Show that there are two abelian groups of order 108 that have exactly one subgroup of order 3.
$108 = 2^ 2 \times 3 ^ 3$
Using the fundamental theorem of finite abelian groups, we have
Possible abelian groups of order 108 are: $\mathbb{Z}_{108}$, $ \mathbb{Z}_4 + \mathbb{Z}_{27}$, $\mathbb{Z}_2+\mathbb{Z}_2+\mathbb{Z}_{27}$, $\mathbb{Z}_4+\mathbb{Z}_9+\mathbb{Z}_3$, $\mathbb{Z}_2+\mathbb{Z}_2+\mathbb{Z}_9+\mathbb{Z}_3$, $\mathbb{Z}_4+\mathbb{Z}_3+\mathbb{Z}_3+\mathbb{Z}_3$, $\mathbb{Z}_2+\mathbb{Z}_2+\mathbb{Z}_3+\mathbb{Z}_3+\mathbb{Z}_3$.
It seems to me that all three $\mathbb{Z}_{108}$, $\mathbb{Z}_4 + \mathbb{Z}_{27}$, $\mathbb{Z}_2+\mathbb{Z}_2+\mathbb{Z}_{27}$, have exactly one subgroup of order 3. Please suggest where I am going wrong.
Is it because $\mathbb{Z}_{108}$ is isomorphic to $\mathbb{Z}_4 + \mathbb{Z}_{27}$?