Given that $G$ is a group of order $45$, we can analyze the number and structure of Sylow 3-subgroups of $G$. Let’s break down the problem step by step.

Problem Setup

The order of $G$ is $|G| = 45$. The prime factorization of 45 is: $45 = 3^2 \times 5.$ By Sylow’s theorems, the number of Sylow $p$-subgroups (where $p$ is a prime dividing the order of $G$) must satisfy specific properties.

Part 1: Determine the Number of Sylow 3-Subgroups

Let $n_3$ be the number of Sylow 3-subgroups of $G$. According to Sylow’s theorems:

1. $n_3 \equiv 1 \pmod{3}$.
2. $n_3$ divides $\frac{|G|}{3^2} = \frac{45}{9} = 5$.

So $n_3$ is a divisor of 5, meaning $n_3 \in \{1, 5\}$. Additionally, $n_3 \equiv 1 \pmod{3}$, which implies that $n_3$ must be 1 (since 5 does not satisfy this congruence).

Thus, $n_3 = 1$. Therefore, there is exactly one Sylow 3-subgroup of $G$.

Part 2: Let $H \subseteq G$ be a Sylow 3-subgroup of $G$.

Since $G$ has only one Sylow 3-subgroup $H$, it follows that $H$ is normal in $G$.

Part 2a: Show that $H$ Has an Element of Order 3

The order of $H$ is $9$ (since $H$ is a Sylow 3-subgroup and $9 = 3^2$). By Lagrange’s theorem, the possible orders of elements in $H$ are divisors of 9, i.e., $1, 3,$ and $9$.

By Cauchy’s theorem, since $H$ is a group of order $9$ and $3$ divides $9$, $H$ must contain an element of order 3.

Part 2b: Use the Above Element to Show that $H$ Cannot Be Simple

To show that $H$ cannot be simple, recall that a group is simple if it has no proper, nontrivial normal subgroups.

Notice that $H$ is a group of order 9, which is a $p$-group (where $p = 3$). Every $p$-group has a nontrivial center, so the center $Z(H)$ is nontrivial.

In fact, for a group of order $9 = 3^2$, the center $Z(H)$ must have order 3 or 9. If $|Z(H)| = 9$, then $Z(H) = H$, implying that $H$ is abelian. If $|Z(H)| = 3$, then $Z(H)$ is a nontrivial proper subgroup of $H$.

In either case, $H$ cannot be simple because either:

• $H$ is abelian, and every abelian group of order 9 has nontrivial proper subgroups (e.g., cyclic subgroups generated by elements of order 3).
• $Z(H)$ is a proper nontrivial normal subgroup of $H$, contradicting simplicity.

Therefore, $H$ cannot be simple.

Conclusion

1. The number of Sylow 3-subgroups of $G$ is $1$.
2. $H$ contains an element of order 3.
3. $H$ cannot be simple because it has a nontrivial proper normal subgroup.

Follow-Up Questions

1. How does the structure of $G$ change if it has 5 Sylow 3-subgroups?
2. Can you describe the structure of a group of order 45 if it is cyclic?
3. How many Sylow 5-subgroups does $G$ have?
4. What can you say about the center of $G$ based on its Sylow subgroups?
5. What are the possible isomorphism classes for a group of order 45?

Tip

When analyzing $p$-groups, always consider their center, as it plays a crucial role in determining whether the group is simple and in understanding its subgroup structure.