**Problem**

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

**Solution**

Possibilities: $108=2^{2} \cdot 3^{3}$

(1) $\mathbb{Z_{4}} \times \mathbb{Z_{27}}$

(2) $\mathbb{Z_{2}} \times \mathbb{Z_{2}} \times \mathbb{Z_{27}}$

(3) $\mathbb{Z_{2}} \times \mathbb{Z_{2}} \times \mathbb{Z_{3}} \times \mathbb{Z_{3}} \times \mathbb{Z_{3}}$

(4) $\mathbb{Z_{4}} \times \mathbb{Z_{3}} \times \mathbb{Z_{3}} \times \mathbb{Z_{3}}$

(5) $\mathbb{Z_{2}} \times \mathbb{Z_{2}} \times \mathbb{Z_{3}} \times \mathbb{Z_{9}}$

(6) $\mathbb{Z_{4}} \times \mathbb{Z_{3}} \times \mathbb{Z_{9}}$

Possibilities (3),(4),(5),(6) all have multiple subgroups of order $3$.

$\mathbb{Z_{4}} \times \mathbb{Z_{27}}$ has one subgroup of order $3$, $\langle(0,9) \rangle$, and $\mathbb{Z_{2}} \times \mathbb{Z_{2}} \times \mathbb{Z_{27}}$ has only one subgroup of order $3$, $\langle(0,0,9) \rangle$.