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.