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.