To prove that the group $G$ of order 45 has a subgroup isomorphic to $\mathbb{Z}/5\mathbb{Z}$ (also denoted $\mathbb{Z}_5$), we can use Cauchy's Theorem from group theory.

Step 1: Apply Cauchy's Theorem

Cauchy's Theorem states that if $G$ is a finite group and $p$ is a prime number dividing the order of $G$, then $G$ contains an element of order $p$. Moreover, the cyclic subgroup generated by this element has order $p$.

Step 2: Determine the relevant prime factor

The order of the group $G$ is $|G| = 45$. We can factorize 45 as: $45 = 3^2 \times 5$ So, the prime factors of 45 are 3 and 5.

Step 3: Apply Cauchy's Theorem for $p = 5$

Since 5 is a prime factor of 45, by Cauchy's Theorem, there exists an element $g \in G$ of order 5. The cyclic subgroup generated by this element $g$ is $\langle g \rangle$, which has order 5.

Step 4: Identify the subgroup isomorphic to $\mathbb{Z}_5$

The subgroup $\langle g \rangle$ is a cyclic group of order 5. A cyclic group of order 5 is isomorphic to $\mathbb{Z}_5$, because any cyclic group of prime order $p$ is isomorphic to $\mathbb{Z}_p$.

Conclusion

Therefore, $G$ has a subgroup that is isomorphic to $\mathbb{Z}_5$, as required.

Would you like more details or have any questions about this proof?

Here are 5 related questions:

1. How does Cauchy's Theorem apply to other prime factors, like 3, in this context?
2. Can a group of order 45 have a subgroup isomorphic to $\mathbb{Z}_3$?
3. What are the possible structures of the group $G$ of order 45?
4. Is the subgroup $\mathbb{Z}_5$ unique in $G$?
5. How would the Sylow theorems apply to a group of order 45?

Tip: When dealing with groups of composite order, always consider the prime factors and apply Cauchy's Theorem to determine the existence of cyclic subgroups of prime order.