Every abelian group of order $9p^2$ where $p\equiv 2\bmod 3$, can be written as the direct product of two cyclic subgroups.

Question

Let $G$ an abelian group of order $9p^2$, where $p$ is an odd prime such that $p\equiv 2\bmod 3$, I have to show that $G$ can be written as the direct product of two cyclic subgroups. In this case, I know that $G$ admits a unique subgroup of order $p^2$. Can the structure theorem help in this case?

Answer 1

The only things needed here are the classification theorem of finitely generated abelian groups and the fact that, if $\text{gcd}(n,m)=1$, then $C_n\times C_m\cong C_{nm}$, where $C_n$ denotes the cyclic group of order $n$.

Since $\text{gcd}(3,p)=\text{gcd}(3,p^2)=1$ we have that $C_{3p}\cong C_3\times C_p$ and $C_{3p^2}\cong C_3\times C_{p^2}$.

By the classification theorem, the possible groups are:

1. $C_{9p^2}\cong C_9\times C_{p^2}$, since $\text{gcd}(9,p^2)=1$ (why: divisors of $9$ are $1,3,9$. as we mentioned, $3$ does not divide $p^2$. If $9$ divides $p^2$ then so does $3$ which is not true, so the only common divisor is 1)
2. $C_9\times C_{p^2}$, nothing to do here
3. $C_9\times C_p\times C_p\cong C_{9p}\times C_p$ since $\text{gcd}(9,p)=1$ (why?)
4. $C_3\times C_3\times C_{p^2}\cong C_3\times C_{3p^2}$ by our observation
5. $C_3\times C_3\times C_p\times C_p\cong C_{3p}\times C_{3p}$ by our observation