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.