Classify finitely generated abelian groups of exponent 18 Exponent of an abelian group is defined to be the smallest $n$ such that $n g = 0,\forall g$.
A group $G$ of exponent $18$ could be the following:
having an element of order $18$, and other elements of order dividing $18$.
$\frac{\mathbb Z}{2\mathbb Z} \oplus \frac{\mathbb Z}{3^2\mathbb Z} \oplus \frac{\mathbb Z}{2\mathbb Z} \oplus \frac{\mathbb Z}{3^{m_1}\mathbb Z}\oplus \ldots \oplus \frac{\mathbb Z}{3^{m_k}\mathbb Z} \oplus \frac{\mathbb Z}{2\mathbb Z} $, with $m$ indices $\leq 2$ .
Is this correct?
 A: By the fundamental theorem of finitely generated abelian groups we know that if $G$ is a finitely generated abelian group, then $G \cong \mathbb{Z}^r \times \prod_{i=1}^n\mathbb{Z}_{p_i}^{k_i}$ where the $p_i$ are prime and the $k_i$ are positive integers. There cannot be any infinite part, since if there were an infinite element it would have infinite order and the group could not have exponent $18$.
As you noted, we need every element to have order $18$ or order dividing $18$, but we also need an element of exactly order $18$ otherwise $18$ won't be the smallest such $n$. We can factor $18 = 2 \cdot 3^2$, so we know the building blocks of our abelian group must be $\mathbb{Z}_{18}$, $\mathbb{Z}_9$, $\mathbb{Z}_6$, $\mathbb{Z}_3$ and $\mathbb{Z}_2$. But we know that $\mathbb{Z}_{18} \cong \mathbb{Z}_9 \oplus \mathbb{Z}_2$ and $\mathbb{Z}_6 \cong \mathbb{Z}_3 \oplus \mathbb{Z}_2$. So we can write any such abelian group with exponent $18$ as $G \cong (\mathbb{Z}_2)^k \oplus (\mathbb{Z}_9)^l \oplus (\mathbb{Z}_3)^m$ for integers $k, l$ and $m$ where $k, l \geq 1$. Since the order of elements in a direct sum is the least common multiple of their orders, the order of any element in a sum of this form will divide $18$. This is because nontrivial elements in $\mathbb{Z}_2$ have order $2$, in $\mathbb{Z}_9$ have order $3$ or $9$ and in $\mathbb{Z}_3$ have order $3$. Any combination of least common multiples taken from these choices yield something that divides $18$. We know that we have exhausted all possible groups because there are no other $\mathbb{Z}_n$ where $n$ divides $18$ that have not been accounted for.  
A: Using the invariant factor decomposition, we have
$$
G \cong C_{d_1} \times C_{d_2} \times \cdots \times C_{d_n}
$$
where $d_1 \mid d_2 \mid \cdots \mid d_n=18$.
Consider the Hasse diagram of the divisors of $18$:
$$
\begin{array}{ccc}
2 & \to & 6 & \to & 18 \\
\uparrow & & \uparrow & & \uparrow  \\
1 & \to & 3 & \to & 9 \\
\end{array}
$$
The abelian groups of exponent $18$ correspond to the paths from $1$ to $18$ that can omit a vertex or repeat a vertex arbitrarily.
Therefore, the groups have the form
$$
C_2^a \times C_6^b \times C_{18}^c,
\quad
C_3^a \times C_9^b \times C_{18}^c,
\quad
C_3^a \times C_6^b \times C_{18}^c
$$
where $a,b,c \in \mathbb N$, including $0$.
