sylow 5-subgroup, and it must be normal. The order of this subgroup is 5, which is cyclic. Therefore, \( G \) has a subgroup isomorphic to \( \mathbb{Z}/5\mathbb{Z} \).

### Summary of Results:
1. The number of Sylow 3-subgroups of \( G \) is \( 1 \).
2. \( H \) has an element of order 3, and \( H \) cannot be simple because it is abelian and has a proper, non-trivial normal subgroup.
3. \( G \) has a subgroup isomorphic to \( \mathbb{Z}/5\mathbb{Z} \).

Would you like further details or have any questions? Here are some related problems to consider:

1. How many Sylow 5-subgroups does \( G \) have?
2. Can \( G \) be non-abelian if \( n_3 = 1 \) and \( n_5 = 1 \)?
3. What is the structure of a group of order 9?
4. What are the possible isomorphism classes for a group of order 45?
5. How does the existence of a normal Sylow 5-subgroup influence the structure of \( G \)?

**Tip:** Sylow theorems are a powerful tool in determining the structure of finite groups and the existence of certain subgroups.

Explore the structure of a group of order 45 using Sylow theorems. Learn about Sylow 3-subgroups, abelian groups, and the existence of subgroups isomorphic to Z/5Z. Detailed step-by-step solutions provided.

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