Show that there are two Abelian groups of order $108$ that have exactly one subgroup of order $3$

Question

Show that there are two Abelian groups of order $108$ that have exactly one subgroup of order 3$3$

Attempt: Since $108=2^2 .3^3$ , Hence the possible finite abelian groups of order $108$ will be isomorphic to $Z_{108},~~ Z_2 \oplus Z_{54},~~ Z_3 \oplus Z_{36},~~ Z_3 \oplus Z_3\oplus Z_{12},~~Z_{18} \oplus Z_6,~~ Z_6 \oplus Z_6 \oplus Z_3$

Now, since $Z_{108}$ is a cyclic group, it will have only one subgroup of order $3$.

(a) Now, how do we check and confirm that the remaining groups have only one subgroup of order $3$?

(b) Since, none of the groups : $Z_2 \oplus Z_{54},~~ Z_3 \oplus Z_{36},~~ Z_3 \oplus Z_3\oplus Z_{12},~~Z_{18} \oplus Z_6,~~ Z_6 \oplus Z_6 \oplus Z_3$ are cyclic, how can we be sure that there cannot exist other non cyclic subgroups of order $3$ in addition to cyclic subgroups of order $3$ in each of these groups?

Thank you for you help.

Answer 1

To address your second question, all subgroups of order $3$ regardless of what group we are in will be cyclic since all groups of prime order are cyclic.

My hint to you for the first question is the following: Consider the direct product of two groups $G_1 \times G_2$. For some $a \in G_1$ and $b \in G_2$, the order of $(a, b) \in G_1 \times G_2$ will be $lcm(|a|, |b|)$, where $|a|$ denotes the order of $a$ as an element of $G_1$, and likewise for $b$. Of course, you might want to take some time out and prove this fact.