Fundamental Theorem of Finite Abelian Groups: Subgroups 
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  $\mathbb{Z}_4 + \mathbb{Z}_{27}$?
 A: Note that $\mathbb{Z}_4\oplus\mathbb{Z}_{27}\cong \mathbb{Z}_{108}$. In general, if $p$ and $q$ are relatively prime, then $\mathbb{Z}_p\oplus\mathbb{Z}_q \cong \mathbb{Z}_{pq}$.
So you should not list them separately: you are listing isomorphism types, not different ways of writing them. So, yes, the $3$-part of $A$ must be cyclic, as otherwise it has at least two subgroups of order $3$, which means $A$ must be isomorphic to either $\mathbb{Z}_{4}\oplus\mathbb{Z}_{27}$, or to $\mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_{27}$.
(You don't need to check all possibilities: just remember that $A$ is the direct sum of its $p$-parts, so you only need to worry about the $3$-part; this is either $\mathbb{Z}_{27}$, $\mathbb{Z}_3\oplus\mathbb{Z}_{9}$, or $\mathbb{Z}_3\oplus\mathbb{Z}_3\oplus\mathbb{Z}_3$). 
A: Yes. For any natural numbers $n$ and $m$ that are coprime (i.e. have no common factors),
$$\mathbb{Z}_{mn}\cong\mathbb{Z}_m\times \mathbb{Z}_n$$
so your list of all abelian groups of order 108 (up to isomorphism) actually has a duplicate: $\mathbb{Z}_{108}$ and $\mathbb{Z}_4\times\mathbb{Z}_{27}$ are isomorphic, and so should not be counted as different for these purposes.
So, up to isomorphism, there are only two abelian groups of order 108 with exactly one subgroup of order 3, namely $\mathbb{Z}_{108}$ and $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_{27}$. 
