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.